- #1

zenterix

- 692

- 83

- Homework Statement
- I've been studying thermodynamics and though I am studying topics like chemical potential, I still cannot be satisfied with my understanding of certain concepts from the 1st and 2nd laws.

- Relevant Equations
- One concept is entropy. Another, which I want to tackle in this post, is the first law and the in particular the concepts of reversible and irreversible processes.

Consider a hydrostatic system in the form of an ideal gas in a container with a movable piston.

First let's consider an irreversible isothermal expansion from state a to state b as depicted below

We are given ##P_1, V_1, T_a##, and ##P_2##.

We can easily compute ##V_2## from the ideal gas law

$$V_2=\frac{nRT_a}{P_2}\tag{1}$$

Then

$$w_{ab,isoth}=-P_2(V_2-V_1)<0\tag{2}$$

$$q_{ab,isoth}=-w_{ab,isoth}>0\tag{3}$$

The area of the shaded region in the picture above is the work and the negative of the heat exchanged with an external reservoir.

$$\Delta U_{ab,isoth}=0\tag{4}$$

For the surroundings, we have similar expressions but with opposite signs.

Now, consider the task of recompressing the gas to state a.

If we compress reversibly, then we have

$$w_{ba, rev}=-nRT\int_{V_2}^{V_1}\frac{dV}{V}=nRT\ln{\frac{V_2}{V_1}}>|w_{ab,isoth}|>0\tag{5}$$

$$q_{ba,rev}=-w_{ba,rev}\tag{6}$$

Consider the following two alternative ways of recompressing

I won't do the precise calculations here, but in the both cases above the shaded area represents both total heat outflow from the system and work done on the system. That is, in both cases, qualitatively we have ##w_{ext}>0## and ##q<0## and the sum is zero (no change in internal energy).

My question is: what is a didactical way to explain what it means that we cannot reverse the initial irreversible expansion?

The question above is posed naively on purpose. Of course we can return to the initial state (I just did it in three different ways above). I am not using the word "reverse" correctly I think.

If I knew how to pose the question conceptually correctly, I would probably not need to ask it in the first place. Not being cheeky here, I legitimately don't know how to pose the question.

Here is another attempt: in terms of the ability to go back to the initial state, what makes the reversible compression different from the irreversible processes shown above?

First let's consider an irreversible isothermal expansion from state a to state b as depicted below

We are given ##P_1, V_1, T_a##, and ##P_2##.

We can easily compute ##V_2## from the ideal gas law

$$V_2=\frac{nRT_a}{P_2}\tag{1}$$

Then

$$w_{ab,isoth}=-P_2(V_2-V_1)<0\tag{2}$$

$$q_{ab,isoth}=-w_{ab,isoth}>0\tag{3}$$

The area of the shaded region in the picture above is the work and the negative of the heat exchanged with an external reservoir.

$$\Delta U_{ab,isoth}=0\tag{4}$$

For the surroundings, we have similar expressions but with opposite signs.

Now, consider the task of recompressing the gas to state a.

If we compress reversibly, then we have

$$w_{ba, rev}=-nRT\int_{V_2}^{V_1}\frac{dV}{V}=nRT\ln{\frac{V_2}{V_1}}>|w_{ab,isoth}|>0\tag{5}$$

$$q_{ba,rev}=-w_{ba,rev}\tag{6}$$

Consider the following two alternative ways of recompressing

I won't do the precise calculations here, but in the both cases above the shaded area represents both total heat outflow from the system and work done on the system. That is, in both cases, qualitatively we have ##w_{ext}>0## and ##q<0## and the sum is zero (no change in internal energy).

My question is: what is a didactical way to explain what it means that we cannot reverse the initial irreversible expansion?

The question above is posed naively on purpose. Of course we can return to the initial state (I just did it in three different ways above). I am not using the word "reverse" correctly I think.

If I knew how to pose the question conceptually correctly, I would probably not need to ask it in the first place. Not being cheeky here, I legitimately don't know how to pose the question.

Here is another attempt: in terms of the ability to go back to the initial state, what makes the reversible compression different from the irreversible processes shown above?

Last edited: