Match the $\mathrm {List-I}$ with $\mathrm {List-II}$ $\begin{array}{|l|l|l|l|}\hline & \text{List-I} & &…
$\begin{array}{|l|l|l|l|}\hline & \text{List-I} & & \text{List-II} \\ \hline \text{A.} & \text{Triatomic rigid gas} & \text{I.} & \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3} \\ \hline \text{B.} & \begin{array}{l} \text{Diatomic non-rigid} \\ \text{gas} \end{array} & \text{II.} & \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{7}{5} \\ \hline \text{C.} & \text{Monoatomic gas} & \text{III.} & \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{4}{3} \\ \hline \text{D.} & \text{Diatomic rigid gas} & \text{IV }& \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{9}{7} \\ \hline\end{array}$
Choose the $\mathrm {correct}$ answer from the options given below :
- A-III, B-IV, C-I, D-II
- A-III, B-II, C-IV, D-I
- A-II, B-IV, C-I, D-III
- A-IV, B-II, C-III, D-I
Solution
$f=6$, Triatomic rigid gas
$\mathrm{f}=7$, Diatomic non-rigid gas
$f=5$, Diatomic rigid gas
$\mathrm{f}=3$, monoatomic rigid gas
$\gamma=1+\frac{2}{6}=\frac{4}{3}$ (Triatomic)
$\gamma=1+\frac{2}{7}=\frac{9}{7}($ Diatomic, non-rigid $)$
$\gamma=1+\frac{2}{5}=\frac{7}{5}($ Diatomic, rigid $)$
$\gamma=1+\frac{2}{3}=\frac{5}{3}($ Monoatomic, rigid $)$
A-III, B-IV, C-I, D-II
Asked in: JEE Main 2025 (07 Apr Shift 1)