An ideal gas is expanded from p 1 , V 1 , T 1 to ( p 2 , V 2 , T 2 ) under different conditions. The incorrect statement among the following is?

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TestHub · Accepted Answer

Area under curve in reversible isothermal is more. So, more work will be done by gas.

$T_{1} = T_{2}$

$Δ U = n C_{V} Δ T = 0$

In reversible adiabatic expansion, $T_{2} < T_{1}$

So $Δ T = - v e$ , $Δ U = - v e$

In free expansion, $P_{e x t} = 0$

So $W = 0$

If carried out isothermally $(Δ U = 0)$

$q = 0$ (adiabatic); from I law

If carried out adiabatically $(q = 0)$

$Δ U = 0$ (isothermal); From I law

During irreversible compression, maximum work is done on the gas (corresponding to shaded area).

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