The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to…

Given here, $∆Q=0$

From first law of thermodynamics, $∆Q=∆U+W$, where, $∆U$ is change in internal energy and $W$ is work done.

We have, $∆U=-W$

$W=\mathrm{Area} \mathrm{under} \mathrm{P}-\mathrm{V} \mathrm{curve}=\int PdV$

Here, volume decreases, hence work done by system will be negative.

$∆U=-W=30\times {10}^3\times 150\times {10}^{-6}$

$=4500\times {10}^{-3}$

$=4.5 \mathrm{J}$