# Is the total entropy change of any isothermal process always zero?

I was asked to solve this following problem:

Two moles of a perfect diatomic gas at 300 K is expanded isothermally from an initial pressure of 3.00 atm to a final pressure of 1.00 atm, against a constant external pressure of 1.00 atm. Determine the values of $\Delta U, \Delta H, \Delta S,$ $\Delta S_\mathrm{surroundings}$ and $\Delta S_\mathrm{total}$.

First, I'm assuming $\Delta S$ means $\Delta S_\mathrm{system}$. My plan to solve this problem was to find $q_\mathrm{rev}$ for this system, but realized that I'm not exactly sure how total entropy is calculated for a non-reversible process. My question is:

We define total entropy as: $$\Delta S_\mathrm{total}=S_\mathrm{system}+S_\mathrm{surroundings}$$

I know that for an isothermal AND reversible expansion: $S_\mathrm{total}= 0$

But what about an isothermal expansion at a constant pressure? Does $S_\mathrm{total}$ also equal 0? If not, then would the equation look something like this: $$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$

where $q_\mathrm{surroundings}$ is the heat lost or gained by the surroundings by a process and $q_\mathrm{rev}$ is the heat lost or gained by the system IF that process was to become reversible.

• If $P$ and $T$ are constant, $V$ must remain constant if there has to be a change. (or you change $n$). Please clarify what you mean by isothermal expansion at constant pressure. Jun 14, 2017 at 9:06

Here are the steps:

1. Use the first law of thermodynamics to establish the final thermodynamic equilibrium state of your system for the irreversible process process you are considering.

2. Forget about the irreversible process path entirely.

3. Devise a reversible process path between the same pair of initial- and final thermodynamic equilibrium states that were obtained with the irreversible process (i.e., devise a path consisting of a continuous sequence of thermodynamic equilibrium states for the system). This reversible process path does not have to bear any resemblance whatsoever to the actual reversible path. There are an infinite number of reversible paths that will get you between the two states, so choose one that is easy to work with for step 4.

4. Calculate the integral of dQ/T for the reversible path you have focused on in step 3. This is the change in entropy for the system.

In my opinion the problem is ill-posed. The total entropy change depends on what happens to the work that is potentially provided by the expanding gas,

$$w = \int \mathrm{d}V \, p_{\mathrm{int}},$$

and how much of it is converted into heat. (Here I'm already assuming that the expansion is somehow slow enough so the gas remains in equilibrium, with a well-defined pressure and temperature. See also this answer about the roles of external and internal pressure in gas laws.)

A part of the work is surely transferred to the surrounding gas, namely $\Delta V \cdot p_{\mathrm{ext}}$. This part does not increase the total entropy, since it's reversible. You can get it back if you let the surrounding gas compress something. The question is where the remaining work $w_{\mathrm{rem}}$ goes,

$$w_{\mathrm{rem}} = \int \mathrm{d}V (p_{\mathrm{int}}-p_{\mathrm{ext}}).$$

Here are some scenarios:

1. Maybe there is strong friction between the cylinder and the piston, so it slowly glides into its final position while the gas expands. Then $w_{\mathrm{rem}}$ would be transformed into heat through the friction and the total entropy would increase by a corresponding amount $\Delta S_{\mathrm{total}} = w_{\mathrm{rem}}/T$. (Is that what you mean by the term $q_{\mathrm{surroundings}}/T$ in your last equation?) I guess this is probably the intended solution, in one way or another.

2. However, one may also let the gas do actual work, lift weights, compress springs, set a flywheel in rotation. Then no entropy would be created, $\Delta S_{\mathrm{total}} = 0$.

3. Or, if there is no friction and nothing else to receive the energy, it may just go into the kinetic energy of the piston. It would be accelerated outwards, overshoot its equilibrium position and start oscillating around it. In principle entropy is not increased. In reality, of course, there would be some friction, between the piston and the walls and in the gas itself, which would damp the oscillation. When in the end all the energy has been converted to heat, the total entropy change is again $\Delta S_{\mathrm{total}} = w_{\mathrm{rem}}/T$.