Why is work done on weight only -mg×h in irreversible expansion?

[zanick]

No. In an irreversible process, entropy change can be due to both mechanisms, while for a reversible process, it is only due to entropy transfer by heat across the border.
In the free expansion you described, if the container is insulated, there is no heat flow across the border, and all the entropy generation takes place within the gas. In the alternate reversible process, we expand the gas against a restraining force, and heat transfer across the border is now allowed (in order for us to get to the same final state). The work done by the gas W and the heat transferred to the gas Q in the reversible process are equal, and the entropy change is $\Delta S=\frac{Q}{T}$. The initial and final states of the gas are the same in both processes, so the entropy changes are the same also. But, for the irreversible path, $\frac{Q}{T}=0$, The difference is that, for the irreversible path, all the entropy change has been generated within the system (for this particular process) and, for the reversible path, all the entropy has been transferred to the system. For any arbitrary process on a closed system (whether reversible or irreversible), the change in entropy entropy change is given by $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$where $T_B$ is the temperature at the border with the surroundings (through which the heat flows) and $\sigma$ is the entropy generated during the process due to irreversibilities.

If the entropy is taking place in the gas, in the irrev example, how it is easily calculated? if you can't use S=dQ/T, can you use a Boltzmann equation instead? S=k.logW ?

 

[Chestermiller]

that was one of my questions to you. if no friction, I am leaning toward it never coming to rest, but i really don't know. It should act like a spring mass system. i suppose that the entropy calculation would have to assume it comes to rest at some point.

It comes to rest as a result of viscous damping (dissipation) by the gas. This is an irreversible effect, and is a major contributor to entropy generation. In a reversible process, it is negligible.

 

[zanick]

that makes sense..."viscous damping" and so does it contribution to entropy of both a reversible and Irrev process.

 

[Chestermiller]

If the entropy is taking place in the gas, in the irrev example, how it is easily calculated? if you can't use S=dQ/T, can you use a Boltzmann equation instead? S=k.logW ?

To calculate the amount of entropy generation, you need to first calculate the overall entropy change from the reversible path. Then you subtract $$\left[\int{\frac{dq}{T}}\right]_{at\ boundary}$$ for the irreversible path. This indirectly determines for you the amount of entropy that was generated.

To get the entropy generation directly, you would need to solve the partial differential equations for fluid flow and heat transfer within the system to determine detailed temperatures and velocities of the gas as a function of time and spatial position throughout the irreversible change. This is a very complicated calculation.