Q 11-11-155JEE MainJEE Main 2019 (11 Jan, Shift 2)Medium
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):
Answer: (A) $\dfrac12R\Delta T$
From $V = K/T$: $dV = -\dfrac{K}{T^2}dT$. With $P = \dfrac{RT}{V} = \dfrac{RT^2}{K}$:
$$dW = P\,dV = -R\,dT \Rightarrow W = -R\Delta T$$
$$Q = \Delta U + W = \frac32R\Delta T - R\Delta T = \frac12R\Delta T$$
Solution by Sreeraj P, M.Sc Physics