Question

A 2.00 mole sample of a perfect gas is heated from 41.2L and 255K to 42.2 L and 317K against a constant external pressure of 1.00 bar. What is delta H, delta U, q, and w for this process? (initial ---> intermediate---> final state). Intermediate state is V=41.2L, T=317K

Hint: first calculate delta h, delta u, q, and w for each arm. (Initial to intermediate being one arm, then intermediate to final being second arm)

Assume that Cp,m=3.5R and Cp,m-Cv,m=R

Answer 1

For initial (41.2L and 255K) to intermediate state (41.2L, 317K)

This is a constant volume process

Enthalpy change H1 = n Cp (T2-T1) = 2 x 3.5 x 8.314 x (317 - 255) = 3608.276 J

Internal energy change

U1 = n Cv (T2-T1) = n x (Cp - R) x (T2-T1)

= 2 x (3.5*8.314 - 8.314) x (317 - 255)

= 2577.34 J

Work done W1 = P(V2-V1) = 0

First law of thermodynamics

dU = q + W

q1 = 2577.34 J

For intermediate (41.2L and 317K) to intermediate state

(42.2L, 317K)

This is a constant temperature process

Enthalpy change H2 = n Cp (T2-T1) = 0

Internal energy change

U2 = n Cv (T2-T1) = 0

Work done W2 = n x RT ln(Vf/Vi)

= 2 x (8.314 x 317) * ln (42.2/41.2)

= 126.41 J

First law of thermodynamics

dU = q + W

0 = q2 + ( 126.41)

q2 = - 126.41 J

Overall process

Change in enthalpy H = H2 - H1 = 0 - 3608.276 = - 3608.276 J

Change in Internal energy U = U2 - U1 = 0 - 3608.276

= - 3608.276 J

W = 126.41 J

q = - 126.41- 2577.34 = - 2703.75 J