$\frac{1}{2}$ 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 = $3 J g m^{- 1} K^{- 1}$ ) is

3276 J

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1638 J

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819 J

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409.5 J

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Solution

The correct option is B
1638 J

If Pressure is doubled. temperature is doubled. Hence dT = 273K.

$m = M x n = 0.5 x 4 = 2 g$

$d U = n C_{V} d T = m c_{V} d T = 2 x 3 x 273 = 1638 J$

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12 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

The correct option is B 1638 J If Pressure is doubled. temperature is doubled. Hence dT = 273K. m=Mxn=0.5x4=2g dU=nCVdT = mcVdT=2x3x273=1638J

$\frac{1}{2}$ 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 = $3 J g m^{- 1} K^{- 1}$ ) is

The correct option is B
1638 J

If Pressure is doubled. temperature is doubled. Hence dT = 273K.

$m = M x n = 0.5 x 4 = 2 g$

$d U = n C_{V} d T = m c_{V} d T = 2 x 3 x 273 = 1638 J$