(C) Given: $\frac{R}{C_{v}} = 0.67$.We know that the gas constant $R = C_{p} - C_{v}$.Substituting this into the given equation: $\frac{C_{p} - C_{v}}

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(C) Given: $\frac{R}{C_{v}} = 0.67$.We know that the gas constant $R = C_{p} - C_{v}$.Substituting this into the given equation: $\frac{C_{p} - C_{v}}{C_{v}} = 0.67$.$\frac{C_{p}}{C_{v}} - 1 = 0.67$.Since the adiabatic index $\gamma = \frac{C_{p}}{C_{v}}$,we have $\gamma - 1 = 0.67$,which implies $\gamma = 1.67$.For a monoatomic gas,the degrees of freedom $f = 3$,so $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = 1 + 0.666... \approx 1.67$.Therefore,the gas is monoatomic.

Class 11 Physics Kinetic Theory of Gases Molar Specific Heat of gas and relation between them (Mayer's formula)

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For a gas,$\frac{R}{C_{v}} = 0.67$. This gas is made up of molecules which are

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diatomic.
B
polyatomic.
C
monoatomic.
D
mixture of diatomic and polyatomic.

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Class 11

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Kinetic Theory of Gases

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Molar Specific Heat of gas and relation between them (Mayer's formula)

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For a gas,$\frac{R}{C_{v}} = 0.67$. This gas is made up of molecules which are

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