CHM 3410 – Problem Set 3 Due date: Monday, September 21st (Note: Our first exam is Friday, September 25th). Do all of the following problems. Show your work. "Entropy, in spite of one hundred years of history, is not very well understood yet and so as such you will win every time you use entropy in an argument." - John von Neumann 1) The Joule-Thompson coefficient for nitrogen (N2) at T = 300.0 K and p = 1.000 atm is JT = + 0.27 K/atm. a) In a particular experiment nitrogen gas undergoes a Joule-Thompson expansion from an initial pressure of 130. atm and an initial temperature T = 300.0 K to a final pressure of 1.0 atm. Estimate the temperature change for the gas. b) Is your estimated temperature change in part a likely to be larger, smaller, or about the same magnitude as the actual temperature change that will occur if the above experiment is carried out (Hint - Look at Fig 2.32 in Atkins.) 2) Consider one mole of an ideal monatomic gas (recall that for a monatomic gas C V,m = 3R/2 = 12.472 J/mol-K, Cp,m = 5R/2 = 20.786 J/mol.K). The gas is initially at a pressure p = 4.000 atm and a temperature T = 400.0 K. Find Ssyst, Ssurr, and Suniv for the following processes: a) An isothermal reversible compression of the gas to a final pressure p f = 10.000 atm. b) An reversible cooling of the gas from T i = 400.0 K to Tf = 300.0 K, at a constant pressure p = 4.00 atm. c) An adiabatic reversible compression of the gas to a final pressure p f = 10.000 atm. d) An adiabatic irreversible expansion of the gas against a constant pressure pex = pf = 1.000 atm. Note that these are the same processes you considered in problem 2 of problem set 2. You can use any information you previously obtained for these processes (or any information in the problem set 2 solutions) to do this problem. 3) A "supercooled" liquid is a liquid that has been cooled to below its normal freezing point yet which is still in the liquid state. Supercooled liquids will spontaneously freeze to form a solid. Consider the following irreversible process involving 1.000 mol of supercooled water. (pi = 1.00 atm, Ti = - 10.0 C) (pf = 1.00 atm, Tf = - 10.0 C) Find Ssyst, Ssurr, and Suniv for the above process. Note that Cp,m (H2O(s)) = 38.09 J/mol .K. HINT: Recall that S can only be calculated for reversible pathways, and so you will need to find such a pathway to do this problem. Y Also do the following from Atkins: Exercises 2.30 a Assume 1.00 mol of gas. Recall that, as derived in class, dU = T dV + CV dT (Eq 2.42). You need to know how the expansion is carried out to find q and w. Assume the expansion is isothermal and irreversible against an external pressure equal to the final pressure of the gas. Problems 2.24 This is actually a simple problem. 2.34 This one is not quite so simple. Start with the equation for internal energy, which is (U/V)T = T (p/T)V - p , and transform it into the "partner" equation. HINT: You need to somehow get a term involving H in your final result. Recall H = U + pV, which can be used to do this. 3.14 You only need to find the value for molar entropy at T = 298.15 K. Below the lowest temperature in the data (14.14 K) you may assume the Debye approximation is valid, that is, that C p,m = aT3, where the value for a is found from the data at T = 14.14 K. EXTRA CREDIT: Problem 2.40. This is actually an easy problem, but I like it because it is an interesting application of thermodynamics. Solutions 1) NOTE: I have changed the problem from the original so that it illustrates what I had actually intended, an example of a Joule-Thompson expansion a) We may estimate the temperature change using JT = (T/p)H (T/p)H If we solve for T we get T = JT p = ( 0.27 K/atm) [ (1 - 130) atm ] = - 34.8 K b) The value for JT used in part a is for p = 1.00 atm and T = 300. K. However, the expansion occurs over a range of pressures from 1 atm to 130 atm. An examination of Fig. 2.32 of Atkins shows that for nitrogen the value for JT becomes smaller as pressure increases since increasing pressure brings you closer to the boundary between cooling and heating, where JT = 0. Our value is therefore larger than it should be in the high pressure region, and so our estimated change in temperature is larger than the true value that will occur for this process. 2) As a general comment, note that we will make extensive use of the results previously found for these processes in problem set 2, and will not rederive them here. a) The process is reversible and isothermal, so Ssyst = if (dq)rev/T = (1/T) if (dq)rev = q/T = ( - 3047. J)/(400.0 K) = - 7.62 J/K. Ssurr = if (dq)rev,surr/T = (1/T) if (dq)rev,surr = - qsyst/T = ( + 3047. J)/(400.0 K) = + 7.62 J/K. Suniv = Ssyst + Ssurr = (- 7.62 J/K) + (+ 7.62 J/K) = 0. b) The process is reversible and carried out at constant pressure, so Ssyst = if (dq)rev/T But for the constant pressure cooling of a substance dq = n Cp,m dT, and so (recalling that Cp,m is constant) Ssyst = if [ n Cp,m/T ] dT = n Cp,m if dT/T = n Cp,m ln(Tf/Ti) = (1.00 mol) (20.786 J/mol.K) ln(300.0/400.0) = - 5.98 J/K Since the process is reversible Suniv = 0 = Ssyst + Ssurr Ssurr = - Ssyst = + 5.98 J/K If you want a good mental picture of how the above process could be carried out reversibly consider the following method. The system is initially in thermal contact with a temperature bath at T = 400.00 K. It is placed in contact with a temperature bath at T = 399.99 K and allowed to come to equilibrium. It is then placed in contact with a temperature bath at T = 399.98 K and allowed to come to equilibrium. This continues until the final temperature, T = 300.00 K, is reached. In the limit of an infinite number of the above steps, each with an infintesimally smaller change in the temperature for the bath, the process will be reversible. c) The process is adiabatic, and so qsurr = 0, which means Ssurr = 0. The process is reversible, so Ssyst = if (dq)rev/T = 0. Therefore Suniv = 0. Life is good. d) The process is adiabatic, and so qsurr = 0, which means Ssurr = 0. Unfortunately, the process is irreversible, and so we cannot use the actual pathway for the process to calculate Ssyst. Instead, we need to find a reversible pathway that connects the initial and final states. The inital and final states for the process are pi = 4.00 atm Ti = 400.0 K pf = 1.00 atm Tf = 280.0 K One possible reversible pathway is the following Step 1 - An isothermal reversible expansion of the gas from pi = 4.00 atm to pf = 1.00 atm, at T = 400.0 K. Step 2 - A constant pressure cooling of the gas from T i = 400.0 K to Tf = 280.0 K, at p = 1.00 atm We have already found expressions for Ssyst for step 1 (problem 2a) and step 2 (problem 2b) and so will simply use the results. Ssyst,1 = q/T = - (nRT ln(pf/pi) = - nR ln(pf/pi) = - (1.00 mol) (8.314 J/mol.K) ln (4.00/1.00) = + 11.53 J/K T Ssyst,2 = n Cp,m ln(Tf/Ti) = (1.00 mol) (20.786 J/mol.K) ln(280./400.) = - 7.41 J/K So Ssyst = Ssyst,1 + Ssyst,2 = + 11.53 J/K + (- 7.41 J/K) = + 4.12 J/K And Suniv = Ssyst + Ssurr = + 4.12 J/K + 0. = 4.12 J/K. 3) To find Ssyst requires that we find a reversible pathway connecting the initial and final states. One such pathway is as follows. Step 1 - Reversible heating of liquid water from T = - 10.0 C to T = 0.0 C. Step 2 - Phase transition from liquid to solid water, which is reversible at T = 0.0 C. Step 3 - Reversible cooling of solid water from T = 0.0 C to T = - 10.0 C. We have previously (problem 2b) found expressions for the entropy change for step 1 and 3 (note that these assume Cp,m is constant). As discussed in class, the entropy change for a phase transition at the normal transition point is Spt = Hpt/Tpt. So S1 = n Cp,m(H2O()) ln(Tf/Ti) = (1.00 mol) (75.291 J/mol.K) ln(273.15 K/263.15 K) = + 2.808 J/K S2 = - n Hfus/Tfus = - (1.00 mol) (6008. J/mol) / (273.15 K) = - 22.00 J/K S3 = n Cp,m(H2O(s)) ln(Tf/Ti) = (1.00 mol) (38.09 J/mol.K) ln(263.15 K/273.15 K) = - 1.421 J/K and so Ssyst = S1 + S2 + S3 = 2.808 J/K + (- 22.00 J/K) + (- 1.421 J/K) = - 20.61 J/K To find the entropy change for the surroundings we need to know what Hfus is at T = 263.15 K. (Remember, however, that our process is freezing, which is the opposite of fusion). Hfus(263.15 K) = Hfus(273.15 K) + 273.15263.15 Cp,m dT = Cp,m T where Cp,m = Cp,m(H2O()) - Cp,m(H2O(s)) = (75.291 J/mol.K) - (38.09 J/mol.K) = 37.20 J/mol.K So Hfus(263.15 K) = 6008. J/mol + (37.20 J/mol .K) ( - 10.0 K) = 5636. J/mol And so Ssurr = - qsyst/T = + Hfus(263.15 K) / 263.15 K = (5636. J/mol) / (263.15 K) = 21.42 J/mol .K Finally, Suniv = Ssyst + Ssurr = - 20.61 J/K + 21.42 J/K = + 0.81 J/K Note that Suniv > 0, as expected, since the freezing of a supercooled liquid is an irreversible process. Exercise 2.30a Recall that dU = T dV + CV dT. For an isothermal irreversible expansion dT = 0, and so dU = T dV Integrating both sides gives U = if T dV But T = a/Vm2 = an2/V2, so U = if (an2/V2) dV = - an2 [ (1/Vf) - (1/Vi) ] For N2 , a = 1.352 L2.atm/mol2 So U = - (1.352 L2.atm/mol2) (1.00 mol) 2 [ (1/24.8 L) - (1/1.0 L) ] = 1.297 L.atm 101.3 J = 131. J L.atm To find the value for work we must first find the final pressure. p = nRT - an2 = (1.00 mol) (0.082057 L.atm/mol.K) (298. K) - (1.352 L2.atm/mol2) (1.00 mol)2 (V - nb) V2 [ (24.8 L) - (1.00 mol) (0.0387 L/mol) ] (24.8 L) 2 = 0.9876 atm - 0.0022 atm = 0.9854 atm and so w = - if pex dV = - pf if dV = - pf (Vf - Vi) = - (0.9854 atm) [(24.8 - 1.0) L] = - 23.45 L.atm 101.3 J = - 2376. J L.atm Finally, since U = q + w, q = U - w = 131. J - (- 2376. J) = 2507. J Problem 2.24 We begin with the relationship Cp - CV = T (p/T)V (V/T)p But (p/T)V (T/V)p (V/p)T = - 1, and so (p/T)V = - (V/T)p (p/V)T = - (V/T)p (V/p)T Substituting this result gives Cp - CV = - T (V/T)p (V/T)p = - T (V/T)p2 (V/p)T (V/p)T (Note: The book has this relationship wrong). For an ideal gas, V = nRT/p, and so (V/T)p = nR/p (V/p)T = - nRT/p2 and so Cp - CV = - T (nR/p)2 = nR , as expected. (- nRT/p2) Problem 2.34 (H/p)T = (H/V)T (V/p)T But H = U + pV, and so (H/V)T = (U/V)T + p (V/V)T + V (p/V)T = (U/V)T + p + V (p/V)T So (H/p)T = [ (U/V)T + p + V (p/V)T ] (V/p)T But (U/V)T = T (p/T)V - p (the thermodynamic relationship given in the problem). Substituting gives (H/p)T = [ T (p/T)V - p + p + V (p/V)T ] (V/p)T = [ T (p/T)V + V (p/V)T ] (V/p)T = T (p/T)V (V/p)T + V (p/V)T (V/p)T = T (p/T)V (V/p)T + V But (p/T)V (T/V)p (V/p)T = - 1 , and so (p/T)V = - (V/T)p (p/V)T If we substitute this result, we get (H/p)T = - T (V/T)p (p/V)T (V/p)T + V = - T (V/T)p + V the final result. Problem 3.14 S(298.15 K) = 0298.15 (Cp,m/T) dT = 014.14 (Cp,m/T) dT + 14.14298.15 (Cp,m/T) dT For temperatures below 14.14 K we assume that Cp,m = a T3. The value for a is found using the heat capacity at 14.14 K (9.492 J/mol.K) = a (14.14 K)3 a = (9.492 J/mol.K) / (14.14 K)3 = 3.36 x 10-3 J/mol.K4 So 014.14 (Cp,m/T) dT = 014.14 (a T3/T) dT = 014.14 (a T2) dT = (a T3)/3 = (3.36 x 10-3 J/mol.K4) (14.14 K)3 / 3 = 3.166 J/mol.K For the second integral, 14.14298.15 (Cp,m/T) dT, we will find the value for the integral numerically using the trapezoid rule. For adjacent temperatures we say T1T2 (T2 - T1) { [ Cp,m(T2)/T2 ] + [ Cp,m(T1)/T1 ] }/2 T (K) 14.14 16.33 20.03 31.15 44.08 64.81 100.90 140.86 183.59 225.10 262.99 298.06 298.15 Cp,m (J/mol.K) 9.492 12.70 18.18 32.54 46.86 66.36 95.05 121.3 144.4 163.7 180.2 196.4 Cp,m/T (J/mol.K2) 0.6713 0.7777 0.9076 1.0446 1.0631 1.0239 0.9420 0.8611 0.7865 0.7272 0.6852 0.6589 0.658 T (K) 2.19 3.70 11.12 12.93 20.73 36.09 39.96 42.73 41.51 37.89 35.07 0.09 S (J/mol.K) 1.587 3.118 10.854 13.626 21.632 35.475 36.026 35.201 31.417 26.758 23.569 0.0593 The data are plotted on the next page. Note that the area under the curve up to any temperature T represents the value for S at that temperature (though the value calculated here for T = 298.15 K was found as described above). 14.14298.15 (Cp,m/T) dT S = 239.322 J/mol.K And so S(298.15 K) = 014.14 (Cp,m/T) dT + 14.14298.15 (Cp,m/T) dT = (3.166 + 239.322) J/mol.K = 242.49 J/mol.K (Cp,m)/T (J/mol.K2) Plot of (Cp,m)/T vs T 1.2 1 0.8 0.6 0.4 0.2 0 0 50 100 150 200 250 300 350 T (K) Problem 2.40 The heat capacity of liquid water is Cp,m(H2O()) = 75.291 J/mol.K The heat capacity of a 65. kg person is Cp = 65000. g 1 mol 18.02 g 0.075291 kJ mol.K = 271.6 kJ/K Since Cp = (dq/dT)p (q/T), it follows that T = q = 10000. kJ = 36.8 K Cp 271.6 kJ/K To find the mass of water that needs to be evaporated each day to maintain normal body temperature we need the value for Hvap at T = 37. C. Since we are making an estimate, and since the enthalpy of vaporization does not change much with temperature, we will use Hvap = 44.02 kJ/mol, the value at T = 25.0 C (Table 2.3). So m = 10000. kJ 1 mol 0.01802 kg = 4.09 kg water (so about 4 liters of sweat per day) 44.02 kJ mol Note that the power output of an average person is power = energy = 10000 x 103 J = 116. J/s = 116. watt time 86400 s or about the same power as used in a 100 watt light bulb.