The temperature of an ideal gas in a sealed rigid $0.20 - \mathrm { m } ^ { 3 }$ container is reduced from $360 \mathrm {~K}$ to $270 \mathrm {~K}$, and the final pressure of the gas is 40 kPA. How much work is done by the gas during this process? $( R = 8.31 \mathrm {~J} / \mathrm { mol } \cdot \mathrm { K } )$
A) $- 12 \mathrm {~kJ}$
B) $9.0 \mathrm {~kJ}$
C) $0 \mathrm {~kJ}$
D) $- 9.0 \mathrm {~kJ}$
E) $12 \mathrm {~kJ}$

Question

Quizplus · Accepted Answer

Since the container is rigid, its volume does not change during the process. Therefore, no work is done by the gas because work (W) is given by W = PΔV, and ΔV = 0 in this case.

The Temperature of an Ideal Gas in a Sealed Rigid 0.20−m30.20 - \mathrm { m } ^ { 3 }0.20−m3

The Temperature of an Ideal Gas in a Sealed Rigid $0.20 - \mathrm { m } ^ { 3 }$