$=333 K, M_{2}$
$=M-\frac{M}{4}=\frac{3 M}{4}$
[As $1/ 4$ th part of air escapes]
If pressure and volume of gas remains constant then $\mathrm{MT}=$ constant
$\therefore \frac{T_{2}}{T_{1}}=\frac{M_{1}}{M_{2}}=\left(\frac{M}{3 M / 4}\right)=\frac{4}{3} \Rightarrow T_{2}$
$=\frac{4}{3} \times T_{1}=\frac{4}{3} \times 333=444 \mathrm{K}$
$=171^{\circ} \mathrm{C}$
(image)
$(A)$ Process $I$ is an isochoric process $(B)$ In process $II$, gas absorbs heat
$(C)$ In process $IV$, gas releases heat $(D)$ Processes $I$ and $III$ are $not$ isobaric

