Heat is applied to a rigid diatomic gas at constant pressure. The ratio $\Delta \mathrm{Q}: \Delta…
Heat is applied to a rigid diatomic gas at constant pressure. The ratio $\Delta \mathrm{Q}: \Delta \mathrm{U}: \Delta \mathrm{W}$ is
- $5: 7: 2$
- $7: 5: 2$
- $2: 5: 7$
- 5: 2: 7
Solution
We know,
$\begin{array}{l}
\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T} \\
\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \\
\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T} \\
\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=\mathrm{R}: \mathrm{C}_{\mathrm{v}}: \mathrm{C}_{\mathrm{p}}
\end{array}$
For a diatomic gas $f=5$
$\begin{aligned}
\therefore \quad & C_{v}=\frac{f}{2} R=\frac{5}{2} R \\
& C_{p}=C_{v}+R=\frac{7}{2} R \\
\therefore \quad R: C_{v}: C_{p}=1: \frac{5}{2}: \frac{7}{2}=2: 5: 7
\end{aligned}$
.
Asked in: MHT CET 2020 (14 Oct Shift 1)
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