Question:

An ideal gas is expanded from (p1,V1,T1) to (p2,V2,T2) under different conditions. The correct statement(s) among the following is(are):
The work done on the gas is maximum when it is compressed irreversibly from (p2,V2) to (p1,V1) against constant pressure p1.
The work done by the gas is less when it is expanded reversibly from V1 to V2 under adiabatic conditions as compared to that when expanded reversibly from V1 to V2 under isothermal conditions.
If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic.
The change in internal energy of the gas is (i) zero, if it is expanded reversibly with T1=T2, and (ii) positive, if it is expanded reversibly under adiabatic conditions with T1≠T2.

The work done by the gas is less when it is expanded reversibly fromV1toV2under adiabatic conditions as compared to that when expanded reversibly fromV1toV2under isothermal conditions.

The change in internal energy of the gas is (i) zero, if it is expanded reversibly withT1=T2, and (ii) positive, if it is expanded reversibly under adiabatic conditions withT1≠T2.

The work done on the gas is maximum when it is compressed irreversibly from(p2,V2)to(p1,V1)against constant pressurep1.

If the expansion is carried out freely, it is simultaneously both is isothermal as well as adiabatic.

Solution:

(A) ω=−PextΔV
Work done on gas when compressed ω is positive ∴ equation becomes ω=PextΔV
(B) under adiabatic conditions ω=(P2V2−P1V1)(γ−1)/γ
since equation V2>V1 ∴ ω=+ve (work done on system)
for reversibly condition (Isothermal) ω=−nRTlnV2/V1=V2>V1 ω=−ve (work done on system)
ω=−ve (for adiabatic reversibly work done by gas)
ω=+ve (for isothermal reversibly work done by gas)
(C) As free expansion in there (irreversible) ∴Pext=0
ω=(−Pext×ΔV)=0 (Isothermal irreversible process)
In adiabatic ΔU=−ve
ΔU=q+ω
−ve=0+ω
ω=−ve
similarly, ΔG=ΔH−TΔS
gas expand ∴ΔS=+ve
ΔG=ΔH−Ve=ΔV+pΔV−ve=q+w+pΔc−ve=o+w+w−ve=−ve+(−ve)−ve
ΔG=−ve ∴It is spontaneous process ∴both take place simultaneously