Chem 5624 Quiz 2 (3/13/09)
Name_____KEY_____________
R = 8.314 J/(mol-deg) = 0.0821 (L-atm)/(mol-deg)
1.
(6) In this problem, you are asked to determine the entropy change
for the expansion of one mol of methane from 1.00 L to 2.00 L
using two different equations of state. [Hint: Use the Maxwell
Relations].
a. Calculate ΔSIGL when methane is treated as an ideal gas.
b. Calculate ΔSvdW when methane is treated as a van de Waals gas:
p = RT/(V - b) - a/V2
with a = 2.253 (L/mol)2atm and b = 0.04278 L/mol
c. Compare your answers in Parts a. and b. and propose a
justification for the result.
a. IGL says p = nRT/V. MR IVc says (∂p/∂T)V = (∂S/∂V)T
so (∂p/∂T)V = nR/V and dS = (nR/V) dV or
ΔSIGL = R ln (V2/V1) = R ln (2.00L/1.00L) = 5.76 J/(mol-K)
b. vdW  (∂p/∂T)V = nR/(V – b) and dS = (nR/(V -b) dV or
ΔSvdW = R ln [(V2-b)/(V1-b)] = R ln (1.957/0.957L) = 5.94 J/(mol-K)
c. ΔSvdW > ΔSIGL because the ratio of final to initial volumes for the
vdW gas > the ratio for the IGL due to non-zero vdW gas molecule
volumes.
(3) Start with dU = TdS – pdV + fdℓ
a. Find the function Z = f(S, p, ℓ) Transform from variable V to
variable p.
dU + d(pV) = TdS + Vdp + fdℓ or Z = U + pV
b. What is Z? Z = H
c. What is (∂Z/∂ℓ)T,p? This partial is f
2.
(6) Consider the freezing (liq  sol) of 1.00 mol of water at 1
atm. Caution: make sure the signs are correct.
a. Calculate the entropy of freezing (ΔSfre,273 ) at 273K given
ΔHfus = + 6.025 kJ/mol at 273K.
ΔSfre,273 = - ΔHfre,273 /T = -22.07 J/mol Note entropy decreases when a
liquid freezes.
3.
b. Using these data, calculate the entropy of freezing (ΔSfre,268 ) at
268K.
Cp,liq = 75.3 J/(mol-K) and Cp,sol= 2.09 J/(mol-K) [Hint: a cycle might help]
273K
263K
liq
↑
(1)
↑
liq
→(2)→
→(4)→
sol
↓
(3)
↓
sol
(1) = Cp,liq ln (273/268) = 1.392 J/(mol-K)
(2) = -22.07 J/(mol-K)
(3) = Cp,sol ln(268/273) = -0.0386 J/(mol-K)
(4) = ΔSfre,268 = sum = -20.72 J/(mol-K)