KCET · Physics · Thermodynamics
If \(\gamma\) is the ratio of specific heats and \(R\) is the universal gas constant, then the molar specific heat at constant volume \(\mathrm{C}_{\mathrm{V}}\) is given by
- A \(\gamma \mathrm{R}\)
- B \(\frac{(\gamma-1) R}{\gamma}\)
- C \(\frac{\mathrm{R}}{\gamma-1}\)
- D \(\frac{\gamma \mathrm{R}}{\gamma-1}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{\gamma-1}\)
Step-by-step Solution
Detailed explanation
From the Mayer's formula
\(C_{p}-C_{V}=R\)
and
\(\begin{gathered}\gamma=\frac{C_{p}}{C_{V}} \\C_{V}=C_{p}\end{gathered}\)
\(\Rightarrow \gamma \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{p}}\)
Substituting Eq. (ii) in Eq. (i), we get
\(\begin{aligned}\gamma C_{V}-C_{V} &=R \\C_{V}(\gamma-1) &=R \\C_{V} &=\frac{R}{\gamma-1}\end{aligned}\)
\(C_{p}-C_{V}=R\)
and
\(\begin{gathered}\gamma=\frac{C_{p}}{C_{V}} \\C_{V}=C_{p}\end{gathered}\)
\(\Rightarrow \gamma \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{p}}\)
Substituting Eq. (ii) in Eq. (i), we get
\(\begin{aligned}\gamma C_{V}-C_{V} &=R \\C_{V}(\gamma-1) &=R \\C_{V} &=\frac{R}{\gamma-1}\end{aligned}\)
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