1/2 mole of helium gas is contained in a container at S.T.P. The heat energy (in Joule) needed to double the pressure of the gas, keeping the volume constant (heat capacity of the gas = $3 {Jg}^{- 1} K^{- 1}$ ) is

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Detailed Solution

$\begin{array}{l} Here, n = \frac{1}{2}, \\ H e a t c a p a c i t y = C_{v} = 3 J g^{- 1} K^{- 1}, M = 4 g m o l^{- 1} \\ ∴ C_{v} = M c_{v} = 4 \times 3 = 12 J m o l^{- 1} K^{- 1} \\ At constant volume P \propto T \\ ∴ \frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = 2 \\ \Rightarrow T_{2} = 2 T_{1} \\ Rise in temperature Δ T = T_{2} - T_{1} = 2 T_{1} - T_{1} = T_{1} = 273 K \\ Heat required, Δ Q = n C_{v} Δ T = \frac{1}{2} \times 12 \times 273 = 1638 J \end{array}$