MHT CET · Physics · Kinetic Theory of Gases
The ratio of the specific heats \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\), in terms of degrees of freedom ( n ) is
- A \(\left(1+\frac{1}{n}\right)\)
- B \(\left(1+\frac{\mathrm{n}}{3}\right)\)
- C \(\left(1+\frac{2}{n}\right)\)
- D \(\left(1+\frac{\mathrm{n}}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(1+\frac{2}{n}\right)\)
Step-by-step Solution
Detailed explanation
We know
\(\mathrm{C}_{\mathrm{v}}=\mathrm{n} \times \frac{\mathrm{R}}{2}\)
From \(C_P-C_V=R\), we get \(C_p=C_v+R=\frac{n R}{2}+R=\left(\frac{n}{2}+1\right) R\)
\(\therefore \quad \frac{C_p}{C_v}=\frac{\left(\frac{n}{2}+1\right) R}{\frac{n R}{2}}=\left(1+\frac{2}{n}\right)\)
\(\mathrm{C}_{\mathrm{v}}=\mathrm{n} \times \frac{\mathrm{R}}{2}\)
From \(C_P-C_V=R\), we get \(C_p=C_v+R=\frac{n R}{2}+R=\left(\frac{n}{2}+1\right) R\)
\(\therefore \quad \frac{C_p}{C_v}=\frac{\left(\frac{n}{2}+1\right) R}{\frac{n R}{2}}=\left(1+\frac{2}{n}\right)\)
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