The temperature of an ideal gas in a sealed rigid 0.20- $m ^ { 3 }$ container is reduced from 360 K to $270 \mathrm {~K} \text {, }$ and the final pressure of the gas is $40 \mathrm { kPA } .$ How much work is done on the gas during this process? (R = 8.31 J/mol ∙ K)

Question

Quizplus · Accepted Answer

Since the container is rigid, its volume does not change during the process. Work done on a gas is given by $W = P\Delta V$, where $P$ is pressure and $\Delta V$ is the change in volume. Because the volume remains constant ($\Delta V = 0$), no work is done on the gas ($W = 0$).

The Temperature of an Ideal Gas in a Sealed Rigid m3m ^ { 3 }m3

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