Work done in an isothermal irreversible process

Question

In calculating the work done in an isothermal irreversible process, why do we consider the external pressure only (using formula: $W_{\text{ext}} = P_{\text{ext}} \mathrm{d}V)$ but not the internal pressure? When the system expands, the internal pressure does the positive work, but even in this scenario, we use the equation as mentioned above. Why is this happening?

Answer 1

The ideal gas law (or any other equation of state) can only be applied to a gas at thermodynamic equilibrium. In an irreversible process, the gas is not at thermodynamic equilibrium, so the ideal gas law will not apply. The force per unit area exerted by the gas on the piston is comprised of two parts in an irreversible process: the local pressure and viscous stresses. The latter depend, not on the amount that the gas has been deformed, but on its rate of deformation. Of course, at thermodynamic equilibrium, the rate of deformation of the gas is zero, and the force per unit area reduces to the pressure. In this case the ideal gas law is recovered.

So, you are correct in saying that, for a reversible process, the internal pressure is equal to the external pressure. But, for an irreversible process, even though, by Newton's 3rd law, the force per unit area exerted by the gas on its surroundings is equal to the force per unit area exerted by the surroundings on the gas, the force per unit area exerted by the gas on its surroundings includes viscous stresses; thus, the force can cannot be determined by simply applying the ideal gas law. In this case, our only alternative (aside from solving the complicated gas dynamics equations for the gas, which include the viscous stresses) is the explicitly specify the external force per unit area imposed by the surroundings.

Answer 2

The key word here is irreversible. Irreversible processes are more like sudden jerks or a quick change in the system and therefore the ${P_{\text{ext}}}$ is not same as the ${P_{\text{int}}}$ throughout the process.

Why do we NOT use the internal pressure (${P_{\text{int}}}$) to find the work done?
It is very difficult to find the ${P_{\text{int}}}$ exerted by the gas in the container since ${P_{\text{int}}}$ is varying with time. For this, we would have to use integration. Something like this:$${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$$ (where $P(V)_{\text{int}}$ is a function of $V$)
But we know the value of ${P_{\text{ext}}}$ and this is constant throughout the process. This is why we use ${P_{\text{ext}}}$ instead of troubling ourselves with integration. Something like this: $$W_{ext}= P_{\text{ext}} \mathrm{d}V$$

So if a system undergoes a decrement in volume suddenly, it is the external pressure that does the positive work throughout this decrement. The equation would be: $$W_{\text{ext}}= P_{\text{ext}}\mathrm{d}V$$ (However we can use ${P_{int}}$ as well, please read the NOTE at the bottom)

And if a system undergoes an increment in volume suddenly, ${P_{\text{int}}}$ is doing the positive work. However, to find the $W_{int}$ we would have to use integration since ${P_{\text{int}}}$ is varying with time. It is much easier if we just find the negative work done by ${P_{\text{ext}}}$ and since we know that $${W_{\text{int}} = - W_{\text{ext}}}$$ we can find the work done. This is why we consider the ${P_{\text{ext}}}$ and according to the situation in the question, we apply the signs.

NOTE: Magnitude of work done by the internal pressure and external pressure is always same. We only use ${P_{\text{ext}}}$ because it is constant whereas ${P_{\text{int}}}$ varies with time. If we use integration to find ${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$, where $P(V)_{\text{int}}$ is a function of $V$, we will get the same magnitude as we would from $W_{\text{ext}}= P_{\text{ext}} \mathrm{d}V$. Only signs are opposite.

Answer 3

Expansion

Work is always done against the external pressure, because that's what opposes the expansion of the gas (and not internal pressure). It's just that in case of a reversible process, the internal pressure (or pressure of the gas in other words) is equal to the external pressure - so we can use it to find the work done. This is not the case for an irreversible process - the gas does not get time to make its pressure equal to the external pressure since the process is not being performed slowly. So, we need to consider the external pressure.

Compression

In this case, the opposing force is the internal pressure. But, since it's an irreversible process, I think we can't know the internal pressure. But, we know the temperature, so I'd use the FLOT to indirectly find the work done.

PS: I'm not sure about the compression part; it'd be nice if someone good at thermodynamics clarified this in the comments. :)