CHM 3410 – Problem Set 4
Due date: Monday, September 21st (The first hour exam is Friday, September 25th. It will cover the following material - Chapter 1, all; Chapter 2, all; Chapter 3, sections 3A, 3B, 3C; handouts).
Do all of the following problems. Show your work.
"Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you any more."
- Arnold Sommerfeld
1) The Joule-Thompson coefficient for nitrogen gas (N2) is m = 0.27 K/atm at T = 300. K and p = 1.00 atm. Estimate the temperature change when 1.000 mol of N2 is expanded by a constant enthalpy process from an initial pressure and temperature pi = 100.0 atm, Ti = 300.K to a final pressure pf = 1.00 atm. Note that in reality the value for m depends on both temperature and pressure, so your answer here is only approximate.
2) Consider one mole of an ideal diatomic gas (recall that for a diatomic gas CV,m = (5/2) R = 20.786 J/mol K, Cp,m = (7/2) R = 29.101 J/mol K). The gas is initially at a pressure p = 4.000 atm and a temperature T = 300.0 K. Find q, w, DU, DH, DSsyst, DSsurr, and DSuniv for the following processes:
a) An isothermal reversible expansion of the gas to a final pressure pf = 1.000 atm.
b) An reversible heating of the gas from Ti = 300.0 K to Tf = 400.0 K, at a constant pressure p = 4.00 atm.
c) An adiabatic irreversible expansion of the gas against a constant pressure pex = pf = 2.000 atm.
3) Consider the following process. An insulated container holds 60.00 g of liquid water at T = 80.0 °C. 40.00 g of liquid water at T = 20.0 °C is added to the container. When the system reaches equilibrium there will be 100.00 g of liquid water in the container at a final temperature Tf. Find Tf and DS for the above process. Cp,m(H2O(l)) = 75.29 J/mol K, and may be assumed constant over the range of temperatures in the problem.
4) Consider the four step cyclic process illustrated below: A ® B ® C ® D ® A. pA = pB = 2.00 atm; pC = pD = 1.00 atm; VA = VD = 20.0 L; VB = VC = 40.0 L. All steps in the process are carried out reversibly. Find DUcycle, DHcycle, DScycle, qcycle, and wcycle, or explain why they cannot be found from the information given.
5) The following heat capacity data are for solid silver (Ag(s)).
T(K) Cp,m (J/mol K) T(K) Cp,m (J/mol K) T(K) Cp,m (J/mol K)
15.0 0.67 130.0 22.13 230.0 24.73
30.0 4.77 150.0 22.97 250.0 25.03
50.0 11.65 170.0 23.61 270.0 25.31
70.0 16.33 190.0 24.09 290.0 25.44
90.0 19.13 210.0 24.42 300.0 25.50
110.0 20.96
For temperatures below 15. K, the heat capacity can be assumed to fit the equation Cp,m = aT3 (using the Debye extrapolation), where the value for a is found using the value for Cp,m at 15. K.
Using the above data and the third law of thermodynamics, find the value for S°(Ag(s)), the absolute entropy of silver metal, at T = 298. K. Compare your result to the value found in the thermochemical tables in the appendix of Atkins.
Solutions.
1) For N2(g) at p = 1.00 atm and T = 300. K, m = 0.27 K/atm. If we assume that m is approximately constant for the pressure and temperature range in the problem, then
m = (¶T/¶p)H @ [(DT)/(Dp)]H
Since the process is constant enthalpy
DT = m (Dp) = (0.27 K/atm) (1.00 atm - 100.0 atm) = - 26.7 K
So a single constant enthalpy pass through a system like a Linde refrigerator would cool nitrogen down by slightly more than 25. K. If the cooled gas were recirculated, brought back to the initial pressure, and expanded several times one could liquify nitrogen by this process.
2) As a general comment, note that we will make extensive use of the results previously found for these types of processes.
a) The process is reversible and isothermal, and the gas is ideal, and so
DU = 0. DH = 0.
q = - w
w = - nRT ln(Vf/Vi) = nRT ln(pf/pi) = (1.000 mol)(8.3145 L atm/mol K)(300.0 K) ln(1.00/4.00)
= - 3458. J
So q = + 3458. J
DSsyst = òif (đq)rev/T = (1/T) òif (đq)rev = q/T = ( 3458. J)/(300.0 K) = 11.53 J/K.
The process is reversible, and so DSuniv = DSsyst + DSsurr = 0.
DSsurr = - DSsyst = - 11.53 J/K
b) The process is reversible and carried out at constant pressure, the gas is ideal, and the gas has a constant value for heat capacity over the range of temperatures of the problem, so
DU = òTiTf n CV,m dT = n CV,m òTiTf n CV,m dT = n CV,m (Tf - Ti)
= (1.000 mol)(20.786 J/mol K)(400.0 K - 300.0 K) = 2079. J
DH = òTiTf n Cp,m dT = n Cp,m òTiTf n Cp,m dT = n Cp,m (Tf - Ti)
= (1.000 mol)(29.101 J/mol K)(400.0 K - 300.0 K) = 2910. J
Since the process is constant pressure, q = DH = 2910. J
From the first law, DU = q + w
so w = DU - q = (2079. J) - (2910. J) = - 831. J
Now DSsyst = òif (đq)rev/T
For a constant pressure reversible change in temperature, đq = n Cp,m dT, and in this problem Cp,m is constant, so
DSsyst = òif (nCp,m/T) dT = n Cp,m òif (1/T) dT = nCp,m ln(Tf/Ti)
= (1.000 mol)(29.101 J/mol K)ln(400.0/300.0) = 8.372 J/mol
The process is reversible, and so DSuniv = DSsyst + DSsurr = 0.
DSsurr = - DSsyst = - 8.372 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 = 300.00 K. It is placed in contact with a temperature bath at T = 300.01 K and allowed to come to equilibrium. It is then placed in contact with a temperature bath at T = 300.02 K and allowed to come to equilibrium. This continues until the final temperature, T = 400.00 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. In practice, if the heating is not carried out rapidly it is, to a very good approximation, reversible.
c) The process is adiabatic, and so q = 0, which also means DSsurr = 0.
Unfortunately, the process is irreversible, and so we cannot use the actual pathway for the process to calculate DSsyst. 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 pf = 2.00 atm
Ti = 300.0 K Tf = (?)
So the first thing we need to do is find the value for Tf. This will allow us to find DU, DH, and DSsyst.
Since q = 0, it follows that DU = w
But DU = òTiTf n CV,m dT = n CV,m òTiTf n CV,m dT = n CV,m (Tf - Ti)
w = - ò pex dV = - pex ò dV = - pf (Vf - Vi)
But V = nRT/p
Vi = (1.000 mol)(0.082057 L atm/mol K)(300.0 K) = 6.154 L
(4.000 atm)
So DU = w
nCV,m(Tf - Ti) = - pf [ (nRTf/pf) - Vi ]
nCV,mTf - nCV,mTi = Vipf - nRTf
nCV,mTf + nRTf = Vipf + nCV,mTi
But nCV,m + nR = nCp,m, and so
nCp,mTf = Vipf + nCV,mTi
or, finally Tf = Vipf + nCV,mTi
nCp,m
= [ (6.154 L)(2.000 atm)(101.3 J/L atm) + (1.000 mol)(20.786 J/mol K)(300.0 K) ]
(1.000 mol)(29.101 J/mol K)
= 257.1 K
We can now find DU, DH, and w.
DU = òTiTf n CV,m dT = n CV,m òTiTf n CV,m dT = n CV,m (Tf - Ti)
= (1.000 mol)(20.786 J/mol K)(257.1 K - 300.0 K) = - 892. J
DH = òTiTf n Cp,m dT = n Cp,m òTiTf n Cp,m dT = n Cp,m (Tf - Ti)
= (1.000 mol)(29.101 J/mol K)(257.1 K - 300.0 K) = - 1248. J
q = 0, and so w = DU = - 892. J
We now need a reversible pathway with the same initial and final state as the actual process to calculate DSsyst. One possible reversible pathway is the following (Note there are other pathways we could use. If they are reversible and have the same initial and final states they will give the same final results.)
Step 1 - An isothermal reversible expansion of the gas from pi = 4.00 atm to pf = 2.00 atm, at T = 300.0 K.
Step 2 - A constant pressure cooling of the gas from Ti = 300.0 K to Tf = 257.1 K, at p = 2.00 atm
We have already found expressions for DSsyst for step 1 (problem 2a) and step 2 (problem 2b) and so will simply use the results.
DSsyst,1 = q/T = - (nRT ln(pf/pi) = - nR ln(pf/pi) = - (1.00 mol) (8.314 J/mol.K) ln (2.00/4.00) = + 5.763 J/K
T
DSsyst,2 = n Cp,m ln(Tf/Ti) = (1.00 mol) (29.101 J/mol.K) ln(257.1/300.) = - 4.491 J/K
So DSsyst = DSsyst,1 + DSsyst,2 = + 5.763 J/K + (- 4.491 J/K) =1.272 J/K
And DSuniv = DSsyst + DSsurr = + 1.272 J/K + 0. = 1.272 J/K.
3) We need to first find Tf, the final temperature of the water. We may find Tf by using the condition that the process is adiabatic. We will use the subscripts “cw” for the cold water and “hw” for the hot water.
q = 0 = qcw(heating the cold water from 20.0 °C to Tf)
+ qhw(cooling the hot water from 80.0 °C to Tf)
We may find q for each of the above steps. Note that we assume Cp,m = constant.
qcw(heating the water from20.0 °C to Tf) = ncw Cp,m (Tf – 20.0 °C)
q(cooling the hot water from 80.0 °C to Tf) = nhw Cp,m (Tf – 80.0 °C)
(Note that since the second and third terms involve differences in temperature it does not matter whether we use °C or K as our temperature unit.)
Substituting for q we get
q = 0 = ncw Cp,m (Tf – 20.0 °C) + nhw Cp,m (Tf – 80.0 °C)
0 = ncw Cp,m Tf - ncw Cp,m (20.0 °C) + nhw Cp,m Tf – nhw Cp,m (80.0 °C)
There is a factor of Cp,m appearing in each term that we can cancel out. If we then rearrange the above equation, we get
(ncw + nhw)Tf = [ ncw(20.0 °C) + nhw(80.0 °C) ]
If we define ntotal = ncw + nhw, then
Tf = [ ncw(20.0 °C) + nhw(80.0 °C) ] = (ncw/ntotal)(20.0 °C) (nhw/ntotal)(80.0 °C)
ntotal
But (ncw/ntotal) = (mcw/mtotal) = 40.0 g/100.0 g) = 0.400
(nhw/ntotal) = (mhw/mtotal) = 60.0 g/100.0 g) = 0.600
Tf = (0.400)(20.0 °C) + (0.600)(80.0 °C) = 56.0 °C
Now, M(H2O) = 18.01 g/mol
ncw = 40.00 g/(18.01 g/mol) = 2.2210 mol nhw = 60.00 g/(18.01 g/mol) = 3.3315 mol
We can now find DS by finding the entropy change for each step in the above three process, using
DS = òif (dq)rev/T
heating the cold water
DScw = ncw Cp,m(H2O()) ln(Tf/Ti) = (2.2210 mol) (75.291 J/mol.K) ln (329.15 K/293.15 K) = 19.36 J/K
cooling the hot water
DShw = nhw Cp,m(H2O()) ln(Tf/Ti) = (3.3315 mol) (75.291 J/mol.K) ln (329.15 K/353.15 K) = - 17.65 J/K
Note that we converted temperature from °C to K in the above calculations.
And so
DSsyst =(19.36 J/K) + (- 17.65 J/K) = 1.71 J/K.
Since DSsurr = 0, then DSuniv = 1.71 J/K.
4) Since this is a cyclic process the change for any state function must be zero. Therefore
DU = 0. DH = 0. DS = 0.
From the first law, DU = q + w, and so q = - w.
w = - ò pex dV However, since all steps in the process are reversible.
w = - ò p dV
This integral simply represents the area under the curve in a p vs V diagram (other than a difference in sign).
Process B ® C and process D ® A are both carried out at constant volume, and so wBC = wDA = 0.
For process A ® B, wAB = - (2.00 atm) (40.0 L - 20.0 L) = - 40.0 L atm.
For process C ® D, wCD = - (1.00 atm) (20.0 L - 40.0 L) = 20.0 L atm.
So wcycle = wAB + wBC + wCD + wDA = (- 40.0 L atm) + 0 + (20.0 L atm) + 0.
= - 20.0 L atm (101.3 J) = - 2026. J
1. L atm
Note that we could have simply sad that the work was equal to the area contained within the cyclic pathway (other than for a difference in sign). This might be useful for more complicated cyclic paths.
Finally, qcycle = 2026. J
5) Below 15.0 K we assume Cp,m = aT3.
Therefore a = Cp,m/T = (0.67 J/mol K)/(15.0 K)3 = 1.985 x 10-4 J/mol K4
The contribution to the absolute entropy for temperatures from 0. K to 15.0 K is then
DS = ò015 (aT3)/T dT = ò015 (aT2) dT = (aT3/3) |015
= (0.67 J/mol K)/3 = 0.223 J/mol K
For temperatures between 15.0 K and 298.0 K we numerically integrate, using the area ofthe quadralateral formed from the two adjacent temperatures and their values for Cp,m/T. We have also included a value for data at 298. K, found from interpolation. A plot of the data is given below, and the calculations are on the next page.
T (K) Cp,m (J/mol K) Cp,m/T (J/mol K2) DS (J/mol K)
Contribution from 0.0 K to 15.0 K 0.223
15.0 0.67 0.04467 1.528
30.0 4.77 0.15900 3.922
50.0 11.66 0.23320 4.665
70.0 16.33 0.23329 4.458
90.0 19.13 0.21256 4.031
110.0 20.96 0.19055 3.608
130.0 22.13 0.17023 3.234
150.0 22.97 0.15313 2.920
170.0 23.61 0.13888 2.657
190.0 24.09 0.12679 2.431
210.0 24.42 0.11629 2.238
230.0 24.73 0.10752 2.076
250.0 25.03 0.10012 1.939
270.0 25.31 0.09374 1.815
290.0 25.44 0.08772 0.693
(298) (25.488) 0.08553
300.0 25.50
______
S°(Ag(s, 298. K)) = 42.44 J/mol K
This is close to the value given in the Appendix of Atkins, S°(Ag(s, 298. K)) =42.55 (0.2% difference).
(Yes, I am surprised...but that is really what I got for my calculation!)