Question

Consider 1.00 mol of an ideal gas (C_{V} = 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.

Answer #1

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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