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Question

# Two mole of ideal diatomic gas (Cv,m=5/2 R) at 300 K and 5 atm expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate change in internal energy △U

A
864.28J
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B
1052.1 J
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C
1247.1 J
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D
2164.1 J
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Solution

## The correct option is C −1247.1 JFor Adiabatic irreversible process nCV(T2−T1)=−pextnR[T2P2−T1P1] .....eq (1) Where, Cv=52R T1=300K Pext=1 atm, P2=2atm P1=5 atm n=2 so, 2×52R(T2−300)=−2×R[T22−3005] 5(T2−300)=−2[T22−3005] Solving we get, T2=270 K Now, From first law, △U=q+w Since q =0, △w=△U=nCv△T=2×52×8.314(270−300) =−1247.1 J

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