Thermodynamics

JEE Main level, single correct, medium

10 of 190

Question 10 of 190

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\dfrac{{\gamma }_{1}}{{\gamma }_{2}} is:

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Answer: (C) 2521\dfrac{25}{21}

As we know, Cv=fR2{C}_{v}=f\dfrac{R}{2}
and Cp=Cv+R=fR2+R{C}_{p}={C}_{v}+R=f\dfrac{R}{2}+R
Therefore, γ=CpCv=1+2f\gamma =\dfrac{{C}_{p}}{{C}_{v}}=1+\dfrac{2}{f}
For monatomic gas, f=3f=3 and hence γ1=53{\gamma }_{1}=\dfrac{5}{3}
For diatomic gas, f=5f=5 and hence γ2=75{\gamma }_{2}=\dfrac{7}{5}
Required ratio, γ1γ2=2521\dfrac{{\gamma }_{1}}{{\gamma }_{2}}=\dfrac{25}{21}

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