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Question

# One mole of an ideal gas âŸ®Cv,m=52RâŸ¯ at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is:

A
270 K
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B
273 K
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C
248.5 K
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D
200 K
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Solution

## The correct option is C 248.5 KSolution:- (C) 248.5KGiven that:-Pext.=2atmP1=5atmP2=2atmT1=300KT2=T(say)=?n=1CV=52RFor an adiabatic process,W=nCVdT=−PΔV∴nCV(T2−T1)=−Pext(V2−V1)⇒1×52R(T−300)=−2(nRT2P2−nRT1P1)⇒52R(T−300)=−2R(T2−3005)⇒5T−1500=−2T+240⇒7T=1740⇒T=248.57K∴T2=T=248.57K

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