b
Here, $n =\frac{1}{2}, c _{ v }=3 Jg ^{-1} K ^{-1}, M =4 g mol ^{-1}$
$\therefore C _{ V }= M _{ CV } 4 \times 3=12 \; J mol ^{-1} K ^{-1}$
At constant volume $P \propto T$.
$\therefore \frac{ P _{2}}{ P _{1}}=\frac{ T _{2}}{ T _{1}}=2, T _{2}=2 T _{1}$
Rise in temperature $\Delta T = T _{2}- T _{1}=2 T _{1}- T _{1}= T _{1}=273 \; K$
Heat required, $\Delta Q = NC V \Delta T =\frac{1}{2} \times 12 \times 273=1638 \; J$