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JEE PHYSICS MCQ: THERMODYNAMICS

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Analyzing +4 | -1
(A): The adiabatic exponent (γ=CpCv)\left(\gamma=\frac{C_p}{C_v}\right) of adiatomic gas is less than that of a mono atomic gas
(R): The molecules of mono atomic gas more degrees of freedom.
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Correct option: C
(A) is true but (R) is false
Option C ((A) is true but (R) is false) is correct because the adiabatic exponent is indeed less for diatomic gases compared to monatomic gases, but this is because diatomic gases have more degrees of freedom, not less.

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