1.

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

Answer»

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

Solution :(a)
Area under the curve of reversible isothermal expansion is more thanthat 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.


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