An ideal gas is expand from `(p_(1),V_(1),T_(1))` to `(p_(2),V_(2),T_(2))` under different conditions. The correct statement(s) among the following is (are):

Options

A. The work done by the gas is less when it is expanded reversibly from `V_(1)` to `V_(2)`under adiabatic conditions as compared to that when expanded reversibley from `V_(1)`to `V_(2)` under isothermal conditions.
B. The change in internal energy of the gas is (i) zero, if it isexpanded reversibly with `T_(1) =T_(2)`, and (ii) positive, if it is expanded reversibly under adaibatic conditions with `T_(1) cancel(=) T_(2)`
C. If the expansion is carried out freely , it is simulataneously both isothermal as well as adiabatic
D. The work done on the gas is maximum irreveribly from `(p_(2),V_(2))` to `(p_(1),V_(1))` against constant pressure`p_(1)`.

Answer

a,c,d

Explanation

(a)
Area under the curve of reversible isothermal expansion is more than that under the adiabatic curve. Hence, work done in reversible isothermal expansion is more.
(b) For reversible expansion with `T_(1) = T_(2)` means isothermal expansion.For reversible isothermal expansion of a gas, `DeltaU = 0 (:' DeltaU = n C_(v) DeltaT)` In reversible adiabatic expansion, `T_(2) lt T_91)`
i.e., `DeltaT = -ve, :. DeltaU = n C_(v) DeltaT = -ve ` ( and not `+ve)`
(c ) In free expansion , `P_(ext) = 0 :. w=0`
If process is carried out isothermally , `DeltaU =0`. Hence, by 1st law , `DeltaU= q + w` or `q= DeltaU -w=0-0=0` ( i.e. adiabatic).
If process is carried out adiabatically , `q=0`.
As `w=0`, therefore,by 1st law, `DeltaU = q+w =0` ( isothermal).
(d) During irrevesible compression,maximum work is done on the gas, corresponding to the shaded area.