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Q.

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 = 3Jg-1 K-1) is

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answer is 1638.

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Detailed Solution

Here, n=12 ,Heat capacity=Cv=3Jg−1K−1,M=4gmol−1∴Cv=Mcv=4×3=12Jmol−1K−1 At constant volume P∝T∴P2P1=T2T1=2⇒T2=2T1 Rise in temperature ΔT=T2−T1=2T1−T1=T1=273K Heat required, ΔQ=nCvΔT=12×12×273=1638J
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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 = 3Jg-1 K-1) is