MHT CET · Physics · Kinetic Theory of Gases
For a gas, \(\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}=0.4\), where \(\mathrm{R}\) is universal gas constant and \(\mathrm{C}_{\mathrm{v}}\) is molar specific heat at constant volume. The gas is made up of molecules which are
- A rigid diatomic
- B monoatomic
- C non-rigid diatomic
- D polyatomic
Answer & Solution
Correct Answer
(A) rigid diatomic
Step-by-step Solution
Detailed explanation
Given: \(\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=0.4\)
\(\begin{aligned}
\mathrm{C}_{\mathrm{V}} & =\frac{\mathrm{R}}{0.4}=\frac{5 \mathrm{R}}{2} \\
\mathrm{C}_{\mathrm{P}} & =\mathrm{C}_{\mathrm{V}}+\mathrm{R} \\
\therefore \quad \mathrm{C}_{\mathrm{P}} & =\frac{7 \mathrm{R}}{2} \\
\gamma & =\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2}}{\frac{5}{2}} \\
\gamma & =\frac{7}{5}
\end{aligned}\)
\(\therefore \quad\) The gas is made up of rigid diatomic molecules.
\(\begin{aligned}
\mathrm{C}_{\mathrm{V}} & =\frac{\mathrm{R}}{0.4}=\frac{5 \mathrm{R}}{2} \\
\mathrm{C}_{\mathrm{P}} & =\mathrm{C}_{\mathrm{V}}+\mathrm{R} \\
\therefore \quad \mathrm{C}_{\mathrm{P}} & =\frac{7 \mathrm{R}}{2} \\
\gamma & =\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2}}{\frac{5}{2}} \\
\gamma & =\frac{7}{5}
\end{aligned}\)
\(\therefore \quad\) The gas is made up of rigid diatomic molecules.
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