One mole of a monatomic gas at 433 K is reversibly taken to half of its original volume by an isobaric process.How much work is done by the gas?

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Quizplus · Accepted Answer

The work done by the gas during an isobaric (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. Since the volume is halved, $\Delta V = V_f - V_i = \frac{1}{2}V_i - V_i = -\frac{1}{2}V_i$, indicating the work is done on the gas, hence negative. Using the ideal gas law $PV = nRT$, where $n = 1$ mole, $R = 8.314$ J/mol·K, and $T = 433$ K, we can find $P$ as $P = \frac{nRT}{V}$. The work done can also be expressed as $W = -nRT \ln\frac{V_f}{V_i}$ for an isothermal process, but for an isobaric process, it simplifies to $W = P\Delta V$. Given the information does not directly match the conditions needed for a straightforward calculation and the lack of specific numerical values for initial volume or pressure, we must rely on the principle that work done in an isobaric process depends on the volume change and the sign indicates work done on the gas if negative. Among the options provided, without the exact numerical calculation from the initial data, the correct choice involves understanding the process's nature (isobaric and volume reduction) and recognizing that work is done on the gas, hence a negative value. However, without the exact volume or pressure, the precise numerical answer cannot be directly calculated from the given information, suggesting a potential misunderstanding in the explanation process. The correct approach involves applying the ideal gas law and the work formula for isobaric processes correctly with given or inferred values.

One Mole of a Monatomic Gas at 433 K Is