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QJEEM23Pha96d: JEE Main 2023 Physics - Class 11 JEEM23Pha96d Solution: Let be the ratio of molar specific heat at constant pressure…

Let γ1\gamma_1 be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and γ2\gamma_2 be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, γ1γ2\frac{\gamma_1}{\gamma_2} is :

Multiple Choice Options

  • A 3527\frac{35}{27}
  • B 2521\frac{25}{21}
  • C 2125\frac{21}{25}
  • D 2735\frac{27}{35}

Step-by-Step Verified Solution

For monoatomic gas γ1=53\gamma_1=\frac{5}{3}

For diatomic gas at low temperatures

amp;γ2=75amp;∴γ1γ2=5375=2521\begin{aligned}& \gamma_2=\frac{7}{5} \\\\& \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}\end{aligned}

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