JEE Main 2022 — Thermodynamics Question with Solution

 In the part $A\to B$, the volume remains constant. Thus, the work done by the gas is zero. It is given that the heat absorbed by the gas is $40 \mathrm{J}$. From first law of thermodynamics, $Q=\mathrm{\Delta }U+W$

$\Rightarrow 40=\mathrm{\Delta }U+0\Rightarrow \mathrm{\Delta }U=40 \mathrm{J}$

or $\mathrm{\Delta }U=U_B-U_A=40 \mathrm{J}$

The increase in internal energy from $A$ to $B$ is $40 \mathrm{J}$. The internal energy is $1560 \mathrm{J}$ at $A$, then

$U_B-1560=40 \mathrm{J}\Rightarrow U_B=1600 \mathrm{J}$

In the part $B\to C$, the work done by the gas is $W=50 \mathrm{J}$ and no heat is given to the system. Again using $Q=\mathrm{\Delta }U+W$

$0=\mathrm{\Delta }U-50$ or $\mathrm{\Delta }U=U_C-U_B=50 \mathrm{J}$

$U_C-1600=50 \mathrm{J}\Rightarrow U_C=1650 \mathrm{J}$

 The change in internal energy from $C\to A$,

$∆U=1560-1650=-90 \mathrm{J}$

The heat given to the system is $Q=-60 \mathrm{J}$

Using $Q=\mathrm{\Delta }U+W$, we get $W=Q-∆U=-60-\left(-90\right)=30 \mathrm{J}$