Thermodynamics

Thermodynamics Lecture 2Thermal equilibrium can be reached when two objects have been in contact for a long time.Relaxation time is the time required for a system to come to thermal equilibrium/Diffusive equilibrium between two substances (say , coffee-cream mixture) is reached when molecules of each substance are free to movearound but no longer has the tendency to one way or the other.Mechanical equilibrium takes place when large-scale motions (expansion of a balloon) can take place but no longer do.Keep in mind that energy is exchanged when two systems are in thermal equilibrium, volume is exchanged between two systems inmechanical equilibrium, and particles are exchanged when two systems are in diffusive equilibrium.Absolute zero is approached when the volume of a gas is held fixed, then its pressure will approach zero as the temperature approaches– 273 oC. A absolute temperature scale or a Kelvin scale is of the same size as a degree Celsius, but the former is measured up fromabsolute zero.Q. 1. What is the absolute zero on the Fahrenheit scale?Q. 2. Determine the Kelvin temperature for each of the following:i) human body temperatureii) the b. p. of water at standard pressure of 1 atmiii) the coldest day you can rememberiv) -196 oC, the b. p. of liquid nitrogenv) 327 oC, the m. p. of Pb.Q. 3. Linear thermal expansion coefficient α= (∆L/L)/∆Ti) For steel α=1.1E-5 K-1, estimate the total variation in length of a 1-km steel bridge between a cold winter night and a hot summerday. Ans 77 cmIdeal Gas: Many of the properties of a low density gas can be summarized in the ideal gas equation.PV= nRTWhere P = pressure, V= volume, n = number of moles of the gas, R is a universal constant= 8.31 J/mol.K (when you measure pressure inN/m2 = Pa and volume in m3), and T is the temperature in kilvinsA mole of molecules is Avogadro’s number (NA = 6.02E23)of them.Another form of ideal gas equation isPV = NkTwhereN = n x NAk is called the Boltzmann’s constant = 1.381E-23 J/KRemember that nR = NkQ. 4. What is the volume of one mole of air at room temperature and 1 atm pressure?N =n RT/P = 1mol x 8.31 J/mol.K x 300 K/(1E5 N/m2) Ans. 0.025 m3 = 25 litersQ. 5. Estimate the number of molecule in your classroom.N = PV/kT = (1E5 N/m2) x (Average room volume = sq area x height)/(1.38E-23 J/K x 300 K)For 4 m x 4 m x 3m room size, it is 1.2E27 = !E27 = (2000 moles)Q. 6. Calculate the mass of a mole of dry air which is a mixute on 78% N2, 21% O2, and 1% argon.M = 0.78 x 28 g + 0.21 x 32 g + 0.01 x 40 g = 28.96 g = 29.0 gQ. 7. Estimate the average temperature of the air inside a hot air balloon. Assume that the total mass of the unfilled balloon and thepayload is 500 kg. What is the mass of the air inside the balloon?Let ρo = density of surrounding air, V = volume of the balloon, M = mass of the unfilled balloon and the payload, m = mas of 1 mole ofair, and ρ = density of air inside the balloon. Assume, outside temp T = 290 K, balloon volume = 1770 m3, mass of unfilled balloon andpayload = 500 kgIn flight, the buoyancy = weight of the displaced air is assumed to be equal to the weight of the payload, balloon, and the air insidethe balloon.i) Mass of the displaced air = density x volume = ρo V = ? Weight of the displaced air = ρo V g. = ?ii) Mass of air inside balloon = ρ V = ?iii) Weight of this air = ρ V g = ?iv) Weight of payload, balloon, and air = Mg + ρ V g = ?v) From ρo V g = Mg + ρ V g , what is ρo – ρ = ?vi) From gas equation, ρ = mn/V = mP/RT from ideal gas eqn., and ρo = mP/RTo. Substitute them in (v) above to get mP/RTo – mP/RT= M/Vvii) What is 1/T = ? (simplify to find)viii) Substitute the known values to find the numerical answer, Ans. T = 370 Kix) M air = mn = mPV/RT = 0.029 kg x 1E5 N/m2 x 1770 m3 / (8.31 J/K x 379 K) = 1600 kgThermodynamics Lecture 3Q. 4. What is the volume of one mole of air at room temperature and 1 atm pressure?N =n RT/P = 1mol x 8.31 J/mol.K x 300 K/(1E5 N/m2) Ans. 0.025 m3 = 25 litersQ. 5. Estimate the number of molecule in your classroom.N = PV/kT = (1E5 N/m2) x (Average room volume = sq area x height)/(1.38E-23 J/K x 300 K)For 4 m x 4 m x 3m room size, it is 1.2E27 = !E27 = (2000 moles)Q. 6. Calculate the mass of a mole of dry air which is a mixute on 78% N2, 21% O2, and 1% argon.M = 0.78 x 28 g + 0.21 x 32 g + 0.01 x 40 g = 28.96 g = 29.0 gQ. 7. Estimate the average temperature of the air inside a hot air balloon. Assume that the total mass of the unfilled balloon and thepayload is 500 kg. What is the mass of the air inside the balloon?Let ρo = density of surrounding air, V = volume of the balloon, M = mass of the unfilled balloon and the payload, m = mas of 1 mole ofair, and ρ = density of air inside the balloon. Assume, outside temp T = 290 K, balloon volume = 1770 m3, mass of unfilled balloon andpayload = 500 kgIn flight, the buoyancy = weight of the displaced air is assumed to be equal to the weight of the payload, balloon, and the air insidethe balloon.i) Mass of displaced air = density x volume = ρo V = ? Weight of the displaced air = ρo V g. = ?ii) Mass of air inside balloon = ρ V = ?iii) Weight of this air = ρ V g = ?iv) Weight of payload, balloon, and air = Mg + ρ V g = ?v) From ρo V g = Mg + ρ V g , what is ρo – ρ = ?vi) From gas equation, ρ = mn/V = mP/RT from ideal gas eqn., and ρo = mP/RTo. Substitute them in (v) above to get mP/RTo – mP/RT= M/Vvii) What is 1/T = ? (simplify to find)viii) Substitute the known values to find the numerical answer, Ans. T = 370 Kix) M air = mn = mPV/RT = 0.029 kg x 1E5 N/m2 x 1770 m3 / (8.31 J/K x 379 K) = 1600 kgMicroscopic Model of an Ideal GasImagine a cylinder of length L, and piston cross-sectional area A containing just one gas molecule. Cylinder inside volume = ?Let the velocity of the molecule be v having the horizontal component vx. The molecule bounces off the cylinder walls and changes itsvelocity, but not the speed. Assume the collisions are elastic so that the molecule does not lose any energy.Pressure = force/area. Ideal gas law gives PV = NkT.The molecule periodically clashes into the piston and bounces off.Average pressure on the pistion = average force on the piston/AAvxv V = LAL= – average force on the molecule/A (justify how?)= molecule mass x average acceleration/A= m (∆vx/∆t)/A, where ∆t = time the molecule takes to undergo one round-trip from the left to the right andback again. ∆t = total distance/velicity component = 2 L/vxDuring the time the molecule undergoes exactly one collision with the piston, the change in its x-velocity = ∆vx = vx, final – vx,initial = (- vx) – (vx) = -2vxWith this value of the velocity change, average pressure , P = – (m/A) (-2vx)/2L/vx) = mvx2/AL = mvx2/V PV = mvx2 If there are N identical molecules with randon positions and directions of motion, PV = Nmvx2 Putting NkT = PV, NkT = Nmvx2 kT = mvx2 1/2kT =1/2mvx2The same kind of equation should hold for y and z directions also. Summing all of them, 1/2mvx2 + 1/2mvx2 + 1/2mvx2 = 1/kT + ½ kT, +!/2 kT =3/2 kTQ. What is KT= how many Joules for an air molecule at room temperature?= how many electrons volts? (1 eV = 1.6E-19 J)Average speed of the molecules in a gas:1/2mv2 =3/2 kT = > v2 average = 3kT/m.Taking square roots of both sides, we get vrms = (3kT)1/2 = (3RT/M)1/21. Calculate rms speed on nitrogen molecule at room temp.2. vrms = (3kT/m)1/2 = [(3T)(k/m)]1/2= [(3T)(k.NA/mNA)]1/2 = (3RT/M)1/2 = ? Ans. 517 m/sM= molar mass = 0.028 kg3. Suppose a container has hydrogen and oxygen molecules in thermal equilibrium. Which molecule moves faster, on average? By whatfactor?4. Uranium isotopes of atomic masses 238 and 235 form uranium hexafluoride UF6. Calculate the rms speed of each type of moleculeat room temperature, and compare them.Uranium hexafluoride with U-238 has a molecular weight of 352 = 0.352 kg, and the other isotope will give a molecular weight of 349.Use vrms = (3kT/m)1/2 = (3RT/M)1/2Ans. heavier isotope has 145.8 m/s and lighter one has 146.4 m/sTHERMODYNAMICS LECTURE 4Equipartition of energy. Equipartition theorem concerns all forms of energy for which the formula is a quadratic function of acoordinate (viz. elastic potential energy 1/2ksx2 as a function of spring constant ks and the amount of displacement x from theequilibrium) or velocity component (viz., translational motion in the x, y, and z directions – ½ mvx2, ½ mvy2, ½ mvz2, and rotationaland vibrational motions ½ Iωx2, ½ Iωx2). Each of these forms of energy is called a degree of freedom. Equipartition theorem says thatfor each quadratic degree of freedom, the average energy will be ½ kT at temperature T. If a system has N molecules, each with fdegrees of freedom, and there are no other (non-quadratic) temperature-dependent forms of energy, then the total thermal energy isUthermal = N. f. 1/kTThis is just the average total thermal energy. For large N, fluctuations away from the average will be negligible. Uthermal is almostnever the total energy of a system because of the presence of some static energy like chemical bond energy or rest mass energy (mc2) ofthe particles in the system.It is safest to apply equipartition theorem only to changes in energy following temperature variation toavoid phase transformations and other reactions involving breaking bonds between particles.Learning degrees of freedom in connection with equipartition theorem is important. In monatomic gas molecules like helium and argon,only translational motion counts, so each molecule has three degrees of freedom, f=3. In diatomic gas molecules like O2 or N2, eachmolecule can also rotate about two different axes at right angles to each other and perpendicular to the length of the molecule. Thesame is true for CO2. Most polyatomic molecules can rotate about all three axes. Vibrations about the length-wise axis is prohibited byquantum mechanics.1.23 Calculate the total thermal energy of a liter of helium at room temperature and atmospheric pressure. Then repeat the calculationfor a liter of airi) Is He monatomic or diatomic?ii) What is the degrees of freedom per molecule?iii) Write down the Uthermal formula =iv) Put the value of f in (iii)v) What is 3 N. (½)kT = ? in terms of PVvi) P = 1 atmosphere = ? 1E5 N/m2vii) V= 1 liter = ? m3viii) What is the product PV = ?x) What is the Uthermal = ?Thermodynamics Lecture #5Heat. Spontaneous flow of energy from one object to another because of difference of temperature.Work. Any other transfer of energy, which is not heat, into or out of a system when an extensive mechanical variable changes.Heat and work refers to energy in transit. Meaning ones are total energy inside a system, how much heat entered a system, and how muchwork was done on a system. Meaningless ones are how much or how much work is in a system.Symbols. U = Total energy inside a system, Q = amount of energy that entered a system as heat, W= workQ + W = total energy that enters the system, and, by conservation of energy, this is the amount by which the system’s energy changes.=> ∆U = Q + W which is known as the first law of thermodynamics.= > Change in energy = heat added + work done. We have taken + sign for heat Q and work W entering the system. When they leave, wewill use – sign∆U = Q + W WQCompression work. Work = F∆x = PA∆x= -P∆V, the negative sign takes care of the fact that volume decreases as the piston moves in.Piston area, AForce, F∆x ∆V = -A∆xFig. 1.8If the pressure is constant, work done is just minus the area under a graph of pressure vs volume. If the pressure is not constant, wedivide the process into a bunch of tiny steps, compute the areaP P Area = ʃPdVArea = P(V2-v1)V1V2 VVV1 V2Fig. 1.9Under the graph for each step, then add up all the areas to get the total work. The work is still minus the total area under the graphof P vs V. If the pressure is known as a function of volume, then= > W = – ʃ P (V) dV (quasistatic) (1-29)Evaluated between the limits of Vi to Vf.Q. 1.31. Some helium in a cylinder with an initial volume of 1 L and an initial pressure of 1 atm is made to expand to a final volumeof 3 L in such a way that the pressure rises in direct proportion to its volume.a) Sketch the graph of pressure vs volume for this process.i) What is V i = ? ii) What is Vf = ? iii) What is Pi = ? iv) What is Pf = ?v) Draw the graph with (Vi, Pi) and (Vf, Pf)b) calculate the work done on the gas during the process, assuming that there are no “other” type of work done.i) Write down the work done formula.ii) What is the average pressure? Iii) What is the change in volume?iv) What is the work done? Ans. – 400 Jc) Calculate the change in the helium’s energy content during the process.i) Is helium diatomic or monatomic? Ii) Degrees of freedom per helium atom = ?iii) What is the internal energy formula in terms of f, N, k, and Tiv) What is internal energy formula in terms of PV?v) What is the change in the product of PV i. e. PfVf – PiVivi) What is the change in the internal energy? Ans. 1200 Jc) Calculate the amount of heat added to or removed from the helium during the process?i) Write down the first law of thermodynamicsii) What is Q = ? in terms of ∆U and W.iii) From (b) and (c) find out what is Q = ? Ans. 1600 JP P BC BACA DV V1 VV2Fig. 1.10 (a) Fig. 1.10(b)Q. 1.33. An ideal gas is made to undergo the cyclic process shown in Fig. 1.10 (a). For each of the steps A, B, and C, determinewhether each of the following is positive, negative, or zero: (a) the work done on the gas, (b) the change in the energy content of thegas, (c) the heat added to the gas. (d) then determine the sign of each of these three quantities for the whole cycle. (e)What doesthis process accomplish?Hints. a) i) In Fig. 1.10 (a), is the initial volume higher or lower than the final volume in path A?ii) So, is work done positive or negative? iii) Is Vi higher or lower then Vf in path C?iv) Is the work done positive or negative? v) What is the change in V in path B?vi)So, what is the work done?vii) Out of the work done on different paths, does the volume change remain the same?viii) Does the pressure or average remain higher in any path? Ix) What is the sign of the net work done forthe whole cycle?b) i) What is the expression for energy from equipartition theorem?Ii) What is the energy expression in terms of P and V?iii) So, does any increase in P or V indicate energy increase?iv) Does P and/or V increase or decrease in path A? So, what happens to energy?v) Does P and/or V increase or decrease in path C? So, what happens to energy?vi) For the whole cycle does P and/V increase or decrease, or remain the same? Or are the final values the same or different from theinitial values at the end of the cycle? So, what happens to energy for the whole cycle?c) i) What is Q = ? in terms of ∆U and Wii) Tabulate the result for W and ∆U by + and – signs. Then determine the sign of Q.W ∆U QPath APath BPath CWhole cyclee) Apparently, the net result of the cycle is to absorb energy as work and emit energy as heat.Q. 1.34. An ideal diatomic gas in a cylinder with a movable piston undergoes the rectangular cyclic process shown in Fig 1.10 (b).Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are frozen out. Assumethat the only type of work done on the gas is quasistatic compression-expansion work.(a) For each of the four steps A through D, compute the work done on the gas, the heat added to the gas, and the change in the energycontent of the gas. Express all answers in terms of P1, P2, V1, and V2Hint. i) What is the work done formula?ii) What is the change in V in path A and what is the work done?iii) What is the change in V and work done in path C?iv) What is the change in V work done in path D?v) What are the degrees of freedom for each molecule?vi) What is the energy formula from equipartition theorem?vii) What is the energy formula in terms of P and V?viii) What is ∆U for path A in terms of P’s and V’s?viii) What is ∆U for path B in terms of P’s and V’s?viii) What is ∆U for path C in terms of P’s and V’s?viii) What is ∆U for path D in terms of P’s and V’s?ix)What is the formula for Q in terms of ∆U and W?x) From ∆U and W, what is Q for path A in terms of P’s and V’s?xi) From ∆U and W, what is Q for path B in terms of P’s and V’s?xii) From ∆U and W, what is Q for path C in terms of P’s and V’s?xiii) From ∆U and W, what is Q for path D in terms of P’s and V’s?xiv) Tabulate the result belowW ∆U QPath APath BPath CPath DWhole cycleCompression of an Ideal Gas.Isothermal compression. Very slow compression so that the temperature of the gas does not rise al all. Adiabatic compression. Very fastso that no heat escapes from the gas during the process. Most real compressions are between these two extremes.Ideal gas has P = NkT/VUse of Eqn. (1-29) yieldsW = – NkTʃ(1/V)dV between the limits of Vi and Vf yields= -NkT. ln V| = -NkT(ln Vf – ln Vi) = NkT ln (Vi/Vf) (1-30)Work done is positive if Vi > Vf that is if the gas is compressed. If the gas expands isothermally, Vi heat input = – work done. For compression Q is negative because heatleaves the gas; for isothermal expansion, heat must enter the gas so Q is positive.PIsothermVf Vi VFig. 1.11.Adiabatic compression. No heat escapes. That is no change of heat. So, Q remains the same which causes the internal energy to rise.Internal energy increase causes the temperature to increase for ideal gas.= > ∆Q = 0; ∆U = Q + W = WFrom equipartition theorem: U = ½ N f k T (1.33)Whence the energy change along any infinitesimal segment of the curve isdU = ½ N k f dT (1-34)Work done during a quasistatic compression is – PdV= > ½ N k f dT = – P dV (1-35)For ideal gas, P = N k T/VPAdiabaticTfTiVFig. 1.12. Adiabatic curve connects to isothermalsSubstitute in (1-35), simplify to get f/2. dT/T = – dV/V (1-36)Integrate both sides between Ti, Tf and Vi , Vf to getf/2 ln (Tf/Ti) = – ln Vf/Vi (1-37)= > ln (Tf/Ti)f/2 = – ln Vf/Vi(Tf)f/2/(Ti)f/2 = (Vi/Vf)After cross-multiplication VfTf f/2 = ViTif/2 (1-38)Or VTf/2 = constant (1-39)PV = NkT = > T = PV/Nk = > V(PV / Nk)f/2 = constant= > Pf/2(V)V)f/2 = constant. x Nk)f/2Raise both sides of the equation to the power 2/f, = > [Pf/2(V)V)f/2 ]2/f= [constant. x Nk)f/2 ]2/f= > P V2/f V = another constant, = > PV(2 + f)/f = constantThe final result is VγP = constant (1.40)where γ is called the adiabatic exponent= (f +2)/fQ. 1.36. In course of pumping up a bicycle tire, a liter of air (which has γ = 7/5) at atmospheric pressure is compressed adiabaticallyto a pressure of 7 atm(a) What is the final volume of this air after compression?(b) How much work is done in compressing the air?© If the temperature of the air is initially 300 K, What is the temperature after compression?Hint. (a) i) Write down the relation for adiabatic compression.ii) What is Vf = ? in terms of Vi, Pf, and Piiii) Vi= ? iv) Pf = ? v) Pi = ? vi) Vf = ?Ans. 0.25 L(b) i) What is P = ? in terms of V and γ from the adiabatic compression eqn.ii) Write down the work done formula.iii) Integrate between the limits of Vi and Vf to get W= [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1]iv) Substitute for P’s and V’s to get W = 1.86 L.atm = 188 J© i) Write down the adiabatic relation in terms of T’s and V’s.ii) Substitute for Ti, Vi, and f to get Tf = 520 K (approximate)Q. 1. 37. In a Diesel engine atmospheric air is quickly compressed to about 1/20 of its original volume. (a) Estimate the temperatureof the air after compression. Assume γ = 5 for air.(b) Explain why a Diesel engine does not require spark plug.Hint. (a) i) Write down the eqn for adiabatic compression in terms of V’s and T’s.ii) Tf = ? in terms of Ti, Vi, Vf and fiii) Substitute the values to get Tf = 1000 K (approximate)(b) The temperature is presumably hot enough to ignite the fuel as soon as it is injected, without the aid of a spark plug.Thermodynamics Chapter Test # 1 Name Date10 atmBP C1 atm A100 600 V (m3)1. An ideal monatomic gas is made to undergo the cyclic process as shown in the Fig. above. For paths A, B, C, answer thefollowing questions:i) WA = ? W = -PdV = – (1E5)(600-100) = -500E5 = -5E7 Jii) ∆UA = ? 3/2∆(PV) =3/2(PfVf – PiVi) =3/2[(1E5)(600) – (1E5)(100)]=(1.5E5)(500) =7.5E7 Jiii) QA = ? ∆U –W = 7.5E7 – (-5E7) = 12.5E7 = 1.25E8 Jiv) WB = ? No change in volume, so W = 0v) ∆UB = ? (3/2)∆PV = 1.5V∆P = (1.5)(600)(10E5 – 1E5) = 1.5 x 600 x 9E5 = 8.1E8 Jvi) QB = ? ∆U –W = 8.1E8 – 0 = 8.1E8 Jvii) WC = ? –PdV = -{(1E5 + 10E5)/2}x (100-600) = -5.5E5 x -500 = 2.75E8 Jviii) ∆UC = ? 3/2∆PV = 1.5(PfVf – PiVi) = 1.5x [1E5 x 100 – 10E5 x 600] = 1.5E7(1 – 60) = 8.85E8ix) QC = ? ∆U –W = 8.85E86-2.75E8= 6.1E82. In course of pumping up a car tire, 3 L of air (which has γ = 7/5) at 27 oC and at 2 atmospheric pressure is compressedadiabatically to a pressure of 10 atm.i) What is the final volume of this air after compression?PfVfγ = PiViγ = Vf = Vi. (Pi/Pf)1/γ = (3 L)(2 atm /10 atm)5/7= 0.95 Lii) How much work is done in compressing the air? W = [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1][(3E-3 x 2E5)/0.4][(3/0.95)0.4 – 1] = 876 Jiii) What is the temperature after compression? 300 (3/0.95)0.4 = 475.20 K3. What is the Kelvin temperature corresponding to night’s low of 60 oF?273 + (5/9)(60 – 32) = 288.5 K4. What is the volume of 10 moles of air at 27 oC and 0.5 atm pressure?V= (10 moles)(8.31 J/mol.K)(300 K)/(0.5xE5 N/m2)=49,860 J.N/m2=49,860 N.m.m2/N= 49,860 m35. Estimate the number of molecule in your classroom of size 4 m x 4 m x 3 m at 1 atm pressure and at 23 oC.N= PV/kT = (1E5 N/m2)(48m3)/[(1.383E-23 J/K). (300K)] = 1.16E276. Calculate rms speed of oxygen molecule at 27 oC.vrms = (3RT/M)1/2 =[3 x (8.31 J/mol. K)(300 K)/(0.032 kg/mol)]1/2= 483 (N. m/kg)1/2 = [kg. m/s2. m/kg]1/2 = 483 [m2/s2]1/2= 483 m/s7. Suppose a container has hydrogen and oxygen molecules in thermal equilibrium. Which molecule moves faster, on average? By whatfactor?vrms H / vrms O == (3RT/MH)1/2/ (3RT/ MO)1/2 = [MO/MH]1/2 = (0.032/0.002)1/2 = 4Hydrogen, 4 times faster than oxygen8. A monatomic gas in a cylinder with an initial volume of 4 L and an initial pressure of 2 atm is made to expand to a finalvolume of 12 L and final pressure 6 atm.i) What is the work done using the average pressure?W= -PdV = – (4 atm)( 8 L) = – (4E5 N/m2)(8E-3m3) = -3,200 Jii) What is the change in the internal energy? , U = 1.5 PV, ∆U = 1.5(PfVf – PiVi)= 1.5[6E5 . 12E-3 – 2E5 . 4E-3] = 1.5[7,200 – 800] = 9,600 Jiii) Calculate the amount of heat added to the gas. Q = ∆U – W = 9,600+3,200 = 12,800 Jvrms = (3RT/M)1/2 ,Mnitrogen = 0.028 kg, Mhydrogen = 0.002 kg, MOxygen = 0.032 kg,W = -P ∆V, U = 1.5 PV, Q = ∆U – W, PfVfγ = PiViγ ; W = [(PiVi)/(γ-1)][(Vi/Vf)γ-1 – 1]; VfTff/2 = ViTif/2 , f = 5.N= PV/kT, 1 atm =1E5 N/m2 , k = 1.38E-23 J/K, R= 8.315 J/mol.K, NA = 6.022E23; PV = nRT;ToF = (9/5) ToC + 32, T K = ToC + 273, ToC = (5/9)(ToF-32) ; 1 L = 1E-3 m3Heat Capacities. An object’s heat capacity C is the amount of heat Q needed to raise its temperature, per degree temperature increase∆T. = > C ≡Q/∆T. (1-41)Specific heat capacity, c, is the heat capacity per unit mass. = > c ≡ C/m = Q/(m ∆T) (1-42)Heat gained for an increase in temperature ∆T or heat loss for a drop in temperature ∆T is given byQ = mc∆TFor water, heat capacity per unit mass is cw = 4.186 J/g.oCC = (∆U – W)/∆T (1-43)Specific heat at constant volume, CV: Since Q =∆U – W, and no change in volume means no work done, W = 0; so, at constant volume Q =∆U, andCV= (∆U/∆T)V= (∂U/∂T)V (1-44)Specific heat at constant pressure, CP: = [(∆U – (-P∆V))/∆T]P = (∂U/∂T)P + P(∂V/∂T)P (1-45)Q. 1. If U = ½ NfkT, CV = ? (1-46)Q. 2. What is f = ? for monatomic gas? What is CV = ?Q. 3. What is CV per mole for a monatomic gas?Q. 4. Solids have six degrees of freedom per atom. What is CV for solids?Q. 5. What is the ideal gas equation in terms of P, V, N, k, and T? What is V= ?Q.6. (∂U/∂T)P = (∂U/∂T)V = ? (1-56)Q. 7. Prove that CP = V + Nk = CV + nR (1-48)Q. 8. To measure the heat capacity of an object, all you usually have to do is put it in thermal contact with another object whose heatcapacity you know. Suppose that a chunk of 100-g metal is immersed in boiling water at 100 deg C, then quickly transferred into aStyrofoam cup containing 250 g of water at 20 deg C. After a minute or so, the temperature of the contents of the cup is 24 deg C.Assume that during this time no significant energy is transferred between the contents of the cup and the surroundings? The heatcapacity of the cup itself is negligible.i) What is the mass of water?ii) What is the heat capacity per unit mass of water?iii) What is the increase in water temperature?iv) What is the heat gained by water? Ans. 4186 Jv) What is the heat lost by the metal?vi) Write down the formula for the heat capacity of the metal.vii) What is the change in temperature of the metal?viii) Calculate the heat capacity for the metal. Ans. 55 J/oCix) What is the specific heat capacity of the chunk of metal? 0.55 J/g.oCQ. 9. The heat capacity of Brookshire’s Rotini Tricolore ia approximately 1.8 J/g.oC. Suppose you toss 340 g of pasta at 25 oC into 1.5L of boiling water. What effect does it have on the temperature of the water (before there is time for the stove to provide more heat)?i) If Tw is the water’s initial temperature and Tf is the final temperature, how do you write in symbols the drop of water temperature?ii) If mw is the water mass and cw is the water’s heat capacity per unit mass, what is the expression for the heat lost by water?iii) How do you write in symbols heat gained by the pasta if Tf is the final temperature and Tp is the initial temperature?9. If mp is the water mass and cp is the water’s heat capacity per unit mass, what is the expression for the heat gain by pasta?10. Equate heat loss expression= heat gain expression11. Solve for Tf12. Evaluate Tf Ans. 93.3 oCLatent heat, L. Amount of heat absorbed during a phase transformation divided by mass.= > L = Q/m (1-50)The latent heat for melting ice is 333 J/g, or 80 cal/g, and that for boiling water is 2260 J/g or 540 cal/g.Q. 10. Your 200-g cup of tea is boiling hot. About how much ice should you add to bring it down to a comfortable sipping temperature of65 oC? Assume that ice is initially at -15 oC. The specific heat capacity of ice is 0.5 cal/g.oC. The heat capacity per unit mass oftea is 1 cal/g.oC.i) What is the tea’s initial temperature? ii) What is the tea’s final temperature?iii) What is the temperature loss of tea? Iv) What is the heat loss formula?v) How much heat is lost by tea? Ans. 700 vi) What is the temperature changein ice to rise to the melting point?vii) What is the heat required for this job if m is the mass of ice? Viii) What is heat required to melt the ice?Ix) What is the temperature rise in water from 0 oC to 65oC? x) What is heat requiredin the ice melting process?xi) From heat loss = heat gain, find the mass of ice. Ans. 46 g.Mathematics for ThermodynamicsVariables. The symbols u, w, x, y, z, t, P, Q, T, V, S, etc. symbols are used for variables i. e. physical quantities that takedifferent values. Out of these, independent variables are the ones that assume their values independently of any other. Time indicatedby the symbol t is independent variable. Our growth, if we represent it by the symbol g, will be a dependent variable.Q. 1. Say you are driving. Your distance covered is represent by the variable x, and time by t. Out of these two, which one isindependent and which one dependent?Q. 2. Say you are heating a kettle of water. The more you rise the temperature T, the more heat water accumulates. Find out dependentand independent variables.Q. 3. Say you cooked your food and then turn off the oven. As the time t passes, the oven becomes cooler by losing its heat Q. Whichones are independent and dependent variables? Do both of them have increasing tendency at the same time in the cooling phase?Function. A function is a relationship between two variables such that to each value of the independent variable there correspondsexactly one value of the independent variable. Functional notation is y = f(x) where f is the name of the function, y is the dependentvariable, x is the independent variable, and f(x) is the value of the function at x.Exponential Function. The exponential function f with base a is denoted byf(x) = ax, where a > 0 but not = 1, and x is any real number (the variable symbol x can take any real number value). If a = 2.134, a isreplaced by the symbol e. And in that case f(x) = ex.Change. A change in a variable’s value is represented by the Greek letter ∆x (delta). Thus, ∆t means a change in time t which is =final time – initial time. Or, ∆t = tf – ti ; or , ∆t = t2 – t1; or, ∆t= tf – to.The suffixes f, i, or 2, 1, f and o are used toindicate the final and the initial values of a variable. If a variable does not change, ∆ of that variable is zero. Binning an object,say to measure its length or any other physical quantity, we use ∆.Q.4. If your height is represented by the variable h, how do you write change in your height by using delta as well as the differentsuffixes?Q. 5. If ∆x is 1 inch, how many ∆x’s are there in 18 inches?Rate of Change. Rate of change is defined as the ratio of the rates of change of the dependent variable and the independent variable.Slope of a Line. A line is drawn by plotting a number of paired values of x and y. Here the ratio of the rate of change of y and x iscalled the slope of the line. Also, you are aware that slope = rise/run. Rise is the change in y value or ∆y and the change in x valueor ∆x. => slope = ∆y/∆x. Keep in mind that the forward slash “/” stands for the mathematical operation of division. Rates of change orslopes of the paired variables are found in the same way.Derivative or the differentiation. When the change in the independent variable is minutely small i. e. almost zero, the rate of changeis then called the derivative of the dependent variable with respect to the independent variable. Derivative of y with respect x isrepresented by dy/dx and is defined asdy/dx = in the limit of ∆x →0, the value of ∆y/∆x.Keep in mind that bits of changes in x and y i. e. ∆x and ∆y in other paired variables are countable or are discrete. In the limit ofincountable or nondiscrete deltas, dy and dx are used. If the ∆x bits are of the sizes of 1 cm, you can count them for your writingpencil. However, if ∆x is of the size of an atom, dx has to be used.Standard Formulas for Finding Derivatives:There are some standard formulas for finding derivatives of functions.d/dx(a) = 0, where a is a constant whose value does not change. => derivative of a constant number is zeroQ. 6. What is d/dx(3.45) = ?d/dx(axn)= naxn-1, => derivative of variable raised to power n is n times the variable exponent reduced by 1. If the variable with theexponent is multiplied by a constant a, the constant also multiplies the derivative.Q. 7. What is d/dx(7×4)= ?d/dx(eax) = a eax, => derivative of e raised to an exponent of a constant times a variable equals the constant times the sameexpressionQ. 8. What is d/dx(e8x) =?Second Derivative or Second Differentiation: Taking derivative twice of a function. Second time differentiation follows the same ruleas the first time differentiation. The second derivative symbol isd2/dx2. Find the second derivative of the function f(x) = 3×3 + 2×2 + x + 1. First time differentiation of the function f(x) followingthe rule d/dx(axn)= naxn-1, yields 9×2 + 4x + 1 + 0 = 9×2 + 4x + 1. Second time differentiation of the first time derivative gives 18x+ 4 + 0 =18x + 4. Check this result. So,(d2/dx2)[f(x)] = (d2/dx2)[3×3 + 2×2 + x + 1]=18x + 4.Partial Differentiation. A function can be of more than one variable. For example, area of a rectangle is a function of both length andwidth, and the volume of a cube is a function is a function of length, width, and height. Inflation of a balloon, or a tire is afunction of both pressure and temperature. Differentiation in this case can be done with respect one variable at a time keeping therest constant. We use del symbol ∂ in the form ∂/∂x instead of d/dx for partial differentiation. Let us consider a function of twovariables f(x,y). The partial derivative of f(x,y) could be written as(∂f/∂x)y and (∂f/∂y)xThe first of these is read as “the partial derivative of f(x,y) with respect to x at constant y”.Q. 9. How to read the second one?Q. 10. (i) How to read β1= (∂V/∂T)P (ii) How to read β= (1/V)(∂V/∂T)PQ. 11. Consider f = x2y3 + 5x and g(x,y,z) = x2 + y2 + z2 – r2 =0(i) What is x2 + y2 + z2 = ? from x2 + y2 + z2 – r2 =0 (ii) What is y2=?(iii) What is y = ?(iv) Substitute y-value in f = x2y3 + 5x and find f(x,z) = ?(v) (∂f/∂x)y = ? (vi) (∂f/∂y)x = ?(vii) (∂f/∂x)z = ? (viii) (∂f/∂z)x = ?(ix) What is z= ? from x2 + y2 + z2 – r2 =0(x) Substitute the z-value in the expression for (∂f/∂z)x , simplify(xi) Substitute the z-value in the expression for (∂f/∂x)z , simplify(xii) Is (∂f/∂x)z = (∂f/∂x)y ?Q. 12. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P(i) Evaluate β= ? Is it 1/T or something else?(ii) Evaluate κ = ? Is it 1/P or something else?Exact Differential. Let u(x,y) be an arbitrary function of x and y. Then the differential of u(x,y) is written asdu = (∂u/∂x)ydx + (∂u/∂y)xdy = Mdx + NdyQ. 13. (i) What is M = ? (ii) What is N= ?The condition for exact differential is (∂N/∂x)y = (∂M/∂y)xLet du = (xy2 + 2x)dx + x2y dy(iii) N = ? (iv) (∂N/∂x)y(v) M = ? (vi) (∂M/∂y)x(vii) Is du an exact differential?Inexact Differentials. Inexact differentials is denoted by the symbol δ (lower case delta).δw= y dx + x dy is inexact and there exists no function w(x,y) which satisfies this equation.Useful Differential Relations for Systems with Two Independent Variables. Three useful mathematical relations can be derived for asystem with three coordinates x, y, and z, which are related by an equation of stateg(x,y,z) =0Here x, y, and z can be P, V, and T, respectively, in thermodynamics relations.Any pair of these variables may be chosen as independent and the equation of state may be expressed in three different forms:x, y independent: z = z(x,y) => z can be written as an explicit function of only x and yx, z independent: y = y(x,z) => y can be written as an explicit function of only x and zy, z independent: x = x(y,z) => x can be written as an explicit function of only y and zQ. 14. (i) What is the differential dx = ?(ii) What is the differential dy = ?(iii) What is the differential dz = ?(iv) Substitute for dy in the expression for dx(v) Simplify, collect like terms with dx and dz(vi) Make sure you get something like[(∂x/∂y)z(∂y/∂x)z – 1]dx + [(∂x/∂y)z(∂y/∂z)x + (∂x/∂z)y]dz= 0(vii) In (vi), the sum of some coefficient times the differentials of independent variables x and z is zero. This implies that thecoefficients of dx and dz must vanish independently. Set the coefficient of dx = 0,Solve for (∂x/∂y)z to get (∂x/∂y)z=1/(∂y/∂x)z(viii) Set the coefficient of dz = 0 => (∂x/∂y)z(∂y/∂z)x + (∂x/∂z)y=0Prove that => (∂x/∂y)z(∂y/∂z)x(∂z/∂x)y= -1Q. 15. Let x, y, z be P, V, and T respectively.(i) Show that => (∂P/∂V)T(∂V/∂T)P = – (∂P/∂T)V(ii) Show that the isothermal pressure coefficient α = (∂P/∂T)V = – (∂V/∂T)P/(∂V/∂P)TEnthalpy of a system = internal energy of the system + work needed to create the system. For example, a magician makes a rabbit out ofnothing and place it on the table. The magician must summon up not only the energy U of the rabbit, but also some additional energy,equal to PV at constant pressure P to push the atmosphere out of the way to make the room of volume V. The total energy required is theenthalpy,H = U + PV (1.51)Change in enthalpy during a constant pressure is∆H = ∆U + P∆V (1.53)According to the 1st law of thermodynamics,Change in energy = heat added to the system, plus expansion-compression work done on it, plus any other work done (e. g. (for example)electrical) on it= > ∆U = Q + (-P∆V) + Wother (1.54)Substituting for ∆U in (1.53) = > ∆H = Q + (-P∆V) + Wother + P∆V= >∆H = Q + Wother (1.55)= > Enthalpy change can be caused by heat and other forms of work and not by compression-expansion work (during constant-pressureprocesses). That is to say, you can forget about expansion-compression work if you deal with enthalpy instead of energy. In no “other”types of work are being done, the change in enthalpy tells you directly how much heat has been added to the system. And that’s why thesymbol H is used.When an object’s temperature is raised, the change in enthalpy per degree, at constant pressure, is the same as the heat capacity atconstant pressure, CP:CP = (∂H/∂T)P (1-56)CP should really be called “enthalpy capacity” as should CV be called “energy capacity”Chemistry deals with ∆H in phase transformations, chemical reactions, ionization, dissolution in solvents, and so on.Q. 1. When you boil 1 mole of water at 1 atm., ∆H = 40,660. What is the change in enthalpy when you boil one gram of water? (One moleof water is about 18 grams)Q. 2. What is the work needed to push the atmosphere away to make room for one mole of water vapor?i) What is the ideal gas equation for n moles?ii) What is the ideal gas equation for 1 mole?iii) What is the Kelvin temperature for water boiling?iv) What is the work done formula for constant pressure P and volume V?v) What is the work done? Ans. 3100 JQ. 3. Consider the combustion of one mole of H2 with ½ mole of O2 under standard conditions. How much of the heat energy comes from adecrease in the internal energy of the system, and how much comes from work done by the collapsing atmosphere. ∆H for this reaction is286 kJ, called the enthalpy of formation of water. (Treat the volume of liquid water as negligible and take the temperature as 25 oC).i) What is the combined mole number of the reactants?ii) What is the Kelvin temperature in question?iii) What is the work done formula for the constant pressure P and the volume V?iv) What is the value of the work done by the atmosphere? Ans. 3717 J = 4 kJ.v) What is the total heat produced in the combustion reaction?vi) How much comes from the internal energy (e. g. chemical energy in the molecular bonds) of the systemQ. 4. Consider the combustion of one mole of methane gasCH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)The system is at standard temperature (298 K) and pressure (105 Pa) both before and after the reaction.(a) Find ∆H in the process of converting a mole of methane into its elemental constituents ( carbon and hydrogen gas) given that ∆Hupon forming one mole of methane from elemental carbon and hydrogen is -74.81 kJ(b) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H toform one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.(c) What is ∆H for the actual reaction in which methane and oxygen form carbon dioxide and water vapor directly?i) What is ∆H in converting the mole of methane into elemental carbon and hydrogen?ii) Add to (i) ∆H to form one mole of CO2iii) Add to (ii) ∆H to form two moles of water Ans. -802.34 kJ(d) How much heat is given off during the reaction, assuming that no “other” forms of work are done?(e) The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy wasproduced by methane combustion given that one mole methane combustion gives off 800,000 J?CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)i) What is one mole of methane mass + two moles of oxygen mass? 0.080 kgii) How many moles of methane would sun contain? 2.5E31iii) How much energy could this mass of the sun give off? 2E37 Jiv) How many years could the fuel last if the sun was giving energy at this rate? 1600 yrsHeat Conduction. The amount of heat Q conducted through a thickness ∆x and area A of a material of thermal conductivity kt and havinga temperature difference of ∆T on either side in time ∆t isQ/∆t = -ktA(dT/dx) (1-60)which is Fourier heat conduction law. The unit of Q is Joule, t is second, kt is watts per meter per Kelvin, T is Kelvin, x is inmeter. Some values of kt are air, 0.026; wood, 0.08; water 0.6; glass 0.8; iron, 80; copper, 400.Q.1. A room window has an area of one square meter and a thickness of 3.2 mm, the temperature inside and outside the room are 20oC and0o C, respectively. If the thermal conductivity of the material of the window is 0.8 W/m.K, find the rate of heat flow through it.Ans. 5000 WattsQ.2. Compute the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 meter square , for atemperature difference of 20oC.Conductivity of an ideal gas.2r 2r2rA molecule of radius r can make a collision with another molecule of radius r if the distance between their centers are 2r. Thecollision can happen at that distance between a molecule of radius 2r and one like a point. When a sphere of radius 2r moves in astraight line of length l, it sweeps out a cylinder of volume 4πr2l. When the volume of this cylinder equals the average volume permolecule in a gas, we are likely to get a collision. The average distance between two collisions is called the mean free path.= > Volume of cylinder = average volume per molecule= > π(2r)2l ≈ V/N= > l ≈ (1/4πr2) (V/N) (1.62)Average time between collisions is ∆t = mean free path, l/vrms (1.63)Q. 3. Find the average time between collisions of nitrogen molecules in the air, given that r = 1.5E-10 m for nitrogen. Treat air as anideal gas.i) What is V/N = ? in terms of k, T, and Pii) What is V/N= ? at room temperature and at atmospheric pressure. Ans. 4E-26 cubic meteriii) What is the mean free path, l = ? Ans. 150E-9 miv) Find vrms for nitrogen molecule = ?v) Average time between collisions is l/vrms = ? Ans. 3E-10 sQuantity of heat conducted in a gas medium through an imaginary barrier having difference of temperatures d = T2 – T1 isQ =1/2 (U1 – U2) = -1/2(U2 – U1) = -1/2 Cv(T2- T1) = -1/2 CvldT/dx (1.64Thermal conductivity is given bykt = ½ CVl/(A∆t) = ½ (CV/Al) (l2/∆t) = ½( CV/V) lvavr (1-65)where vavr is the average speed of the molecules.Heat capacity of the gas per unit volume = CV/= 1/2Nk/V = f/2 P/T (1-66)Q. 4. Air at room temperature and atmospheric pressure has f = 5. (a) Find CV/V = ? Ans. 800 J/m3.K(b) Find the thermal conductivity for air. Ans. 0.031 W/m.KThermal conductivity of a given gas depends only on its temperature through vavr∞√T and possibly through f, and kt is proportional tothe square root of the absolute temperature.Viscosity. Fluids have internal friction called viscosity. F = Aη(dux/dz) (1.69)Where η is called the coefficient of viscosity = F(dz/dux)A . The unit of η is N.s/m2 = Pa.sQ. 5. Find η = ? if F is 3 N, A = 10 cm2, and (dz/dux) = 0.1 s∆z FluidArea ADiffusion. Fick’s law Jx = – Ddn/dx (1-70)Where Jx is the particle flux, D is the coefficient of friction. D for CO at room temperature and pressure is 2E-5 m2/s.Q. 6. Find Jx for Co at room temperature and atmospheric pressure is dn/dx = 100/cmThermodynamics Mid-term Test Name—————————————————–DateNo going out during testing. Finish your outside jobs before the test starts. Empty your desk before sitting for the test. All cellphones must be on the instructor’s desk. No goggles. No hats. No communication devices.1. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P(i) Evaluate β= ? Is it 1/T or something else?(ii) Evaluate κ = ? Is it 1/P or something else?2. Fractional increase in ocean volume due to global warming can be taken to be equal to the fractional rise of ocean water. Giventhat the coefficient of volume expansion β for ocean water is 87.5E-6/oC, what will be the sea level rise of the original depth of 1000m for 1oC rise of global temperature? Ans. 9 cm3. You have 2.00 kg of water at a temperature of 20.0 oC. How much energy is required to raise the temperature of water to 95 oC. If you pay just 10.0 cents for heating each 3.60E6 J, what is the cost of warming the water, given that the specific heat of cwateris 4.19 kJ/kg. K.Ans. Heat required is 629,000 J, and the cost is 1.75 cents4. Suppose 10.0 g of water at a 100.0 oC is in an insulated cylinder equipped with a piston to maintain a constant pressure of101.3 kPa. Enough heat is added to vaporize water to steam at 100.0oC. Water volume is 10.0 cc and steam volume 16,900 cc. (a) What isthe work done by the water as it vaporizes? (b) How much heat water needs to vaporize? (c) What is the change in internal energy ofwater? Ans. 1710 J Q= 22,600 J ∆E = 20,900 J5. Suppose you insulate above the ceiling of a room with an insulation material having a thermal resistance R = 5.2836 m2. K/W.The room is 5.00 m by 5.00 m. The temperature inside the room is 21.0 oC and the temperature above the insulation is 40.0 oC. How muchheat enters the room through the ceiling in a day if the room temperature is maintained at 21.0 oC? Ans. 7.77E6 J6. Suppose your classroom has a temperature of 22oC. What is the average kinetic energy of the molecules (and atoms) in the air?Ans. 6.11E-21 J7. What is the mean free path l of nitrogen molecules in the air at normal atmospheric pressure of 101.3kPa and a 293.15 K, giventhat the radius of nitrogen molecule is 0.150 nm = 0.150E-9 m? Ans. 9.99E-8 m8. The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy wasproduced by methane combustion given that one mole methane combustion gives off 800,000 J?CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)i) What is one mole of methane mass + two moles of oxygen mass? 0.080 kgii) How many moles of methane would sun contain? 2.5E31iii) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H toform one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.iv) How much energy could this mass of the sun give off? 2E37 Jv) How many years could the fuel last if the sun was giving energy at this rate? 1600 yrs∆V = βV∆T; ∆V/V = ∆d/d, ∆d/d = β∆T; Q = cwatermwater∆T; 1 kW h =3.60E6 J; W = PʃdV integrated between Vi and Vf; Q = mLvaporization,Lvaporization = 2.260E6 J/kg; ∆U = Q – W;Q = ∆t. A. (Thigh – Tcool)/R; KEave = 1.5kT; k = 1.38E-23 J/K; l = kT/(17.77r2P); one mole of methane has 0.016kg and one mole ofoxygen has 0.064 kg ; ∆H = Q + Wother.Thermodynamics Mid-term Test Name—————————————————–DateNo going out during testing. Finish your outside jobs before the test starts. Empty your desk before sitting for the test. All cellphones must be on the instructor’s desk. No goggles. No hats. No communication devices.1. Isothermal compressibility is defined as κ = -(1/V)(∂V/∂P)T , β= (1/V)(∂V/∂T)P , and V = (nRT)/P(i) Evaluate β= ?(ii) Evaluate κ = ?2. Fractional increase in ocean volume ∆V/V due to global warming can be taken to be equal to the fractional rise of ocean water∆d/d. Given that the coefficient of volume expansion β for ocean water is 87.5E-6/oC, what will be the sea level rise of the originaldepth of 1000 m for 1oC rise of global temperature?3. You have 2.00 kg of water at a temperature of 20.0 oC. How much energy is required to raise the temperature of water to 95 oC .If you pay just 10.0 cents for heating each 3.60E6 J, what is the cost of warming the water, given that the specific heat of cwater is4.19 kJ/kg. K.4. Suppose 10.0 g of water at a 100.0 oC is in an insulated cylinder equipped with a piston to maintain a constant pressure of101.3 kPa. Enough heat is added to vaporize water to steam at 100.0oC. Water volume is 10.0 cc and steam volume 16,900 cc. (a) What isthe work done by the water as it vaporizes? (b) How much heat water needs to vaporize? (c) What is the change in internal energy ofwater?5. Suppose you insulate above the ceiling of a room with an insulation material having a thermal resistance R = 5.2836 m2. K/W.The room is 5.00 m by 5.00 m. The temperature inside the room is 21.0 oC and the temperature above the insulation is 40.0 oC. How muchheat enters the room through the ceiling in a day if the room temperature is maintained at 21.0 oC?6. Suppose your classroom has a temperature of 22oC. What is the average kinetic energy of the molecules (and atoms) in the air?7. What is the mean free path l of nitrogen molecules in the air at normal atmospheric pressure of 101.3kPa and a 293.15 K, giventhat the radius of nitrogen molecule is 0.150 nm = 0.150E-9 m?8. The sun has a mass of 2E30 kg and gives off energy at a rate of 3.9E26 watts. How long could the sun last if its energy wasproduced by methane combustion given that one mole methane combustion gives off 800,000 J?CH4 (gas) + 2O2 (gas) → CO2 (gas) + 2H2O (gas)vi) What is one mole of methane mass + two moles of oxygen mass?vii) How many moles of methane would sun contain?viii) Determine ∆H in the formation of a mole of CO2 and two moles of water vapor from their elemental constituents given that ∆H toform one mole of CO2 is – 393.51 kJ , and ∆H to form one mole of H2O vapor is – 241.82 kJ.ix) How much energy could this mass of the sun give off?x) How many years could the fuel last if the sun was giving energy at this rate?∆V = βV∆T; ∆V/V = ∆d/d, ∆d/d = β∆T; Q = cwatermwater∆T; 1 kW h =3.60E6 J; W = P∫dV integrated between Vi and Vf; Q = mLvaporization,Lvaporization = 2.260E6 J/kg; ∆U = Q – W;Q = ∆t. A. (Thigh – Tcool)/R; KEave = 1.5kT; k = 1.38E-23 J/K; l = kT/(17.77r2P); one mole of methane has 0.016kg and one mole ofoxygen has 0.064 kg ; ∆H = Q + Wother.The Second LawIrreversible processes are not inevitable but are overwhelmingly probable. Temperature is a way of quantifying the tendency of energyto enter or leave an object during the course of random rearrangements.Q. 1. Take three coins – a penny, a nickel, and a dime. Flip them. (a) How many possible outcomes are there? Represent head by H andtail by T. [ Probability of any event = how many of that event/out of how many total number] (b) What is the probability of gettingthree heads? (c) What is the probability of getting three tails? (d) What is the probability of getting two heads and a tail? (e) Whatis the probability of getting exactly one head and two tails?Penny Nickel DimeH H HEach of the different outcomes in the flipping of the coins is called a microstate. In general, the microstate of a system refers tothe state of each individual particles. In this case, it refers to the state of each coin. For example, HHT or TTH, etc. Macrostaterefers to how many heads or tail. In HHT, there are two heads. Multiplicity refers to the number of microstates corresponding to amacrostate. It is denoted by Greek letter capital omega Ω.Q. 2. In the flipping of the coins in #1, fill out the chart.Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities3H Ω.(3)=?2H Ω.(2) =?1H Ω.(1) = ?0H Ω.(0) = ?Q. 3. In the above example, number of coins was 3 with each having 2 possibilities giving a total of 8 microstate possibilities. (a)Write a mathematical formula to get 8 out of 2 and 3. (b) If you flip 100 coins, what is going to be the total number of microstateswritten in the form as in (a)? (c) What is the total number of macrostates of 0 through 100 heads?Ω(0) = ? when every coin shows tails-upΩ( 1) = ? When only one coin shows heads-up. Any one can show heads-up – first one, second one, third one, etc. up to the 100thone.Ω(2)= ? When two coins, a pair, show heads-up. 100 choices for the first one, and for each of these choices, there are 99 remainingchoices for the second coin. Order of the pair is immaterial – #1 in the first and # 2 in the second as well as # 2 in the first and #1 in the second. The total number of ways of selecting will be 100 . 99. The number of distinct pairs will be (100 . 90)/2.Ω(3) = ? When three coins, a triplet, show heads-up. 100 choices for the first one, 99 choices for the second one , and 98 forthe third. And any triplet could be chosen in several ways – 3 choices for which one to flip first, and for each of these, 2 choicesfor which to flip second. Thus the number of distinct triplets is (100 . 99 . 98)/(3. 2)Ω (n) = ? [100. 99. 98. …….. (100- (-1))]/[n. (n-1). (n-2) … … 3. 2. 1]= 100!/[n!. (100-n)!If there are N coins, the multiplicity of the macrostate with n heads isΩ (n) = N!/n! (N-n)!is the of ways of choosing n objects out N.Q. 4. Suppose you flip four fair coins.(a) Make a list of all possible outcomes as in the table below.Coin #1 Coin #2 Coin #3 Coin #4H H H H(b) Make a list of all the different “macrostates” of 0 heads to 4 heads and their probabilities© Compute the multiplicity of each macrostate using the combinational formula Ω (n) = N!/n! (N-n)!Two-State Paramagnet.The two outcomes of a coin-flipping example can be applied to two-state paramagnets. A paramagnet is a material in which theconstituent particles act like tiny compass needles that tend to align parallel to any externally applied magnetic field. If theconstituent particles interact strongly enough with each other, the material can magnetize even without any externally applied field,we call it a ferromagnet. Paramagnetism is a magnetic alignment that lasts only as long as an external field is applied. Individualmagnetic particles can be called dipoles because each has its own magnetic dipole moment vector. Each dipole could be an electron, agroup of electrons in an atom, or an atomic nucleus. For any such microscopic dipole, quantum mechanics allows the component of thedipole moment vector along any given axes to take on only certain discrete values – intermediate values are not allowed. In thesimplest case only two values are allowed – one positive and the other one negative. In a two-state paramagnet, each elementarycompass needle can have only two possible orientations – either parallel or anti-parallel to the applied field.We define N↑= # of elementary dipoles that point up at some particular timeN↓ = # of dipoles that point down at some particular timeThe total number of dipoles is then N = N↑ + N↓ which will be taken to be fixed.B ↑↓↓↓↓↑↑↑↓↑↓↑↓↑↓Fig. 2.1. A symbolic representation of a two-state paramagnet, in which each elementary dipole can point either parallel orantiparallel to the externally applied magnetic field.This system has one macrostate for each possible value of N↑ from 0 to N. The multiplicity of any macrostate is given by the sameformula of coin-tossing:Ω(N↑) = N!/(N↑! N↓!) (2-7)The external magnetic field exerts a torque on each little dipole, trying to twist it to point parallel to the field. If the externalfield points up, then an up-dipole has less energy than a down-dipole, since it needs additional energy to twist from up to down. Thetotal energy of the system is determined by the total number of up- and down-dipoles. So specifying which macrostate this system is inis the same as specifying its total energy. In fact, in nearly all physical examples, the macrostate of a system is characterized, atleast in part, by its total energy.Q. 1. How do you justify the example of coin-tossing for paramagnetism?Einstein Model of a Solid.Consider a collection of microscopic systems that can each store any number of energy “units” all of the same size. Equal-size energyunits occur for any quantum-mechanical harmonic oscillator whose potential energy function has the form 1/ksx2. The size of the energyunit is hf where h is Planck’s constant (6.63E-34 J.s) and f is the natural frequency of the oscillator (1/2π√(ks/m). Examples ofquantum mechanical oscillators include the vibrational motions of diatomic and polyatomic molecules. In 3-D, each atom can oscillate in3 independent directions. So, if there are N oscillators, there are only N/3 atoms. The model of a solid as a collection of identicaloscillators with quantized energy units was proposed by Einstein in 1907.Let us consider a solid with N= 3 oscillators.Oscillator #1 #2 #3 Oscillator #1 #2 #3Energy 0 0 0 Energy 3 0 01 0 0 0 3 00 1 0 0 0 30 0 1 2 1 02 0 0 2 0 10 2 0 1 2 00 0 2 1 0 21 1 0 0 1 21 0 1 1 1 10 1 1Each row in the table corresponds to a different microstate. There are 1 microstate with energy 0, 3 with 1, six with two, and ten withthree. The general formula for the multiplicity of an Einstein solid with N oscillators and q energy units isΩ(N, q) = (q + N – 1)!/(q!(N-1)!))The following is a graphical representation of the microstate of an Einstein solid. Each energy unit is represented by a dot. Avertical line represents a partition between one oscillator and the next. A solid with 4 oscillators is represented by this sequence.The sequence represents the microstate in which the first oscvillator has one unit of energy, the second oscillator has three, thethird oscillator has none, and the fourth oscillator has four. Any microstate can be represented uniquely in this way, and everypossible sequence of dots and lines corresponds to a microstate. There are always q dots and N -1 lines, for a total of q + N – 1symbols. Given q and N, the number of possible arrangements is just the number of ways of choosing q of the symbols to be dots, that is(q + N – 1)!/(q!(N-1)!))Q. 2. For an Einstein solid with each of the following values of N and q, list all of the possible microstate, count them, and verifythe formula Ω(N, q) = (q + N – 1)!/(q!(N-1)!))(a) N=3, q = 4; (b) N = 3, q = 5; (c) N = 3, q = 6; (d) N = 4, q = 2; (e) N=4, q =3; (f) N= 1, q = anything; (g) N= anything, q = 1Soln. (a) 400 310 031 220 211040 301 103 202 121? ? ? ? ?How many does the formula predict? How many did you get here?(b) 500 410 041 320 032 311 221? ? ? ? ? ? ?? ? ? ? ? ? ?How many does the formula predict? How many did you get here?© 600 501 015 ? ? ? ?060 150 420 ? ? ? ?006 051 402 ? ? ? ?510 105 240 ? ? ? ?How many does the formula predict? How many did you get here?(d) 2000 0020 1100 ? ?0200 002 ? ? ?How many does the formula predict? How many did you get here?e) 3000 2100 ? ? ?0300 2010 ? ? ?0030 2001 ? ? ?? ? ? ? ?How many does the formula predict? How many did you get here?(f) If N = 1, then all the energy must belong to the one and only one oscillator, so there is only one microstate, which we woulddenote by “q”. According to the formula, the multiplicity isq!/q! =1(g) If q = 1, then there is only one unit of energy to distribute among N oscillators, so the allowed states would be 1000…, 0100…,0010…, , and so on up to …0001. There are N places to put the unit of energy , so the number of possible microstates is N. According tothe formulaΩ(N,1) = N!/(1! (N-1)! = NQ. 3. Calculate the multiplicity of an Einstein solid with 30 oscillators and 30 units of energy.Soln. For N = 30 and q= 30, the number of microstates should be Ω (30, 30) = 59!/30!.29! = 5.91E16Q. 4. For an Einstein solid with four oscillators and two units of energy, represent each possible microstate as a series of dots andvertical lines.i) How many vertical lines are needed for four oscillatorsii) How many dots are needed for two energy?? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?There are ten microstates.Paramagnetic Salt.The work done to increase the magnetization of a sample, not including the energy stored in the magnetic field itself, is given byδW = μoH dM (2-8)The total work done is obtained by integrating between the limits of Mi and Mf.= > W = ʃ μoH dM (2-9)where magnetization M has been defined as the magnetic moment (either pole strength times pole separation distance) per unit volume.Magnetic flux density B =permeability of free space, μo times the magnetic field intensity, H.Q. 5. If a uniformly magnetized rod of volume v has N’ elemental magnetic dipoles of moment ∆m, find the magnetization of the bar.i) What is the total dipole moment in volume v?ii) What is the dipole moment per unit volume?Q. 6. Given H = MT/D, what is W = ʃ μoH dM ? integrated between the limits of of Mi and Mf.Also, if the equation of state is M = DH/T, what is W = ? in terms of μoH M. Ans. μoHfMf – μoHiMiParamagnetic Solid.Isothermal ProcessesQ. 7. From the first law of thermodynamics,dU = δQ + δW (2-10)i) Substitute (2-8) for δW in (2-10) and solve for δQ = ?Take the internal energy function as a function of temperature T and magnetism M.= > U = U(T, M) (2-11)ii) Write down the differential for dU = ? (2-12)(Check with your handout on “Mathematics for Thermodynamics”)i) What is δQ = ? Ans. (∂U/∂T)MdT + [(∂U/∂M)T – μoH]dM (2-13)ii) Imagine the process to take place at constant magnetization, (δQ/dT) = ? (2-14)iii) Let CM = (∂U/∂T)M, (2-15)substitute this in your answer to ii). δQ = ? (2-16)Ans. δQ = CMdT + [(∂U/∂T)T – μoH]dMiv) Consider a process at constant H. Differentiate your answer to v) with respect to T:What is CH = (δQ/dT)H = ?Ans, CH = CM + [(∂U∂M)T – μoH](∂M/∂T)H (2-17)Many paramagnetic substances obey Curie lawM= DH/T = > Induced magnetic moment = (Curie constant x magnetic intensity)/Temperature (2-18)and have an internal energy which is a function of temperature only. For such a material, called a perfect paramagnetic substance,answer to vi) simplifies considerably.CH = CM-μoH(∂M/∂T)H (2-19)CH = CM + μoDH2/T2 (2-20)CH = CM μo M2/D (2-21)The internal energy for a perfect paramagnetic solidU = ʃCMdT (2-22)Adiabatic Processes for Perfect Paramagnetic SolidStart with the answer to iii) : δQ = (∂U/∂T)MdT +[(∂U/∂M)T – μoH]dM (2-16)In an adiabatic process, there is no change in the quantity of heat, = > δQ =0; however temperature varies = > (∂U/∂M)T vanishes. Usethe definition of CM, to find 0 = CMdT – μoHdMQ. 8.i) Solve for CMdT = ? (2-17)ii) What is H = ? from Curie law (2-18)iii) Substitute for H on the right hand side of i), CMdT = ? (2-19)iv) Rewrite iii) to find (CM/T)dT = ? (2-20)v) Integrate left hand side from To to T and and right hand side from Mo to MDid you get CMln(T/To) = μo(M2 – Mo2)/2D ? (2-21)Solve for T = ? Ans. T = Toexp[μo(M2 – Mo2)/(2DCM)] (2-22)This relation shows how a paramagnetic material may be used to cool a sample by adiabatic demagnetization. A large magnetic field isimpressed on the sample as it is cooled to as low a temperature as is possible. The external magnetic field is then turned off andtemperature drops exponentially.Simple SolidA simple solid being a hydrostatic system, work is done at constant pressure called isobaric work.W = P(Vi-Vf) (2-23)For a solid well below the melting point and for moderate pressures up to several hundred atmospheres, the equation of state is givenbyV = Vo(1 + βT – κP) (2-24)where β called the thermal volume expansibility is a measure of how much the material expands as its temperature is raised at aconstant pressure:β≡ 1/V . (∂V/∂T)P (2-25)and κ the isothermal compressibility is a measure of how much the solid changes volume as the pressure is changedκ= -1/V . (∂V/∂P)T (2-26)A better approximation for the equation of state for a solid, we writeV = Vo[1 + β(T –To)– κ(P-Po)} (2-27)Where Vo, β, and κ refer to temperature To and pressure Po.Q. 9.i) Following Eqn (2-27), write down Vi = ? Vf = ?ii) Substitute Vi and Vf in (2-23), to get W = ? Ans. W = PVoβ (Ti – Tf) (2-28)iii) From (2-24) , dV = ?iv) For an isothermal process (dT = 0) , dV = ?v) Substitute in the work equation W = – ʃPdV between the limits Pi and Pfvi) Integrate v) Ans. W = Voκ/2(Pf2 – Pi2) (2-29)Hydrostatic equations are CV = (∂U/∂T)V (2-30)CP = CV + [(∂U/∂V)T + P] (∂V/∂T)P (2-31)vii) Solve (2-31) for [(∂U/∂V)T + P] ?viii) Substitute in δQ = CVdT + [(∂U/∂V)T + P]dV and simplify.ix) Did you get δQ = CVdT + [CP – CV)/βVodV (2-32)x) For an adiabatic process, what is dT = ? from ix)xi) Integrate dT to get T≡[(1- γ)/βVo]V + c (2-33)xii) Combine xi) with (2-24) to find P = (-γ/κVo)V + c (2-34)which is the relation between pressure and volume for an adiabatic process.xiii) Work done during an adiabatic compression is W = ʃ – PdV between the limits of Vi and Vf.Take differential of (2-34) to get dP = (-γ/κVo)dV. Solve for dV to substitute in the work integral. Did you get W = (Voκ/2γ) (Pf2 –Pi2) (2-35)Q. 10. The pressure on a 1 cm cube of aluminum of mass 2.7 g was increased from 1 to 1000 atmospheres at a constant temperature of300K. Use the simple solid model to calculate (a) the amount of work done by using Eqn (2-29), (b) the amount of heat transferred ,(iii) the change in internal energy of the sample. Given that the compressibility of Al is 1.4E-11 m2/N and expansivity = 69E-5 /K.a) i) What is the volume Vo of Al sample? Ans. 1E-6 m3ii) W = ? Ans. 0.07 Jb) i) From Eqn (2-32) with dT =0 for isothermal case, δQ = ?ii) use dV = -VoκdP and CP – CV = mass of Al x38 J/kg.K in i) to find Q. Ans. -2.08 Jc) Use the first law of thermodynamics ∆U = Q + W to find ∆U = ?Interacting SystemsTo understand heat flow and irreversible processes, consider a system of two solids A and B. They are very weakly coupled meaningexchange of energy between them is much slower than the exchange of energy among the atoms within each. Their energies are UA and UBwhich can change on longer time scales. The total multiplicity for all allowed values UA and UB counting all possible microstates withonly the sum Utotal = UA + UB held fixed. qtotal = qA + qBThere are seven possible macrostates with qA = 0, 1, 2, 3, 4, 5, 6. The total multiplicity of any macrostateSolid A of energy UA Solid B of energy UB# of HOs NA=3, Units of # of HOs NB = 3, units ofenergy qA = 3 energy qB = 3Ωtotal is just the product of the individual multiplicities, since the systems are independent of each other. For each of the ΩAmicrostates available to solid A, there are ΩB microstates available to solid B.Q. 1. a) Fill out the table. b)What is the total number of microstates accessible to the system over long time scales (sum the lastcolumn)? c) Check the number by applying the standard formula to the entire system of six oscillators and six energy units – ΩB (6,6)= (6 + 6 – 1)!/(6!(6-1)!))qA ΩA= (qA + NA – 1)!/(qA!(NA-1)!)) qB ΩB= (qA + NA – 1)!/(qA!(NA-1)!)) Ωtotal =ΩAΩB Ωtotal/sum0 1 6 28 28 28/4621 3 5 21 63 63/462? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?? ? ? ? ? ?Individual numbers in column 5 represent macrostates, to be read as 1st macrostate has 0 energy in solid A and all the 6 energy unitsin solid B, 2nd macrostate has 1 energy unit in solid A and 5 energy units in solid B, 3rd macrostate has energy 2 units in solid Aand 4 energy units in solid B, and so on. The sum of column 5 represents all microstates. Column 6 represents the chances of findingthe system in the 1st, 2nd, 3rd, etc. microstate.Q. 2. Which macrostate is the most probable?Q. 3. Plot the total multiplicity in a bar graph – qA along x-axis and Ωtotal along y-axis.___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Fundamental Assumption of Statistical Mechanics. In an isolated system in thermal equilibrium, all accessible microstates are equallyprobable.Over long time scales, the energy gets passed around randomly in such a way that all 462 microstates are equally probable.Q. 4. Produce a computer-generated table and graph for a system of two Einstein solids with NA = 300, NB = 200, qtotal =100. You canuse a computer spreadsheet program, or comparable software, or perhaps even a graphing calculator.qA ΩA= (qA + NA – 1)!/(qA!(NA-1)!)) qB ΩB= (qA + NA – 1)!/(qA!(NA-1)!)) Ωtotal =ΩAΩB Ωtotal/sum0 1 100 2.8E81 2.8E81 2.8E81/9.3E1151 300 99 9.3E80 2.8E83 2.8E83/9.3E115? ? ? ? ? ?? ? ? ? ? ?. . . . . .. . . . . .. . . . . .100 1.7E96 0 1 1.7E96 1.7E96/9.3E115The spontaneous flow of energy when a system is at , or very near, its most likely macrostate, that is, the macrostate with thegreatest multiplicity. The “law of increase of multiplicity” is one version of the famous second law of thermodynamics.Q. 5. Following the above statement of the second law of thermodynamics, name the macrostate in Q. 1. for the spontaneous flow ofenergy.Q. 6. Consider a system of two Einstein solids, A and B, each containing 10 oscillators, sharing a total 20 units of energy. Assume thesolids are weakly coupled, and that the total energy is fixed.a) How many different macrostates are available to this system?b) How many different microstates are available to the system? Ω(20,20) = ? Ans. 6.89E10c) Assume that this system is in thermal equilibrium, what is the probability of finding all the energy in solid A? Ω(10,20)/Ω(20,20)= ? Ans. 1.00E7/6.89E10 = 1.45E-4d) What is the probability of finding exactly half of the energy in solid A? Ω(10,10)Ω(10,10)/Ω(20,20) = 8.53E9/6.89E10e) Under what circumstances would this system exhibit irreversible behavior?The probability of energy being evenly distributed is greater than that of all the energy being in A by a factor of 1000. If the systemstarted with all the energy in one solid or the other, we would be pretty sure that it would evolve toward a state with energy moreevenly distributed. And if it started out with the energy evenly distributed, we could be pretty sure that at some later time wewouldn’t find all the energy on one side or the other- this would happen less than one time in a thousand.Large Systems.In the previous section, it was shown that the multiplicity function becomes very sharp with the increase of the number of oscillators.Here we look at what happens when the system is much larger, so that each oscillator contains, say 1020 or more oscillators.Very Large NumbersWe encounter three kinds of numbers in statistical mechanics – small numbers, large numbers, and very large numbers. Examples of smallnumbers are numbers less than one hundred.Large numbers are written in powers of ten. The most important large number is Avogadro’s number. Large numbers do not change whensmall numbers are added to them. For example 10E23 + 23 = 10E23.Very large numbers can be made by exponentiating large numbers. Very large numbers can be multiplied by large numbers without changingthem. For instance 1010exp23 x 1023 = 10(10exp23 + 23) = 1010exp23Q. 7. The natural logarithm function, ln, is defined so that elnx = x for any positive number x(a) Sketch a graph of the natural logarithm function: y = ln x(b) Prove the identities ln ab = ln a + ln b (start with e ln ab = ? = e lna e lnb = ?© Prove that ln ab = b ln a (e ln a exp b ) = ab =(d) Prove that (d/dx)(ln x) = 1/x(e) Derive the approximation ln(1 + x) ≈ x which is valid for |x|<< 1. Use a calculator to check the accuracy of this approximationfor x = 0.1 and x = 0.01Q. 8. (a) Simplify the expression e a ln b (Write it in a way that does not involve logarithm)(b) Assume b < 1023 = x ln 10. Noe solve for x. Ans. 4.34E22Thermodynamics Chapter Test #2Name——————————————————— Date____________________________Q. 1. The temperature of a block of copper in increased from 127 oC to 137oC. Find the change in pressure to keep the volume constant,given that β≈5.2E-5 K-1, and κ ≈ 7.6E-12 Pa-1.i) If volume is to remain constant, what is the relation between V and Vo? V = Voii) Following i), what is the change in pressure expression in terms of β, κ, and temperature change?∆P = (β/κ) ∆Tiii) Substitute the given values to find the change in pressure. Ans. 6.8E7Pa ≈680 atmQ. 2. Take four coins – a penny, a nickel, a dime, and a quarter. Flip them.i) How many possible outcomes are there? 16ii) What is the probability of getting four heads? 1/16iii) What is the probability of getting four tails? 1/16iv) What is the probability of getting two heads and two tails? 6/16v) What is the probability of getting exactly one head and three tails? 4/16Penny Nickel Dime Quarter` H H H HH H H TH H T HH T H HT H H HH H T TH T T HT H T HT T H HH T H TT H H TH T T TT T T HT T H TT H T TT T T TQ. 3. In the flipping of the coins in #1, fill out the chart.Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities4H Ω.(4)=? 13H Ω.(3) =? 42H Ω.(2) = ? 61H Ω.(1) = ? 40H Ω.(0) = ? 1Ω(N,n) = (N!)/[n!. (N-n)!]; V = Vo[1 + β(T –To)– κ(P-Po)} where Vo, β, and κ refer to temperature To and pressure Po. 0! = 1Thermodynamics Chapter Test #2Name——————————————————— Date____________________________Q. 1. The temperature of a block of copper in increased from 127 oC to 137oC. Find the change in pressure to keep the volume constant,given that β≈5.2E-5 K-1, and κ ≈ 7.6E-12 Pa-1.i) If volume is to remain constant, what is the relation between V and Vo? V = Voii) Following i), what is the change in pressure expression in terms of β, κ, and temperature change?∆P = (β/κ) ∆Tiii) Substitute the given values to find the change in pressure.Q. 2. Take four coins – a penny, a nickel, a dime, and a quarter. Flip them.i) How many possible outcomes are there?ii) What is the probability of getting four heads?iii) What is the probability of getting four tails?iv) What is the probability of getting two heads and two tails?v) What is the probability of getting exactly one head and three tails?Penny Nickel Dime QuarterQ. 3. In the flipping of the coins in #1, fill out the chart.Macrostate Multiplicity Probability = State multiplicity/Sum of all multiplicities4H Ω.(4)=?3H Ω.(3) =?2H Ω.(2) = ?1H Ω.(1) = ?0H Ω.(0) = ?Ω(N,n) = (N!)/[n!. (N-n)!]; V = Vo[1 + β(T –To)– κ(P-Po)} where Vo, β, and κ refer to temperature To and pressure Po. 0! = 1A macrostate (the total number in the same state) of a system , or configuration is specified by the number of particles in each of theenergy levels of the system. Thus, let there be Nj the number of particles that occupying the jth energy levels of n energy levels.That is if n = 3, we understand that N1 in the 1st level, N2 in the 2nd level, and N3 in the 3rd level. The total number of particlesis N1 + N2 + N3 = NA microstate (which one is in which state) is specified by the number of particles in each energy state. In general, there will be morethan one energy state (i. e. quantum state) for each energy level, a situation called degeneracy. A microstate is the most specificdescription one can get. In general, there will be many, many differenent microstates corresponding to a given macrostate. The basicpostulate of statistical thermodynamics is that all possible microstates of an isolated assembly are equally probable. We areinterested in the number of microstates but not in their detailed description.The number of microstates leading to a given macrostate is called the thermodynamic probability. It is the number of ways in which agiven configuration can be achieved. This is an “unnormalized probability, an integer between zero and infinity. For the kth macrostate, the thermodynamic probability is taken to be wk. A true probability Pk could be obtained by dividing wk by the total number ofmicrostate Ω available to the system.In the coin-tossing experiment, each macrostate is defined by the number of heads and the number of tails. If we toss N= 4 coins,macrostates are (i) zero head and four tail, (ii) one head and three tails, (iii) two heads and two tails, (iv) three heads and onetail, and (v) four heads and zero tail. A microstate is specified by the state – head or tail of each coin. It is the most detaileddescription possible. Our interest is to find the number of microstates for each macrostate, the thermodynamic probability. The trueprobability is the number divided by the total number of microstates. ThusPk = wk/Ωwhere 5Ω = ∑ wk = 16k = 1We can calculate the average occupancy numbers i.e. the average number of heads and tails. Let j = 1 or 2, where N1 is the number ofheads and N2 is the number of tails. Let Nk be the occupation number for the kth macrostate. Then the average occupation number isNjav={∑Njk wk}/{∑wk } = ∑ Njkwk/Ω = ∑Njk wkk k k kThe average number of heads is thereforeNjav = (1/16){(4×1) + (3×4) + (2×6) + (1 x4) + (0 x1)} = 2Similarly, average N2av = 2, as we would expect. Then N1av + N2av = 4 = N. In a plot of thermodynamic probability w vs the number ofheads N1, the curve becomes symmetric about N1 = 2, the most probable configuration.In a large number N of distinguishable coins, the number of ways to select from N candidates N1 heads and N-N1tails is w = N!/{(N!)(N-N1)!} which is the binomial coefficient.Macrostate Label Macrostate Specifications Microstate Therm. Prob. True Prob.k N1 N2 Coin 1 Coin2 Coin 3 Coin 4 wk Pk____________________________________________________________________________________1 4 0 H H H H 1 1/162 3 1 H H H T 4 4/16H H T HH T H HT H H H3 2 2 H H T T 6 6/16T T H HH T H TT H T HH T T HT H H T4 1 3 H T T T 4 4/16T H T TT T H TT T T H5 0 4 T T T T 1 1/164.1 CyclesThermodynamic cycle. Processes which can continue indefinitely without permanently changing the system.A heat engine is any device that absorbs heat and converts part of that energy into work. The net work done by the engine in some giventime period is denoted by W. The benefit of operating a heat engine is the work Produced. The cost of operation is the heat absorbed,Qh.Hot reservoir. Heat absorbed by the engine comes from a place called the hot reservoir. Its temperature is denoted by Th. Heatabsorbed from a hot reservoir in some given time period is denoted by Qh.Cold reservoir. The waste heat is dumped into a place called the cold reservoir. Its temperature is denoted by Tc. Heat expelled to thecold reservoir in some given time period is denoted by Qc.Reservoir. Anything that is so large that its temperature does not change noticeable when heat enters or leaves. In steam engines,the hot reservoir is the place where fuel is burned and the cold reservoir is the surrounding environment.As a system undergoes a cycle, it goes through a series of transformations, after which the system has returned to its initial state.During this cycle, the system may do work and absorb or exhaust heat to one or many heat reservoirs. The net result is the conversionof heat to work, if the system is a heat engine, or the use of work to remove heat, if the system is a refrigerator.Efficiency. The efficiency of an engine, e, is the benefit/cost ratio. = > e ≡ benefit/cost = W/Qh (4.1)Q. 1. What is emax for given values of Th and Tc?According to the first law, energy absorbed = energy expelled + work done = > Qh = Qc + W (4.2)i) W = ? from Eqn. (4.2)ii) Substitute W from i) in Eqn (4.1)iii) Do you get e = 1 – Qc/Qh ?iv) What has to be Qc = ? if e = 1Coefficient of performance (COP) in heating mode = |Qh|/WCoefficient of performance (COP) in heating mode = |Qc|/WA good refrigerator should have a COP of 5 or 6.Q. 2. A refrigerator has a COP of 5. When the refrigerator is running, it power is 500 W. A sample of water of mass 500 g and attemperature 20 deg Celsius is placed in the freezer compartment. How long does it take to freeze the water to ice at 0 deg Celsius?Hint. i)_ How much heat is to be extracted to reduce the water temperature to 0 deg Celsius?Q1 = mc∆T= mass x specific heat of water x change of temperature = ? cwater = 4186 J/kg. oCii) What is the latent heat of fusion ? Lf =3.33E5 J/kgiii) How much heat is to be extracted to freeze water at 0 oC?, Q2 = m Lf = ?iv) What is the total amount of heat that has to be subtracted, sum of i) and ii). = ?Ans. 2.08E5 J/kg(v) COP = |Qh|/W = ? = > W = ? Ans. 4.17E4 Jvi) Power = Work/Time = > Time = Work/Power = ? Ans. 83.3 sConfiguration space. The coordinates x, y, and z specify coordinate space. A point in configuration space corresponds to one position,while a point in momentum space corresponds to one momentum.Momentum space. Momentum axes px, py, pz define momentum space.Phase space. A combination of configuration and momentum space into a six dimensional is called the phase space. A trajectory in phasespace represents a curve of constant total mechanical energy. Since classically, we can predict No true trajectories can intersect.Classically, the momentum and position of a particle can be determined to as high a precision as desired; however, quantummechanically, the Heisenberg uncertainty principle limits the precision with which we can simultaneously measure y and py-∆y.∆py≥ h (1)In 6-dimentional phase space, we can determine the position of a particle only to within a hyper-volume of h3.Phase space for a system of N particles is 6N-dimensional spanned by the momenta and positions of the particles. Each point in thisspace represents what is called a microstate of the system. Generally, we cannot know the microstate of the system, only themacrostate. If the system is in a given macrostate, it usually can be in any of a very large number of microstates.Ensembes and the Distribution Function. If we know nothing about a system of N particles, the system may be in any microstate. Thepoint representing the system may be anywhere within the phase space. If, on the other hand, one knows some macroscopiccharacteristics, such as the total energy and volume, the system may be in only those microstates which are consistent with the knownvalues. When considering a system with some macroscopic characteristics, we actually consider a set of possible systems, each in adifferent microstate. This collection of systems form an ensemble. An ensemble is represented in phase by a subset of all the possiblepoints. In quantum mechanics, the subset may be a collection of discrete, quantized points. In classical mechanics, this subset ishyper-volume or hyper-surface.The system is in only one microstate, corresponding to one point in phase space, at any instant of time; however, it will becontinuously changing microstates within the allowed set. The probability of finding a system in a given microstate is called is calledthe distribution function.The notation f(xi,pi) represents the distribution which is a function of all positions, xi, and momenta pi, where I = 1 to 3N.Alternatively, we may choose to number each of the microstates , j = 1 to Ω, where Ω is the number of possible microstates.The distribution function is defined so that it describes the probability of finding the system in a particular microstate. If theensemble consists of a continuous region,f(xi,pi).dVx. dVp (2)is the probability of finding the system in the hyper-volume dVx.dVp about the point (xi,pi) in phase space.In rectangular coordinates, dVx = dx1dx2dx3dx4…dx3N-1 dX3N (3)And dVp = dp1dp2dp3dp4…dp3N-1 dp3N (4)The measurement of the observable property , A, of system such as energy or magnetization, three types of averages are important – timeaverage, ensemble average, and the most probable value.Time average is obtained by repeatedly measuring the system as it evolves with time.Ensemble average is found by averaging over all the microstates in the ensemble≡ ∑Ajfj/∑fj (5)or ≡∑A(xi,pi)f(xi,pi)/∑f(xi, pi) (6)If the ensemble consists of a continuous region of phase ∑ will represent an integral ≡∑A(xi,pi)f(xi,pi)/ ∑f(xi, pi) →ʃA(xi,pi)f(xi,pi)dVx dVp /∑f(xi, pi) (7)Q. 1.Transform the sum into integration for continuous x: ∑mixi →The Canonical Ensemble and the Partition FunctionSince systems are not generally isolated but rather in thermal equilibrium, we are much more likely to know a system’s temperature thatits energy. Let us separate the systems and the reservoir (a large system with a known temperature but isolated from the environment)by a diathermal wall so that the number of particles in the system , N, and the number particles of the reservoir, No, are fixed. Thetotal energy of the system plus the reservoir is Uo and the energy of the system alone is ε. Then the energy of the reservoir must beUo – ε. Since energy may flow between the system and the reservoir, ε is not fixed but fluctuates about average value.Boltzmann factor. The probability of finding the system in a single state with energy ε must be proportional to the numberof states Ω(Uo-ε) of the reservoir with energy Uo-ε. That isf(ε) ∞ Ω(N, Uo, ε) (8)Q. 2.What is f(ε1) ∞ ?Q. 3. What is f(ε2) ∞ ?Q. 4. What is f(ε1)/f(ε2) = ?The ratio of the probability of finding the system in a single microstate with one of two different energies ε1 and ε2 is thenf(ε1)/f(ε2) = Ω(N, Uo, -ε1)/ Ω(N, Uo, -ε2) (9)The microscopic definition of entropy isS(E, V, N)≡k ln (Ω(E)) (10)where k is a constant.Q. 5. What is Ω(E) = ? (11)And f(ε1)/f(ε2) = Ω(N, Uo, -ε1)/ Ω(N, Uo, -ε2) (12)= > f(ε1)/f(ε2) = exp[S(N, Uo,- ε1)/k]/exp[S(N, Uo,- ε2)/k] (13)Q. 6. What is a Taylor series? When can you expand a function in a Taylor series?Since the reservoir is much larger than system, ε < f(ε1)/f(ε2) = exp[S(N, Uo)- ε1)/kT]/exp[S(N, Uo)- ε2)/kT] (19)= eS(N, Uo). – eε1/kT/ eS(N, Uo). – eε2/kT) = – eε1/kT/ – eε2/kT (20)= > f(ε1)/f(ε2) = = – eε1/kT/ – eε2/kT (21)This exponential term is called the Boltzmann factor and the ratio of Boltzmann factors represents the ratio of the probability thatthe system will be in a given microstate of energy ε1 to the probability that the system will be in a single microstate of energy ε2.The distribution function for the system must be of the formf(εj) = Aexp(-εj/kT) = Ae-εj/kT, j is a suffix of ε. (22)The constant A is determined by requiring the distribution function to be normalized –∑f(εj)= ∑Ae-εj/kT = 1 (23)j jThus A = (∑e-εj/kT)-1 (24)j= > A = [∑exp(-εj/kT)]-1 (25)jThe Partition Function for the Canonical EnsembleThe sum of Boltzmann terms, called the partition function, is represented by a Z (from German zustandsumme) –Z = ∑exp(-εj/kT) = sum of all Boltzmann factors (26)jQ. 10. Write partition function for four states j = 1, 2, 3, 4Q. 11. The two loest energy states of atomic hydrogen have energies of -13.6 eV and – 3.4 eV. Determine the ratio of the number ofhydrogen atoms in the first excited state to the number in the ground state at T = 300 K. Repeat for T = 6000 K and T = 15 M K. Giventhat k = 8.62E-5 eV/KHints. i) Write down the ration of probabilities for single states with energy ε1 and ε2.f(ε2)/f(ε1) =?ii) What is the formula for the number of electrons in n = 1, 2, 3, states etc.iii) How many electrons are in the n = 1 state with energy ε1 = -13.6 eViv) How many electrons are there in the n = 2 states with energy ε2 = -3.4 eV?v) What is the ratio of atoms in state n =2 to the number in state n = 1 ?vi) (ε1 – ε2)/kT = ? for T = 300 Kvii) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 5.0E-172viii) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans. 2.0E-171Since the ratio of n=2 state to n=1 state is so small, all the hydrogen is in its ground state (n = 1)ix) (ε1 – ε2)/kT = ? for T = 6000 Kx) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 2.7E-9xi) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans.1.08E-8xii) (ε1 – ε2)/kT = ? for T = 15E6 Kxiii) f(ε2)/f(ε1) = e(ε1 – ε2)/kT = ? Ans. 0.992xiv) Ratio r = 4 xf(ε2)/f(ε1) = ? Ans. 3.968xv)Q. 12. Prove that the probability of finding an atom in any particular energy level is P(ε) = (1/Z)e-F/kTWhere F = ε-TS and the “entropy” of a level is k times the logarithm of the number of degenerate states for that level (which defines S= k ln n)Hint. n degeneracy means the probability of the system being in that level is n times the probability of being in any one of the state.P(ε) = nf(ε) = n. 1/Z . e-ε/kTWrite n as eln n = eS/k= > P(ε) = eS/k. (1/Z). e-ε/kT = 1/Z . e –(ε-TS)/kT = 1/Z. e-F/kTQ. 13. Consider a hypothetical atom that has two states: a ground state with energy 0 and an excited state with energy 2 eV. Draw agraph of the partition function (PF) for the system as a function of temperature, and evaluate the PF mathematically at T = 300 K, 3000K, 30,000 K, and 300,000 K, given that k = 8.62E-5 eV/KHints. i) Write down the PF expression, Z = ?ii) Expand it for two-state systemiii) Where does Z → as T → to 0iv) Where does Z → as T → to ∞v) Substitute t = kT/ε, then Z = ?vi) Fill out the tableT t = kT/ε Z300 K ? ?3000 ? ?30,000 ? ?300,000 ? ?vii) Plot kT/ε (along x) vs Z (along y)We rewrite Eqn (22) f(εj) = Aexp(-εj/kT) = Ae-εj/kTas Ƥ(s) = (1/Z) e-E(s)/kT(27)= > Probability Ƥ(s) of finding state s with energy E(s).Z = ∑e-E(s)/kT = the sum of all Boltzmann factors.sZ does not depend on any particular state, but does depend on temperature.To interpretation Eqn (27), let us suppose for a moment that the ground state energy of a one-atom system is zero i. e. Eo = 0, whilethe excited states have positive energies. Then the probability of the ground state is 1/Z, and all other states have smallerprobabilities. States with energies much less than kT have probabilities only slightly less than 1/Z, while states with energies muchgreater than kT have negligible probabilities, suppressed by the exponential Boltzmann factor. Q. 15 Fig. shows a bar graph of theprobabilities for the states of a hypothetical system.Q. 14. If the ground state energy is zero, what is Boltzmann factor = ?Q. 15. The figure below represents a hypothetical system. (a) Estimate partition function for this. (b) Estimate the probability ofthis system being in its ground state.Bars represent relative probabilities of a hypotheticalsystem.Horizontal axis is the energy. The smooth curverepresents theBoltzmann distribution for one particular temperature.At lowerTemperatures it would fall off more suddenly, while athigherTemperatures it would fall off more gradually. BarsrepresentProbability Ƥ(s) Boltzmann factors, tooE(s)kT 2kT 3kTHints. i) What is the ground state energy in the problem?ii) What does it contribute to the partition function?iii) What is the height of all the 9 bars in the figure? Measure and sum them.iv) What is the height of the first bar?v) Relative to the height of the first bar, the total of all of them = ? (sum of heights/first bar height)vi) What is Z = ? (the quotient obtained in # v)vii) The probability of the ground state = ? (reciprocal of Z)Q. 16. Imagine a particle that can be in only three states with energies – 0.05 eV, 0, and 0.05 eV. This particle is in equilibriumwith a reservoir at 300 K.(a) Calculate the PF for the particle(b) Calculate the probability for this particle to be in each of the three states.© Because the zero point for measuring energies is arbitrary, we would just as well say that the energies of the three states are 0,0.05 eV, and 0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn’t.(a) i) kT = (8.62E-5 eV/K) (300 K) = ?ii) Z= Z1 + Z2 + Z3 = e-(-0.05/0.026) + eo + e-(0.05/0.026) = ?(b) Ƥ1 = Z1/Z = ? Ƥ2 = Z2/Z = ? Ƥ3= Z3/Z = ?© i) Z= Z1 + Z2 + Z3 = e0 + e-(0.05/0.026) + e-(0.10/0.026) = ?ii) Ƥ1 = Z1/Z = ? Ƥ2 = Z2/Z = ? Ƥ3= Z3/Z = ?Nature possibly can’t care what we use as our zero-point for measuring energy.November 28, 2011The probability that a system is in any particular one of its microstate s, given that it is in equilibrium with a reservoir attemperature T:Ƥ(s) = (1/Z)e-βE(s) (6-12)where β= 1/kT called the Boltzmann factorand Z = ∑ e-βE(s) (6-13)swhere the sum is carried over all the possible states.Q. 1. In a system of 5 atoms, 2 of them in the ground state, 2 in the 4-eV state, and 1 in the 7-eV state. What is the average energyof all the atoms? Ans. 3 eVFor a large sample of N atoms, and N(s) the number of atoms in any particular state s, the average value of energy is= ∑[E(s).N(s)]/N = ∑ [E(s) N(s)]/N = ∑E(s)Ƥ(s) (6-16)Where Ƥ(s) is the probability of finding an atom in state s.= > Average energy = sum of all energies weighted by their probabilities.In statistical mechanics, each probability is given by (6-12).So, = (1/Z)∑E(s) e-βE(s) (6-17)The sum is similar to the partition function (6-13), but with an extra factor E(s) in each term.The average value of any other variable of interest can be computed in exactly the same way. If a variable X has the values X(s) instate s, then = ∑X(s)Ƥ(s) = (1/Z)∑X(s)e-βE(s) (6-18)Average values are additive. That is the average values of two objects is the sum of their individual average values. In a collectionof many identical, independent particles, the total average energy is just the product of average energy of just the one and theirnumber.Q. 2. Suppose you have 10 atoms: 4 with energy of 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and one with energy 6 eV.(a) Compute the average energy of all the atoms by adding up all the energies and dividing by 10. Ans. 1.7 eV(b) Compute the probability that one of the atoms chosen at random would have energy E, for each of the four values of E that occur.© Compute the average energy using the formula = ∑E(s)Ƥ(s)Hint. (a) You can do it.(b) i) What is the total number of atoms?ii) How many atoms have energy of 0 eV?iii) What is the probability of finding at atom with energy of 0 eV?iv) How many atoms have energy of 1 eV?v) What is the probability of finding at atom with energy of 1 eV?vi) How many atoms have energy of 4 eV?vii) What is the probability of finding at atom with energy of 4 eV?viii) How many atoms have energy of 6 eV?ix)What is the probability of finding at atom with energy of 6 eV?© i) What is the product of 0 eV and iii) of (b)?ii) What is the product of 1 eV and v) of (b)?iii) What is the product of 4 eV and vii) of (b)?iv) What is the product of 6 eV and ix) of (b)?v) What is the sum of i) through iv)? Does it agree with (a)Q. 3. Z = ∑ e-βE(s), i ) find ∂Z/∂β = ?sii) Divide i) by –Z. Is it the same as (6-16)?siii) What is (-∂/∂β)(ln Z) Is it the same as (6-16)?Q. 4. Go to Q. 1. a) i) What is the average value of energy?ii) Compute the deviations of each of the energies from the average energy for all the five atoms.iii) Compute the squares of each of the deviations.iv)Find the sum of (iii)v) Find the averagevi) Find the square root of v). Ans. 2.7 eVThis is called the root-mean-square (rms) deviation or the standard deviation.(b) Prove that σE2 = – 2.i) σE2 = = (1/N) ∑ (∆Ei)2 = (1/N)∑(Ei – )2 = ?Paramagnetism Q. 5. Each elementary dipole in an ideal two-state paramagnet has just two possible states: an “up” state with energy –μBand a “down” state with energy + μB, where B is the strength of an externally applied magnetic field while the component of thedipole’s magnetic moment in the direction of the field is ±μ.i) What is E(up) = ? ii) What is E(down) = ? iii) What is e-βE(up) = ?iv) What is e-βE(down) = ? v) What is Z = ? Express in hyperbolic functions Ans. 2 cosh(βμB)vi) Probability of finding dipole in “up” state = ?vii) Probability of finding dipole in “down” state = ?viii) What is the sum of these two probabilities?ix) Find the average energy of the dipole: Sum the products of i) and vi) and ii) and vii)x) Simplify ix) to find – μBtanh(βμB)xi) For a collection of N such dipoles what is the total energy?xii) Differentiate Z in v)xiii) Multiply xii) by (-1/Z)xiv) Does xiii) equal x)?xv) Find the average value of a dipole’s magnetic moment along the direction of Bxvi) = ∑μz(s)Ρ(s) = (+μ)(Ρ↑) + (-μ)(Ρ↓) = ? Ans. μtanh(βμB)xvii) What is the total magnetization for the sample of N dipoles?Q. 6. Consider a system of N identical harmonic oscillator at temperature T, the allowed energies of each oscillator being 0, hf, 2hs,3hf, and so on.(a) Prove by long division that 1/(1 -x) = 1 + x + x2 + x3 + ……………….For what value of x does this series have a finite sum?(b) Evaluate the partition function for a single oscillator. Use the result in (a) to simplify as much as possible© Use = (-1/Z)(∂Z/∂β) to find the average energy of a single oscillator at temperature T(d) What is the total energy of N oscillator at temperature T?(e) Use heat capacity C = ∂U/∂T expression to find the heat capacity.(f) Use Mathematica to plot the function C/Nk = e1/t//{t2(e1/t – 1)2Hint. (a) Write down 1 = 1 + 0.x + 0. x2 + 0.×3 + ………………. And carry out the division.(b) Take the ground state to have an energy of 0, then sum the states with energies ε (=hf), 2ε (=2hf), 3ε (=3hf), 4ε (=4hf), 5ε(=5hf) and so on to find Z = ? Ans. 1/(1-e-βε) by the result of (a)© ans. = ε e-βε /(1-e-βε)(d) U = N Ans. U = Nε/(eβε- 1)(e) C ={Nε2/kT2}{(eε/kT)/(eε/kT – 1)2When kT>>ε, what is eε/kT= ?, keep the first term in the numerator and up to the second term in the denominator. Simplify further.What do you get? Ans. Nk(f) Plot it.Carnot Engine establishes an upper limit on the efficiencies of all other engines. It is of great importance both from practical andtheoretical viewpoints.Carnot cycle. An ideal reversible cycle operating between two energy reseroirs.Carnot theorem. No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operatingbetween the same two reservoirs.Carnot cycle description. Let the Carnot cycle takes place between two temperatures Tc and Th. Let the working substance be an idealgas contained in a cylinder fitted with a movable piston at the end. The cylinder’s wall and piston are thermally nonconducting. Thecycle consists of two adiabatic processes and two isothermal processes, all reversible:1. Process A→B is an isothermal expansion. The gas placed in thermal contact with an energy reservoir at temperature Th. Duringthe expansion, the gas absorbs energy │Qh│from the reservoir through the base of the cylinder and does work WAB in raising the piston2. In process B →C, the base of the cylinder is replaced by a thermally nonconducting wall, and the gas expands adiabatically. –that is no energy enters or leaves the system by heat. During the expansion, the temperature of the gas decreases from Th to Tc andthe gas does work WBC in raising the piston.3. In process C→D the gas is placed in thermal contact with an energy reservoir at temperature Tc and is compressed isothermallyat temperature Tc. During this time, the gas expels energy │Qc│to the reservoir, and the work done by the piston on the gas is WCD4. In process D→A, the base of the cylinder is replaced by a nonconducting wall, and the gas is conmpressed adiabatically. Thetemperature of the gas increases to Th, and the work done by the piston on the gasis WDAThe net work done is given by the area enclosed by the path ABCDAABCDThe thermal efficiency is given by 1 – │Qc│/│Qh│ 1- Tc/ThQ.1. The highest theoretical efficiency of a Carnot engine is 30%. If this engine uses the atmosohere, which has a temperature of 300K, as its cold reservoir, what is the temperature of its hot reservoir?i) Write down the formula you will use.ii) What is 30% in decimal form?iii) Substitute the values and solve for Th Ans. 429 KQ. 2. A steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water to steam, and this steam thendrives a piston. The cold reservoir’s temperature is that of outside air 300 . What is the maximum thermal efficiency of this engine?i) What is Tc=?ii) What is Th= ?iii) What is efficiency = ? Ans. 40%Quantum StatisticsThe ratio of probabilities of two different microstates with energy ε1 and particles N1 and energy ε2 and particle N2f(ε1, N1)/ f(ε2, N2) = e-[ε1 –μ1)]/kT / e-[ε2—μN2]/kT (1)= > Ratio of probabilities or relative probabilities with energy ε and number NF(ε, N) = (1/Ƶ) e-[ε –μN]/kT (3)Where Ƶ is called the grand partition function. Ƶ = ∑∑ e-[ε –μN]/kT (4)N SQ. 1. Assume a single hydrogen atom can be found in 3 possible states. State is ionized hydrogen and has 0 energy. State 2 containsone electron and has an energy ε1 = -13.56 eV. State 3 contains two electrons and has an energy of ε2 = -14.31 eV. The atom mayexchange electrons with surrounding atoms, bu the hydrogen as a whole is neutral. Determine the grand partition function for the system, treating μ as given. What value must μ have so that the gas is neutral (N) = 1)?Hint. i) Write down the expression for the grand partition functionii) Ƶ =? If ε = 0 and number of electron N =0iii) Ƶ = ? if ε = ε1 and N = 1iv) Ƶ = ? if ε= ε2 and N = 2v) What is the sum of ii), iii), and iv)vi) What is 0 times ii) = ?vii) What is 1 times iii) = ?viii) What is 2 times iv) = ?ix) Sum vi), vii), and viii)x) Divide ix) by v).xi) Set x) equal to 1xii) What is exp-(ε2 – 2μ)/kT = ?xiii) What is μ= ?xiv) Is xiii) ε2/2 ? Ans. -7.16 eVStatistical PhysicsIt considers how the overall behavior of a system of many particles is related to the properties of the particles themselves. Itdetermines the most probable way in which a certain total amount of energy E is distributed among the N particles of a system ofparticles in thermal equilibrium at the absolute temperature T. That is how many of the N particles are likely to have energy ε1, howmany ε2, and so on.The greater the number W of different ways in which the particles can be arranged among the available states to yield a particulardistribution of energies, the more probable is the distribution. The most probable distribution , which corresponds to the system’sbeing in thermal equilibrium, is the one for which W is maximum. The condition is that the system consists of a fixed number N ofparticles whose total energy is some fixed amount E.n(ε) = number of particles with energy = g(ε) f(ε) (1)where g(ε) = number of states with energy= statistical weight corresponding to energyf(ε) = distribution function= average number of particles in each state of energy ε= probability of occupancy of each state of energy εFor a continuous distribution g(ε) is replaced by g(ε) dε, the number of states with energies between ε+ dε.We consider systems of three different kinds of particles.Maxwell –Boltzmann distribution function for identical particles that are sufficiently far apart to be distinguishable , for instance,molecules of a gas. Quantum mechanically speaking, wave function of particles overlap to a negligible extent.Bose-Einstein distribution function is applicable to particles of 0 or integral spin that cannot be distinguished from one another.Quantum mechanically speaking, their wave functions overlap. Photons fall in this category. Particles described by this distributionare called bosons.Fermi-Dirac distribution function is applicable to particles with half-integral spin (1/2, 3/2, 5/2, …) that cannot be distinguishedfrom one another. Particles following this distribution are called fermions. Electrons fall in this category.MAXWELL-BOLTZMANN STATISTICSThe average number of particles fMB(ε) in a state of energy ε in a system of particles at the absolute temperature T is given byfMB( ε)= A e- ε/kT (2)where A depends on the number of particles in the system and is analogous to the the normalization constant of a wave function, and kis Boltzmann constant = 1.381×10-23 J/K = 8.617 x10-5 eV/K.Substituting the value of fMB( ε) in Eqn. (1), we getn( ε) = A g( ε) e- ε/kT (3)which gives the number n( ε) of identical distinguishable particles in an assembly at the temperature T that have energy ε.Law of AtmospheresLet us consider an ideal atmosphere in which gas molecules are in thermal equilibrium, and that there is no turbulence present and nowinds are pushing the molecules around. In such an atmosphere, density of molecules becomes a simple function of height, z.Let m = mass of each moleculeN/V = number per volumeThe force/area caused by the weight of the molecules areΔP = mgNΔz/VIn the limit Δz → 0,dP/dz = -mgN/V = -mgn (4)Using the ideal gas equation PV = NkTn = P/kT (5)Differentiating wrt t, dn/dP = 1/kTWhence dn/dz = (dn/dP)(dP/dz) = -mgn/kT (6)The solution of this eqn isN = n0e-mgz/kT(7)Q. 1. Estimate the thickness of the atmosphereAns. z = kT/mg, m = 28 x 1.66E-27kg, z = ?Q. 2. A cubic meter of atomic hydrogen at 0 degree Celsius and at atmospheric pressutre contains about 2.7E25 atoms. Find the numberof these atoms in the first excited states (n=20 at zero deg Celsius and at 10,000 deg Celsius.Soln. n(ε1) = A g(ε1) e-ε1/kt and n(ε2) = A g(ε2) e-ε2/ktThe ratio of the number of atoms in the n = 1 and n=2 states isn (ε2)/n(ε1) = g(ε2)/g(ε1) e- (ε2 – ε1)/ktSince number of possible states that corresponds to the quantum number n is 2n2. The number of states of energy ε1 corresponding to n =1 is 2. This is the case of 1s electron. It has l=0, ml= 0, ms = -1/ or +1/2.The number of states n = 2 is 8 i. e. the number of states of energy ε2 is g(ε2) = 8; a 2s electron can havel=0, ml= 0, ms = -1/ or +1/2, and 2p electron can have l=1, ml= 0, ±1, in each case of ms = -1/ or +1/2..Since the ground state energy ε1 = – 13.6 eV, and ε2 = ε1/n2 = -3.4 eV, ε1 – ε2 = 10.2eV, T = 0 deg Celsius = 273 K (ε2 – ε1)/kt = 10.2 eV/(8.617E-5 eV/kt)(273 K) = 434 n (ε2)/n(ε1) = (8/2)e-434 = 1.3E-188Thus about 1 atom in every 1xE188 is in the first excited states at 0 deg Celsius. With 2.7E25 atoms in our sample, we are confidentthat all are in the ground state.(b) T = 10,273 K. and (ε2 – ε1)/kT = 11.5 n (ε2)/n(ε1) = (8/2)e-11.5 = 4E-5Now the number of excited atoms is about 1E21, a substantial number even though only a small fraction.Q. 3. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in n=2 energy level?Soln. g(ε2) = 8, g(ε1) = 2, (ε2 – ε1)/kt = -3ε1/4kt, since ε2 = ε1/4;n (ε2)/n(ε1) =1/1000 = (8/2)e3ε1/4kT ; solving fot T: T = 1/k(-3/4ε1)/ln4000) = 1.43E4 KMolecular Energy Distribution in an Ideal GasWe will find the vdistribution of energies among the molecules of an ideal gas. In translational motion, energy is not quantized. Also,number of molecules is quite large. We consider continuous distribution of energies. Let n(ε) dε be the number of molecules whoseenergy lies between ε and ε+dεn(ε) d ε = [g(ε)dε] [f(ε)] = Ag(ε)e-ε/kTdε (8)where g(ε0dε = # of states that have energies between ε and ε+dεLet us find it.A molecule of energy ε has a momentum p = √(2mε) (since ε= p2/2m) = √(px2 + py2 + pz2)Where each set of momentum components px , py, pz specifies a different state of motion. Let us imagine a space with momentumcoordinates px , py, pz. The number of states g(p) dp with momenta whose coordinates lie between p and p + dp is proportional to thevolume of a spherical shell in momentum space p in radius and dp thick. The volume of a sphere of radius p is 4πr2. If we multiply itby the thickness dp, we get the volume. That is the volume is 4π p2dp.So, we can write downg(p) dp = B p2dp (9)where B is a constant and g(p) is not the same as g(ε)Since each momentum magnitude corresponds to a single energy ε, the number of energy states g(ε) dε between ε and ε+dε is the same asthe number of momentum states g(p)dp between p and p + dp, and so g(ε) dε = Bp2dp(10)From p2= 2mε, 2pdp =2 mdε => dp = mdε/p = mdε/(√(2mε)Now Eqn (10) can be written asNumber of energy states g(ε) dε = B . 2mε. mdε/(√(2mε) = 2m3/2B √ε dε CHECK THIS (11)The number of molecules with energies between ε and ε + dε isn(ε) dε = C√ε e-ε/kT dε (12)where C = = 2m3/2ABNormalizationThe constant C can be found in this way called normalization:∞ ∞ ∞N = ∫n(ε) dε = C ∫√εe-ε/kT dε = (C/2)(√π (kT)3/2 ) DO IT Use integral table ∫√x( e-ax)dx = (1/2a)√π/a0 0 0C = ?n( ε) dε = ?AVERAGE MOLECULAR ENERGY∞E = ∫εn(ε) dε Substitute for n(ε) dε and integrate to get 3/2 (NkT)0E/N = ? = average energy of an ideal gas moleculeQ. 4. Show that∞vavrg = 4π(m/2πkT)3/2 ∫v3 exp(-mv2/2kT) dv = √(8kT/πm)0vrms = √(3kT/m)Q.5. Find the speed of oxygen at zero degree CelsiusQ. 6. Diferentiate the following expression wrt v, set it equal to 0, and then solve for v which will be vp.n(v)= 4πN(m/2πkT)3/2 ∫v2 exp(-mv2/2kT)Q. 6. Draw a bell-shaped graph to locate the approximate positions of vrms, vavrg, and the most probable speed vp = √2kT/mFermi-Dirac DistributionThe probability of a state being occupied by nFD phalf-spin particles of energy ε and chemical potential μ is given bynFD = 1/[e(ε-μ)/kT + 1]Q. 1. What is nFD = ? if ε>> μQ. 2. What is nFD = ? if ε < > μQ. 5. What is nFD = ? if ε < 1/1000 =(8/2)e3ε1/4kT ; ε1 = -13.6 eV; k = 8.617E-5 eV/K; T =?6. Given that Z = ∑ e-βE(s)swhat is Z =? Is E(s) =0, 1, 2, and 3. ———————————————-7. In # 7 what is –(∂Z/∂β)/Z =8. If you toss two dice, what is the total number of ways in which you can obtain (a) a 12 and (b) a 7 ———-9. The Fermi-Dirac statistics is governed by nFD = 1/[e(ε-μ)/kT + 1]. For a system of Fermi-Dirac particles at room temperature,compute the probability of a single-particle state being occupied if its energy is ε – μ = 0. —————-10. The Bose-Einstein is given by nBE = 1/[e(ε-μ)/kT -1]. If ε – μ = 0.01 eV, n = ? ————–eC = 1 – Th/Tc; dS= dQ/T; e = {│Qh│-│Qc│}/│Qh│, Lf = 333000 J/kg; Q = mLf ; kT = 0.026 eVMidterm TestThermodynamicsPrint your nameThe main span of a bridge made of steel has a length of 1158 m. Assume the temperature swing between -50oC to 50o C. The linearexpansion coefficient for steel is 13E-6/oC.i) Wghat is ΔT = ?ii) What is ΔL = ?iii) ΔL = αLΔT = ?” 1.5 mIf P1 = 24.9 kPa, V1=0.100 m3, and V2 = 0.900 m3, P2 = ?2.77 kPaA weightlifter raises a mass of 180.0 kg through a height h = 1.25 m. Think him to be a thermodynamic system. How much heat energy doeshe have to give off if his internal energy decreases by 4000.0 J?i) How much mechanical work (mgh) does he do ?ii) Q = ΔE + W = ? -1790 J = -0.428 kcalSuppose this roomful air is at 22.0oC. What is the average kinetic energy of the molecules 9and atoms) in the air?KE avrg = 1.5kT , k = 1.38E-23J/K, T = ? K Ans. 6.11E-21 JWhat is the energy in electron volt, 1eV = 1.602E-19 JMidterm TestThermodynamicsPrint your nameThe main span of a bridge made of steel has a length of 1158 m. Assume the temperature swing between -50oC to 50o C. The linearexpansion coefficient for steel is 13E-6/oC.iv) Wghat is ΔT = ?v) What is ΔL = ?vi) ΔL = αLΔT = ?”vii)If P1 = 24.9 kPa, V1=0.100 m3, and V2 = 0.900 m3, P2 = ?A weightlifter raises a mass of 180.0 kg through a height h = 1.25 m. Think him to be a thermodynamic system. How much heat energy doeshe have to give off if his internal energy decreases by 4000.0 J?iii) How much mechanical work (mgh) does he do ?iv) Q = ΔE + W = ?v)Suppose this roomful air is at 22.0oC. What is the average kinetic energy of the molecules 9and atoms) in the air?KE avrg = 1.5kT , k = 1.38E-23J/K,What is the energy in electron volt? 1eV = 1.602E-19 JThermodynamics: Lecture 1Diathermal wall.Very thin wall (could be metallic such as copper) that prevents particles passing through but allows some energy topassThermal contact. In thermal contact, two objects are separated by a wall that is a good conductor of heat.Thermal equilibrium. Two objects are in thermal equilibrium if they do not change temperature when separated by a diathermal wall.Temperature: A measure of hotness or coldness of an object.Temperature is that property of an object that determines whether it is in thermal equilibrium with other objectsTwo objects in thermal equilibrium with a third object will be in thermal equilibrium with each other.Thermometry.Empirical temperature scale.A reproducible experimental definition of a way to measure temperature.Thermometric properties.Property of a substance that changes with temperature. All other properties must be held as constants.System. A small part of the universeSurroundings. Remaining portion of the universe that can directly interact with the system is called the surroundings or theenvironment.HJeat.Trasnsfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings.Internal energy. All the energy of a system that is associated with its microscopic components – atoms and molecules – when viewed froma reference frame at rest with respect to the center of mass of the system.State variables. The state of a system can be described by specifying pressure, volume, temperature, and internal energy. Themacroscopic state of an isolated system can be specified only if the system is in thermal equilibrium internally. In the case of a gasin a container, internal thermal equilibrium requires that every part of the gas be at the same pressure and temperature.Transfer variables. While state variables are characteristic of a system in thermal equilibrium, transfer variables are characteristicsof a process in which energy is transferred between a system and its surroundings or the environment. Transfer variables are zerounless a process occurs in which energy is transferred across the boundary of the system. Because a transfer of energy across theboundary represents a change in the system, transfer variables are not associated with a given state of the system, but with a changein the state of the system. Heat is a transfer variable. For a given set of conditions of a system, there is no definite value for theheat. We can only assign a value of the heat if energy crosses the boundary by heat, resulting in a change in the system.Work is another transfer variable in thermodynamics. Work = force x displacementImagine air in a syringe. If the cross-sectional area of the piston is A, and a push by a pressure P on the piston moves it through dx,the volume it moves through isdV = A.dx and the force on the piston is force = pressure x area of cross section => F = PAWork done then W = force, PA .distance, dx =>PAdx = PdVIn the syringe, if the gas is compressed, work done is negative. If the gas expands, work done is positiveQ. The piston of a syringe has a diameter of 5 cm. A pressure of 10 N/sq m is applied to the piton to squeeze the gas from 10 cm to 4cm. Find out the work done.Hints.Pressure, P = ?, area A = ?, force F = P.A = ? ; dx = distance moved through = ?; volume change, dV = ?; work done = PdV = ?