# Work done by irreversible and reversible processes

Hamiltonian
I am not able to follow the derivation of work done in a reversible and irreversible process as I don't get why the work done should be different in the two processes.
a reversible process is said to be a process that occurs infinitesimally slowly and an irreversible process goes from initial to the final state in a single step in finite time and cannot be reversed.

I don't understand why an irreversible process can't be reversed if you consider the example of a cylinder containing a gas onto which a piston is placed if the piston is pushed down and taken from a height I to a height F infinitesimally slowly it is said to be reversible but if it is taken in one step it is said to be irreversible.
why can't the process be reversed if the piston is taken down in a single step?

either way shouldn't the work done be the same?

my book says work done by an irreversible process is $$P\Delta V$$ and work done by an isothermal reversible process is $$-2.303nRT\log(Pi/Pf)$$

For the isothermal reversible process, maybe you have a bunch of tiny weights on top of the piston, which you remove one by one to facilitate the reversible expansion of the gas inside. The internal and external forces per unit area on the piston face are always equal and the work done by the gas is$$W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \ln\frac{V_2}{V_1} = nRT \ln \frac{P_1}{P_2} \quad \left( \approx 2.303nRT\log{\frac{P_1}{P_2}}\right)$$The pressures in the above could be the internal or external pressures to the piston, since they're equal at all times. For the irreversible expansion, however, the force per unit area at the inside piston face is not given by the ideal gas law (since we need to take into account other factors, like viscous stresses in the gas), so instead we deal with the external pressure. If you compress the gas, hold the piston in place and suddenly release the piston, the atmospheric pressure on the external face is considered constant and the work done by the gas during the irreversible expansion, between two equilibrium states, is just$$W = \int_{V_1}^{V_2} P_{\text{ext}} dV = P_{\text{ext}} \Delta V$$(N.B. The force on the inside face is some function ##\vec{F}(t)##, but if you apply the work-energy theorem to the piston between the equilibrium states you'll notice that ##\int \vec{F} \cdot d\vec{r} - P_{\text{ext}} \Delta V = 0##, i.e. the work done by the gas inside is ##P_{\text{ext}} \Delta V##).

It might be easier to visualise if you sketch the two paths on the PV plane, and look at the difference in areas between the two paths

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vanhees71
Mentor
(N.B. The force on the inside face is some function ##\vec{F}(t)##, but if you apply the work-energy theorem to the piston between the equilibrium states you'll notice that ##\int \vec{F} \cdot d\vec{r} - P_{\text{ext}} \Delta V = 0##, i.e. the work done by the gas inside is ##P_{\text{ext}} \Delta V##).
For a massless-frictionless piston, the force F the gas exerts on the inside piston face does not only integrate in this way. At all times during the irreversible process, ##F=P_{ext}A##. The viscous stresses and rate of piston movement combine in such a way that this happens.

vanhees71 and etotheipi
For a massless-frictionless piston, the force F the gas exerts on the inside piston face does not only integrate in this way. At all times during the irreversible process, ##F=P_{ext}A##. The viscous stresses and rate of piston movement combine in such a way that this happens.

For the massless piston that has to be true, since the net force on a massless piston must be zero otherwise the acceleration would diverge. But for the massive piston, there must be an imbalance in force per unit area on either side in order to accelerate and then decelerate the piston, so the force per unit area on the inside face would need to vary with time. Is that right?

Mentor
For the massless piston that has to be true, since the net force on a massless piston must be zero otherwise the acceleration would diverge. But for the massive piston, there must be an imbalance in force per unit area on either side in order to accelerate and then decelerate the piston, so the force per unit area on the inside face would need to vary with time. Is that right?
Yes. But, if you specify that your system is the gas, the ##P_{ext}## is the pressure on the inside face of the piston. And, if you specify that your system is the gas plus piston, the ##P_{ext}## is the pressure on the outside face of the piston, but the kinetic energy change of the piston needs to be included. Of course, by the time you reach final steady state, the viscous stresses will have damped the piston motion, and your equation will apply, but with the upper limit of integration being infinite time.

vanhees71 and etotheipi
Thanks, yes I completely agree with all of that. You said it much better than I did