MHT CET · Physics · Thermodynamics
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
- A \(\frac{1}{\gamma-1}\)
- B \(\gamma\)
- C \(\frac{\gamma}{\gamma-1}\)
- D \(\frac{\gamma-1}{\gamma}\)
Answer & Solution
Correct Answer
(D) \(\frac{\gamma-1}{\gamma}\)
Step-by-step Solution
Detailed explanation
For an isobaric process,
\(\begin{aligned}
& \Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T} \text { and } \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} \\
& \mathrm{~W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}
\end{aligned}\)
\(\therefore \quad \frac{W}{\Delta Q}=\frac{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{p}}}\)
\(=\frac{\frac{C_p}{C_v}-1}{\frac{C_p}{C_v}}\)
\(\therefore \quad \frac{W}{\Delta Q}=\frac{\gamma-1}{\gamma} \quad \ldots\left(\because \frac{C_p}{C_v}=\gamma\right)\)
\(\begin{aligned}
& \Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T} \text { and } \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} \\
& \mathrm{~W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}
\end{aligned}\)
\(\therefore \quad \frac{W}{\Delta Q}=\frac{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{p}}}\)
\(=\frac{\frac{C_p}{C_v}-1}{\frac{C_p}{C_v}}\)
\(\therefore \quad \frac{W}{\Delta Q}=\frac{\gamma-1}{\gamma} \quad \ldots\left(\because \frac{C_p}{C_v}=\gamma\right)\)
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