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JEE Main 2020 (5 Sep, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2020 (5 Sep, Shift 2)

Q205JEE Main 2020 · 5 SEP · SHIFT 2Medium

Thermodynamic Processes

In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is:

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Answer: (a)

Density $\rho \propto \frac{1}{V}$ , so adiabatic $PV^{\gamma} =$ const gives $P \propto \rho^{\gamma}$ .
Diatomic: $\gamma = \frac{7}{5}$ . $n = 32^{7/5} = (2^5)^{7/5} = 2^7$
So the answer is (a) 128.

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