$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

William Xu Dis 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

In adiabatic reversible expansions, the amount of work done is greater than the amount of work done in an irreversible expansion as the gas is expanding against greater external pressures. Internal energy, in this case, is given by the equation ΔU = w, and because it is a state property, it should be the same in both reversible and irreversible expansion. How can this be possible when the work done by the two types of expansion is different and there is no heat transfer? Is this even possible, making the equation not applicable to irreversible processes? I know in the textbook isothermic reversible expansions were considered, but I don't recall temperature being mentioned for irreversible processes.

Hazem Nasef 1I
Posts: 51
Joined: Tue Oct 10, 2017 7:13 am

Re: Internal Energy Change in Adiabatic Reversible Expansion vs Adiabatic Irreversible Expansion

You are correct in that reversible expansions do more work than irreversible expansions. Although internal energy is a state property, both work and heat depend on the path taken, so they are not state properties. Work for a reversible expansion is nRT*ln(V2/V1) and is -P*deltaV for an irreversible expansion.

William Xu Dis 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

Re: Internal Energy Change in Adiabatic Reversible Expansion vs Adiabatic Irreversible Expansion

I know that work and heat are different, but what does that mean for the fact that ΔU = w in adiabatic processes. Because internal energy is a state property, it should be the same for both irreversible and reversible expansions, at least that is what the textbook says. So how can work be different for both but the equation ΔU = w still be true?