½ mole of helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (specific heat of the gas $= 3\ J\ gm^{- 1}K^{- 1}$) is

Question

Dr. Abubacker · Accepted Answer

(b) Molecular mass of $He;M = 4g$
$\Rightarrow$ Molar value of $C_{V} = Mc_{V} = 4 	imes 3 = 12\frac{J}{mole - kelvin}$
At constant volume $P \propto T,$ therefore on doubling the pressure temperature also doubles
$i.e.,\ T_{2} = 2T_{1} \Rightarrow \mathrm{\Delta}T = T_{2} - T_{1} = 273K$
Also $(\mathrm{\Delta}Q)_{V} = \mu C_{V}\mathrm{\Delta}T = \frac{1}{2} 	imes 12 	imes 273 = 1638J$

Q. ½ mole of helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (specific heat of the gas $= 3\ J\ gm^{- 1}K^{- 1}$) is

Explanation

Reviewed by Dr. Abubacker

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