Question

2. A 1.00L piston with 1.00mole of an ideal gas at 298.0K and 1.00bar is isothermally and reversibly compressed to a final volume of 0.100L, then irreversibly expanded in 1 step to its original volume with an applied pressure of 1.00bar.

a. (6 pts) Calculate the change in heat for the system (q sys) for the compression and expansion steps.

b. (6 pts) Calculate the change in entropy for the compression and expansion steps.

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c. (6 pts) Calculate the net change in entropy for the compression and expansion steps together.

d. (6 pts) Is there a net change in entropy for the system? If so, why; if not, why not?

e. (6 pts) Is there a net change in the entropy for the surrounding? If so, why; if not, why not?

Answer 1

(a) For isothermal compression work done can be calculated as

w = - nRTxln(Vf/Vi) = - 1 molx8.314 J.mol-1K-1x298Kxln(0.100L / 1.00L) = 5705 J

For isothermal prcess U = nCv​T = 0

From first law of thermodynamica

U = w + q

=> 0 =  5705 J + q

=> q = - 5705 J (answer)

For the second step, no work is done during free expansion. Hence w = 0

Also during free expansion T remains constant. Hence U = 0

Hence from 1st law q = 0 (answer)

(b): Entropy change for compression:

S(comp) = q(rev)/T = - 5705 J/ 298 K = - 19.14 J/K (answer)

Entropy change for expansion:

S(expan) = nRxln(Vf/Vi) = 1.00 molx8.314 JK-1mol-1xln(1.00L / 0.1L) = +19.14 J/K (answer)

(c): net change in entropy, S = S(comp) + S(expan)

=> S = - 19.14 J/K + (+19.14 J/K) = 0 (answer)

(d) Hence there is no net change in the entropy of the system, because the system comes to its original condition after successive compression and expansion.

(e) There would ne no change in net surrounding entropy if both the system and surrounding temperature are same, other wise it will not be non-zero.