Thermodynamics

JEE Main level, single correct, medium

147 of 190

Question 147 of 190

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx=\mathrm{TV}^{\mathrm{x}}= constant, then x\mathrm{x} is:

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Answer: (B) 25\frac{2}{5}

Equation of adiabatic change is

TVγ−1= constant \mathrm{TV}^{\gamma-1}=\text { constant }

Put γ=75\gamma=\frac{7}{5}, we get: γ−1=75−1\gamma-1=\frac{7}{5}-1

∴x=25\therefore \mathrm{x}=\frac{2}{5}

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