Question

$\frac{1}{2}$ mole of helium gas is contained in a container at ${27}^{\circ }C$. Find the heat energy needed to double the pressure of the gas, keeping its volume constant. Given heat capacity of the gas is 3 J/g-K.

Answer 1

Initial temperature, $T_1={27}^{\circ }C=300K$
As volume is constant, $\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{P_1}{T_1}=\frac{2P_1}{T_2}\Rightarrow T_2=2T_1$
Heat capacity =3 J/g-K
At constant volume, $\Delta Q=\Delta U$, as $\Delta W=0$
$\Rightarrow \Delta Q=mC\Delta T=\frac{1}{2}\times 4\times 3\times (2T_1-T_1)=\frac{1}{2}\times 4\times 3\times T_1=\frac{1}{2}\times 4\times 3\times 300=1800J$