Consider 1.00 mol of an ideal gas (CV = 3/2 R) occupying 22.4 L that undergoes an isochoric (constant volume) temperature increase from 298 K to 342 K. Calculate ∆p, q , w, ∆U, and ∆H for the change.
For Units, pressure in atm and the rest in J.
Ans. P1 = n R T1 / V= 1 * 0.082 * 298 / 22.4 = 1.09 atm
P2 = n R T2 / V = 1 * 0.082 * 342 / 22.4 = 1.25 atm
dP = P2 - P1 = 1.25 - 1.09 = 0.16 atm
w = -pdV
for isochoric process, dV = 0
and hence w = -p * 0 = 0
Cv = 3/2 R = 1.5 * 8.314
q = dU = n Cv dT = 1 * 1.5 * 8.314 * (342 - 298) = 548.7 J
dH = dU + d(PV) = dU + PdV + VdP = 548.7J + 0 + (22.4 * 0.16 L atm)
0.082Latm mol-1K-1 = 8.314 JK-1mol-1
1 L atm = 101.39 J
dH = 548.7 J + (3.584 * 101.39)
dH = 912.08 J
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