A compression at a constant pressure of 200 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 on the gas during this process?

Question

Quizplus · Accepted Answer

The work done on the gas during a constant pressure process can be calculated using the formula $W = P\Delta V$, where $W$ is the work done, $P$ is the pressure, and $\Delta V$ is the change in volume. Given that $P = 200 \, \text{kPa}$ and the volume changes from $0.200 \, \text{m}^3$ to $0.120 \, \text{m}^3$, the change in volume $\Delta V = 0.120 \, \text{m}^3 - 0.200 \, \text{m}^3 = -0.080 \, \text{m}^3$. Since work done on the gas is considered positive and the pressure is constant, the work done is $W = 200 \, \text{kPa} \times -0.080 \, \text{m}^3 = -16 \, \text{kPa} \cdot \text{m}^3$. To convert this to kilojoules, note that $1 \, \text{kPa} \cdot \text{m}^3 = 1 \, \text{kJ}$, so $W = -16 \, \text{kJ}$. However, since the question asks for the work done on the gas, and work done on the system is positive, the correct answer is $16 \, \text{kJ}$.

A Compression at a Constant Pressure of 200 KPa Is 0.200 m30.200 \mathrm {~m} ^ { 3 }0.200 m3

A Compression at a Constant Pressure of 200 KPa Is $0.200 \mathrm {~m} ^ { 3 }$