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 $3\, J\, g^{-1}\, K^{-1}$.

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

Answer 1

Answer: (C)

Here, $n = \frac{1}{2}, c_{v} = 3\,J\,g^{-1}\,K^{-1}$, $M = 4g \, mol^{-1}$
$\therefore \quad C_{v} = Mc_{v} = 4 \times 3 = 12 \,Jmol^{-1}\, K^{-1}$
At constant volume $P \propto T$.
$\therefore \quad \frac{P_{1}}{P_{1}} = \frac{T_{2}}{T_{1}} = 2$, $T_{2} = 2T_{1}$
Rise in temperature $\Delta T=T_{2} - T_{1} = 2T_{1} - T_{1} = T_{1} = 273\, K$
Heat required, $\Delta Q = nC_{V} \Delta T = \frac{1}{2}\times12\times273 = 1638\,J$