Consider the following change of state for one mole of a monatomic ideal gas (CV
ID: 921501 • Letter: C
Question
Consider the following change of state for one mole of a monatomic ideal gas (CV= 3R/2, CP= 5R/2): initial pressure 1.00 atm, initial temperature 273.15 K, and final pressure 0.500 atm, final temperature 546.3 K. This change of state may be brought about in an infinite variety of ways. Consider for this problem the following two reversible paths, each consisting of two parts: (1) an isothermal expansion followed by an isobaric temperature rise, and (2) an adiabatic expansion to a volume of 89.6 L. followed by a constant volume temperature rise. a. Determine P, V, and T for one mole of the gas after the initial step of each of each path. Sketch the paths on a P-V diagram. b. Calculate W, Q, change in E, change in H, for each part of each path and for the total path and tabulate the results. c. Verify that E and H are state functions in this example, but that w and q are not.
Explanation / Answer
Initial state: (p1, T1)=(1.00 bar, 272.1 K)
Therefore, V1 = RT1/p1 = 22.624 L
The final state: (p2, T2) = (0.50 bar, 546.2 K). Therefore, V2 = RT2/p2 = 90.828 L.
These are the initial and final states for each of the four processes. The intermediate states for each are given below, labeled by subscripts 0.
(p0, T0) = (0.50 bar, 272.1 K); V0 = 45.248 L.
(V0, T0) = (90.828 L, 272.1 K); p0 = 0.249 bar
p0 = p2 = 0.50 bar. The volume is given by
V0=(p1V1 /p0) 1/ = [1.00 bar × (22.624 L) 5/3 /0.5 bari ]3/5 =34.292 L.
T0=p0V0/R = 206.22 K
V0 = V2 = 90.828 L.
The pressure is given by p0=p1 (V1/V0) =
1.00(22.624/90.828)5/3 =9.8609 × 102 bar.
T0=p0V0/R = 107.72 K
With these points determined, the plots can be made for each process. Please see the last page for the plots.
Isothermal expansion followed by isobaric heating:
wa = -RT1 ln(V1/V0)
= -R (272.1 K) ln(45.248/22.624)
wa = -1568.2 J mol-1
wb=p2(V2 V0) = 0.5 × 105 Pa × (90.828 45.248) × 103 m3
=2279.0 J mol1
. w=3847.2 J mol1.
qa=wa,
since Ua = 0. =1568.2 J mol1.
qb=Cp,m(546.2 272.1) =5697.5 J mol1.
q=7265.7 J mol1.
Ua=0
Ub=CV,m(546.2 272.1) = 3418.5 J mol1.
U=3418.5 J mol1.
Ha=0
Hb=qb = 5697.5 J mol1.
H=5697.5 J mol1.
Sa = Rln (V1/V0) = 5.7632 J K-1 mol-1
Sb=Cp,m ln (T2/ T1 ) = 14.484 J K1 mol1.
S=20.247 J K1 mol1
Isothermal expansion followed by isochoric heating:
wa = -RT1 ln(V2/V1)
= R(272.1 K) ln (90.828 /22.624)
= -3144.6 mol-1
wb=0
w=3144.6 J mol1
.qa=wa = 3144.6 J mol1,
since Ua = 0.
qb=Ub wb =CV,m(546.2 272.1) + 0
= 3418.5 J mol1. q=6563.1 J mol1.
Ua=0
Ub=CV,m(546.2 272.1) = 3418.5 J mol1.
U=3418.5 J mol1.
Ha=0
Hb=Cp,m(546.2 272.1)
= 5697.5 J mol1.
H=5697.5 J mol1.
Sa = Rln(V2/V1) = 11.557 J K-1 mol-1
Sb=CV,m ln (T2 /T1 ) = 8.6905 J K1 mol1.
S=20.248 J K1 mol1.
Adiabatic expansion followed by isobaric heating:
wa=Ua,
since qa = 0. =CV,m(T0 T1) = 1.5R(206.22 272.1) = 821.64 J mol1.
wb=p2(V2 V0) = 0.5 × 105 Pa × (90.828 34.292) × 103 m3 =2826.8 J mol1.
w=3648.4 J mol1.
qa=0.
qb=Cp,m(546.2 206.22) =7066.9 J mol1.
q=7066.9 J mol1.
Ua=wa = 821.64 J mol1.
Ub=CV,m(546.2 206.22) = 4240.1 J mol1.
U=3418.5 J mol1
Ha=Cp,m(206.22 272.1) = 1369.4 J mol1.
Hb=qb = 7066.9 J mol1.
H=5697.5 J mol1.
Sa=0. Sb=Cp,m ln (T2 /T0 ) = 20.247 J K1 mol1.
S=20.247 J K1 mol1.
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