1
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
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
Open in App
Solution
The correct option is C −1247.1 J
For 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
For 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
Suggest Corrections
0
Similar questions
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
