Two mole of ideal diatomic gas (Cv,m=5/2R) at 300K and 10atm expanded irreversibly and adiabatically to a final pressure of 2atm against a constant pressure of 1 atm. Calculate change in internal energy △U
A
−2164.1J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
864.28J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
−1052.1J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
−1662.8J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D−1662.8J For Adiabatic irreversible process nCV(T2−T1)=−pextnR[T2P2−T1P1] .....eq (1)
Where, Cv=52R T1=300K Pext=1atm,P2=2atm P1=10atm n=2
so, 2×52R(T2−300)=−2×R[T22−30010]
5(T2−300)=−2[T22−30010]
Solving we get, T2=260K
Now, From first law, △U=q+w
Since q =0, △w=△U=nCv△T=2×52×8.314×(260−300)=−1662.8J