What is the work performed by a gas, if it expands from 5.00 L to 15.0 L against a constant external pressure of 1.62 atm? Express the final answer in joules.

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Quizplus · Accepted Answer

The work done by a gas during expansion or compression can be calculated using the formula $W = -P\Delta V$, where $W$ is the work done, $P$ is the external pressure, and $\Delta V$ is the change in volume. The negative sign indicates that work is done by the system on the surroundings when the gas expands. Given that the external pressure is 1.62 atm and the volume change ($\Delta V$) is from 5.00 L to 15.0 L, which is a 10.0 L increase, we first convert the pressure to Pascals (since 1 atm = 101.325 kPa = 101325 Pa) and volume to cubic meters (since 1 L = 0.001 m³). Thus, $P = 1.62 \times 101325$ Pa, and $\Delta V = 10.0 \times 0.001$ m³. The work done is then $W = -P\Delta V = -(1.62 \times 101325 \times 10 \times 0.001)$ J = -1,640 J.

What Is the Work Performed by a Gas, If It