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The percentage of heat supplied to a diatomic ideal gas that is converted into work in an isobaric process is
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Verified Answer
The correct answer is:
$71.4$
For isobaric process, $\Delta \mathrm{p}=0$
$\begin{aligned}
& \Delta \mathrm{W}=0 \\
& \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W} \\
& =\Delta \mathrm{Q} \\
& \mathrm{C}_{\mathrm{v}} \mathrm{dT}=\mathrm{C}_{\mathrm{p}} \mathrm{dT} \\
& \frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{p}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{7}{2} \mathrm{R}}=\frac{5}{7}
\end{aligned}$
The percentage of heat supplied into work
$\begin{aligned}
& =\frac{5}{7} \times 100 \\
& =71.4 \%
\end{aligned}$
$\begin{aligned}
& \Delta \mathrm{W}=0 \\
& \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W} \\
& =\Delta \mathrm{Q} \\
& \mathrm{C}_{\mathrm{v}} \mathrm{dT}=\mathrm{C}_{\mathrm{p}} \mathrm{dT} \\
& \frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{p}}}=\frac{\frac{5}{2} \mathrm{R}}{\frac{7}{2} \mathrm{R}}=\frac{5}{7}
\end{aligned}$
The percentage of heat supplied into work
$\begin{aligned}
& =\frac{5}{7} \times 100 \\
& =71.4 \%
\end{aligned}$
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