MHT CET · Physics · Thermodynamics
Heat is applied to a rigid diatomic gas at constant pressure. The ratio \(\Delta \mathrm{Q}: \Delta \mathrm{U}: \Delta \mathrm{W}\) is
- A \(5: 7: 2\)
- B \(7: 5: 2\)
- C \(2: 5: 7\)
- D 5: 2: 7
Answer & Solution
Correct Answer
(B) \(7: 5: 2\)
Step-by-step Solution
Detailed explanation
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 \(f=5\)
\(\therefore \quad C_{v}=\frac{f}{2} R=\frac{5}{2} R \)
\( C_{p}=C_{v}+R=\frac{7}{2} R \)
\( \therefore \quad R: C_{v}: C_{p}=1: \frac{5}{2}: \frac{7}{2}=2: 5: 7\)
\(\begin{array}{l}
\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T} \\
\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \\
\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T} \\
\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=\mathrm{R}: \mathrm{C}_{\mathrm{v}}: \mathrm{C}_{\mathrm{p}}
\end{array}\)
For a diatomic gas \(f=5\)
\(\therefore \quad C_{v}=\frac{f}{2} R=\frac{5}{2} R \)
\( C_{p}=C_{v}+R=\frac{7}{2} R \)
\( \therefore \quad R: C_{v}: C_{p}=1: \frac{5}{2}: \frac{7}{2}=2: 5: 7\)
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