# Why is P_ext different for reversible and irreversible isothermal expansion?

In summary, for a reversible isothermal expansion, the external pressure (Pext) is taken as a variable, while for an irreversible isothermal expansion, Pext is taken as a constant. This is because in a reversible process, the system is at equilibrium at every stage, while in an irreversible process, the system is not at equilibrium and viscous stresses can contribute to Pext. The work done on the surroundings is equal to the integral of Pext dV for both reversible and irreversible processes, but Pext cannot be calculated using the ideal gas law for an irreversible process.

I was able to derive the work done in a reversible isothermal expansion. There as the P changes to P-dP, the volume increases by dV and hence the internal pressure also decreases by dP and equilibrium is maintained. (Thanks to @Nugatory and @Chestermiller fo r explaining it).
Now when we take an irreversible isothermal expansion in the book the ##P_{ext}## is taken constant and then dV is integrated to get
$$W = -P_{ext}(V_2-V_1)$$
But for reversible isothermal expansion, the ##P_ext## is taken as a variable. That is it is taken as ##\frac{nRT}{V}## and then integrated to get
$$W = RTln(\frac{V_2}{V_1})$$
why is ##P_{ext}## different for the two cases?

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.
If ## p_{int}>p_{ext} ## in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process.

On the other hand if ## p_{int}=p_{ext} ## then the process is reversible.
In that case ## p_{int}=p_{ext}=p ##
You have written the other way round.
That that in reversible process the p is variable whereas in irreversible constant.
Can you check on that?

The equation you wrote for an irreversible expansion assumes a constant force per unit area Pext applied to the gas, and also, there should be a plus sign rather than a minus sign.

In an irreversible expansion, viscous stresses can contribute to Pext and, in addition, the pressure and/or the temperature within the gas are not uniform spatially, so we can't use the ideal gas law to determine the force per unit area exerted by the gas on the boundary (Pext). So the ideal gas law cannot be used to calculate the work done on the surroundings.

For both reversible and irreversible processes, the work done on the surroundings is equal to the integral of Pext dV. For an irreversible process, Pext is not equal to the pressure calculated from the ideal gas law P, but for a reversible process, the deformation is slow (such that viscous stresses are negligible), and also Pext = P = nRT/V. For an irreversible process, we can control Pext as a function of time t during the deformation to be whatever we wish.

Chet

Raghav Gupta said:
A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.
If ## p_{int}>p_{ext} ## in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process.

On the other hand if ## p_{int}=p_{ext} ## then the process is reversible.
In that case ## p_{int}=p_{ext}=p ##
You have written the other way round.
That that in reversible process the p is variable whereas in irreversible constant.
Can you check on that?
The expressions are correct

@Raghav Gupta ,take a look at @Chestermiller 's reply. The first paragraph solved my doubt. Second paragraph explains the reason why you can't take P=nRT/V for irreversible process.

Raghav Gupta
@Raghav Gupta ,take a look at @Chestermiller 's reply. The first paragraph solved my doubt. Second paragraph explains the reason why you can't take P=nRT/V for irreversible process.
I was also a bit confused.Maybe I have read a wrong article or not clearly understood.Really these mentor guys explain in a nice detailed way.

Chestermiller

## 1. What is irreversible expansion?

Irreversible expansion is a thermodynamic process in which a system expands and does work on its surroundings, but cannot be returned to its original state without the input of external energy.

## 2. What causes irreversible expansion?

Irreversible expansion is caused by an irreversible change in the internal structure or properties of a system due to external factors, such as changes in pressure or temperature.

## 3. What is an example of irreversible expansion?

An example of irreversible expansion is the expansion of a gas in a closed container due to an increase in temperature. The gas will expand and do work on the surroundings, but it cannot be returned to its original state without removing heat from the system.

## 4. How is irreversible expansion different from reversible expansion?

Irreversible expansion differs from reversible expansion in that reversible expansion is a process in which a system can be returned to its original state without any loss of energy, while irreversible expansion results in a loss of energy due to the irreversible changes in the system.

## 5. What are the implications of irreversible expansion in thermodynamics?

The implications of irreversible expansion in thermodynamics include the idea of entropy, which is a measure of the disorder or randomness of a system. Irreversible expansion increases the entropy of a system, and is a fundamental principle in the second law of thermodynamics.