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JEE Main 2019 (11 Jan, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2019 (11 Jan, Shift 2)

Q219JEE Main 2019 · 11 JAN · SHIFT 2Difficult

Thermodynamic Processes

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $\Delta$ T. The amount of heat absorbed by gas is (R is gas constant):

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Answer: (b)

$VT = K$ and $PV = RT$ give $P \propto \frac{T}{V} \propto \frac{1}{V^2}$ , so $PV^2 =$ constant ( $x = 2$ ).
$C = C_v + \frac{R}{1 - x} = \frac{3}{2}R - R = \frac{R}{2}$
$Q = C\Delta T = \frac{1}{2}R\Delta T$ , so the answer is (b).

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