MHT CET · Physics · Kinetic Theory of Gases
The molar specific heat of an ideal gas at constant pressure and constant volume is \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{C}_{\mathrm{v}}\) respectively. If \(\mathrm{R}\) is universal gas constant and \(\gamma=\frac{C_p}{C_y}\) then \(C_v=\)
- A \(\frac{1-\gamma}{1+\gamma}\)
- B \(\frac{1+\gamma}{1-\gamma}\)
- C \(\frac{\gamma-1}{\mathrm{R}}\)
- D \(\frac{\mathrm{R}}{\gamma-1}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{R}}{\gamma-1}\)
Step-by-step Solution
Detailed explanation
\(
C_p-C_v=R
\)
Dividing both the sides by \(\mathrm{C}_{\mathrm{v}}\),
\(
\begin{array}{ll}
\therefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \\
\therefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{array}
\)
C_p-C_v=R
\)
Dividing both the sides by \(\mathrm{C}_{\mathrm{v}}\),
\(
\begin{array}{ll}
\therefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \\
\therefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{array}
\)
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