Question

The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp/(J/K) = 20.17 + 0.4001 (T/K). Calculate q, w, Delta U, and Delta H when the temperature is raised from 0 degree celsius to 100 degree celsius (a) at constant pressure, (b) at constant volume.

Answer 1

[AI-generated answer — Bot GPT 3.5]

Given: Cp/(J/K) = 20.17 + 0.4001 (T/K)

(a) Constant pressure:
Temperature change = 100 - 0 = 100 K

q = ΔH = ∫Cp dT
= ∫(20.17 + 0.4001T) dT
= 20.17T + 0.4001(T^2)/2
= 20.17(100) + 0.4001(100^2)/2
= 2017 + 2000.5
= 4017.5 J

w = PΔV = 0 (as no volume change at constant pressure)

ΔU = q - w = 4017.5 J

ΔH = 4017.5 J

(b) Constant volume:
q = ΔU = ∫Cv dT
= ∫Cp - R dT
= ∫(20.17 + 0.4001T - 8.314) dT
= (20.17T + 0.4001(T^2)/2 - 8.314T)
= 20.17(100) + 0.4001(100^2)/2 - 8.314(100)
= 2017 + 2000.5 - 831.4
= 3186.1 J

w = 0 (as no work done at constant volume)

ΔU = q = 3186.1 J

Hence, at constant pressure, q = 4017.5 J, w = 0, ΔU = 4017.5 J, ΔH = 4017.5 J
At constant volume, q = 3186.1 J, w = 0, ΔU = 3186.1 J.