A compression at a constant pressure of $200 \mathrm { kPa }$ is performed on $8.00$ moles of an ideal monatomic gas. The compression reduces the volume of the gas from $0.200 \mathrm {~m} ^ { 3 }$ to $0.120 \mathrm {~m} ^ { 3 }$. How much work was done by the gas during this process?
A) $- 40 \mathrm {~kJ}$
B) $0 \mathrm {~kJ}$
C) $40 \mathrm {~kJ}$
D) $16 \mathrm {~kJ}$
E) $- 16 \mathrm {~kJ}$

Question

Quizplus · Accepted Answer

The work done by the gas during a constant pressure process can be calculated using the formula $W = P \Delta V$, where $P$ is the pressure and $\Delta V$ is the change in volume. Here, $\Delta V = V_{\text{final}} - V_{\text{initial}} = 0.120 \mathrm{~m}^3 - 0.200 \mathrm{~m}^3 = -0.080 \mathrm{~m}^3$. Given $P = 200 \mathrm{~kPa} = 200 \times 10^3 \mathrm{~Pa}$, the work done is $W = 200 \times 10^3 \mathrm{~Pa} \times (-0.080 \mathrm{~m}^3) = -16 \times 10^3 \mathrm{~J} = -16 \mathrm{~kJ}$. The work is negative because the gas is compressed, indicating work is done on the gas.

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A Compression at a Constant Pressure Of $200 \mathrm { kPa }$