Two mole of ideal diatomic gas (Cv,m=5/2R) at 300K and 5atm expanded irreversibly and adiabatically to a final pressure of 2atm against a constant pressure of 1 atm. Calculate change in internal energy △U
A
864.28J
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B
−1052.1J
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C
−1247.1J
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D
−2164.1J
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Solution
The correct option is C−1247.1J For Adiabatic irreversible process nCV(T2−T1)=−pextnR[T2P2−T1P1] .....eq (1) Where, Cv=52R T1=300K Pext=1atm,P2=2atm P1=5atm n=2 so, 2×52R(T2−300)=−2×R[T22−3005]
5(T2−300)=−2[T22−3005]
Solving we get, T2=270K Now, From first law, △U=q+w Since q =0, △w=△U=nCv△T=2×52×8.314(270−300)=−1247.1J