Question

$1/2$ mole of helium is contained in a container at $STP$. How much heat energy is needed to double the pressure of the gas, (volume is constant) heat capacity of gas is $3J{g}^{-1}{K}^{-1}$.

• A. $1436J$
• B. $736J$
• C. $1638J$
• D. $5698J$

Answer 1

Answer: (C)

Here, $n=\frac{1}{2},c_v=3J{g}^{-1}{K}^{-1}$, $M=4gmo{l}^{-1}$
$\therefore C_v=M{c}_v=4\times 3=12Jmo{l}^{-1}{K}^{-1}$
At constant volume $P\propto T$.
$\therefore \frac{P_1}{P_1}=\frac{T_2}{T_1}=2$, $T_2=2T_1$
Rise in temperature $\Delta T=T_2-T_1=2T_1-T_1=T_1=273K$
Heat required, $\Delta Q=n{C}_V\Delta T=\frac{1}{2}\times 12\times 273=1638J$