Question

A perfect gas has a constant molar volume heat capacity of Cvm=1.5R and a constant pressure molar heat capacity of Cpm=2.5R. For the process of heating 2.80 mol of this gas with a 120 W heater for 65 seconds, calculate:

a) q, w, delta(T), and delta(U) for heating at a constant volume
b) q, w, delta(T), and delta(H) for heating at a constant pressure

Answer 1

Given Cv = 1.5R

Cp = 2.5R

moles of the gas, n = 2.80 mol

Power of heater, P = 120 W

time of heat supply, t = 65 s

Hence total heat supplied, q = Pxt = 120 W x 65 s = 7800 J

(a) For heating at a constant volume(Isochoric process)

V = 0

Heat supplied, q = 7800 J (answer)

Work done, w = P_ext * V = 0 (answer)

Applying first law of thermodynamics,

U = q + w = 7800 J + 0

=> U = 7800 J (answer)

Hence in isochoric process the total heat supplied is converted to internal energy.

Hence q = U = nCv*T

=> 7800 J = 2.80 mol * 1.5R * T = 2.80 mol * 1.5 * 8.314 J.K-1.mol-1 x T

=> T = 7800 J / (2.80 mol * 1.5 * 8.314 J.K-1.mol-1) = 223.4 K (answer)

(b) For heating at a constant pressure(Isobaric process)

P = 0

Heat supplied, q = 7800 J (answer)

Also for isobaric process, q = nCp*T

=> q = nCp*T = 7800 J

=> T = 7800 J / (nCp) = 7800 J / (2.80 mol * 2.5 R)

=> T = 7800 J / (2.80 mol * 2.5 *8.314 J.K-1.mol-1) = 134.0 K (answer)

Change in internal energy, U = nCv*T = 2.80 mol * 1.5 * 8.314 J.K-1.mol-1 * 134 K

=> U = 4679 J (answer)

Applying first law of thermodynamics,

U = q + w = 7800 J + w

=> 4679 J = 7800 J + w

=> w = 4679 J - 7800 J = - 3121 J (answer)

For isobaric process, heat supplied is equal to enthalpy change. Hence

q = H = 7800 J (answer)