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JEE Main 2019 (11 Jan, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2019 (11 Jan, Shift 1)

Q213JEE Main 2019 · 11 JAN · SHIFT 1Easy

Thermodynamic Processes

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

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Answer: (b)

Adiabatic: $TV^{\gamma - 1} =$ constant, so $x = \gamma - 1$ .
Rigid diatomic: $\gamma = \dfrac{7}{5}$ .
$x = \dfrac{7}{5} - 1 = \dfrac{2}{5}$
So the answer is (b) $\dfrac{2}{5}$ .

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