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Q.

 12 mole of helium gas is contained in a container at S.T.P. The heat energy needed in joule to double the pressure of the gas, keeping the volume constant, is ...(R=8.31J/mole-K)

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answer is 1702.

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Detailed Solution

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At STP, T=273 K

at constant volume, 

T2T1=P2P1=2

⇒T2-T1T1=2-1=1

⇒∆T=T1=273 K

∆Qv=Cv n ΔT=12nfR∆T

⇒∆Qv=12×12×3×8.31×273=1701.5 J

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