An ideal gas is expanded from $V_{1} to V_{2}$ through three different process
a) Reversible adiabatic b) Reversible isothermal
c) Irreversible adiabatic (against a constant external pressure $P_{ext}$ )
the correct option is

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$[{(T_{f})}_{g a s}] Re v .isothermal > [{(T_{f})}_{g}] Irre.adiabatic > [{(T_{f})}_{g}] Rev.adiabatic$

$W_{reversibleisothemal} > W_{Ir.ckiabatic} > W_{rev.atiabctic}$

$[{(T_{f})}_{g a s}] Rev.isothermal > [{(T_{f})}_{g}] Rev.adiabatic > [{(T_{f})}_{g}] Irrev.adiabatic$

$(P)_{f Rev.Isothermal} > P_{f Rev.adiabatic} > P_{fIrreversibleadiabatic}$

answer is B.

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Detailed Solution

$W_{isothermal} > W_{Rev.adiabatic} > W_{Ir.adiabatic}$

In adiabatic expansion, temperature decreases.

Greater the work done by the gas, greater will be the decrease in temperature. Work done is greater when the process is carried out reversibly. During reversible adiabatic expansion, the decrease in the temperature is more than that when the process is carried out irreversibly.

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