JEE Main 2023 — Thermodynamics Question with Solution

(A) Change in internal energy is expressed as$∆U=n{C}_v∆T$, here, $∆T$ is change in temperature.

Since, in an isothermal process temperature remains constant, thus, $∆U=0$

$A\to II$

(B) In an adiabatic process, heat transfer, $Q = 0$.

So, from first law of thermodynamics, 

$Q=∆U +W$

$∆U=-W$

Since, work done by gas is positive, thus, $∆U$ is negative

$B\to I$

(C) In an isochoric process, volume remains constant, so work done by or on the gas $W =P∆V=0$

$C\to IV$

(D) In an isobaric process, pressure remains constant, so work done$W=P∆V\ne 0$ and change in internal energy $∆U=n{C}_v∆T\ne 0$

Thus, heat absorbed goes partly to increase internal energy and partly do work.

$D\to III$