In an isobaric process of an ideal gas, the ratio of work done by the system to the heat supplied…
In an isobaric process of an ideal gas, the ratio of work done by the system to the heat supplied $\left(\frac{W}{Q}\right)$ is
- $\frac{1}{\gamma-1}$
- $\gamma$
- $\frac{\gamma}{\gamma-1}$
- $\frac{\gamma-1}{\gamma}$
Solution
For an isobaric process,
$\begin{aligned}
& \Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T} \text { and } \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} \\
& \mathrm{~W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}
\end{aligned}$
$\therefore \quad \frac{W}{\Delta Q}=\frac{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{p}}}$
$=\frac{\frac{C_p}{C_v}-1}{\frac{C_p}{C_v}}$
$\therefore \quad \frac{W}{\Delta Q}=\frac{\gamma-1}{\gamma} \quad \ldots\left(\because \frac{C_p}{C_v}=\gamma\right)$
Asked in: MHT CET 2024 (04 May Shift 1)
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