# Entropy change in an adiabatic expansion

Question
One mole of ideal gas initially at a pressure of 1 atmosphere and $$T = 298\ \mathrm K$$, is expanded into a volume 50 % larger adiabatically.

In this adiabatic process, no work is done on the surroundings and $$\mathrm dQ=0$$. And entropy is defined as $$\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$$. However, there must be a change in entropy surely?

Is the definition $$\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$$ only at a constant volume? Because I have also seen entropy defined as $$\mathrm dS=\frac{C_V}{T}\,\mathrm dT+\frac{R}{V}\,\mathrm dV$$ and there would be a change in both temperature and volume. However, how do I find what these changes are? Would it make a difference if the expansion was reversible or irreversible (the question didn't make it explicitly clear)

• I am sorry to be posting this exact same comment in every single question about entropy but I believe that proper notation is important. The definition of entropy is $\mathrm{d}S = \mathrm{d}q_{\text{rev}}/T$ and not $\mathrm{d}S = \mathrm{d}q_{\text{rev}}/\mathrm{d}T$. The latter is a ratio of two infinitesimals (i.e. it is a partial derivative) and is not an infinitesimal; in fact it is the definition of heat capacity (whether it is the heat capacity at constant pressure or at constant volume depends on the process for which the quantity $\mathrm{d}q$ is measured). – orthocresol Jul 20 '15 at 8:50
• Is it possible to look at the problem from the other definition of entryopy $S=klnW$? Clearly the number of possible microstates accessible to the gas is changing so S has to change. – Mecury-197 Oct 18 '15 at 0:15

"One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically."

The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.

### If reversible:

If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic). Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant. In this case you can apply both of your equations. But keep in mind that your equation for $$dS$$ is only true for ideal gases.

$$dS = \frac{\delta q_{rev}}{dT} = 0$$

$$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$$

$$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$$

This is now a differential equation you can solve for temperature if you know volume.

As the other answers have indicated, a reversible expansion will involve the gas doing work on the surroundings. One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings. A loss in the potential to do work (of the universe) is an increase in entropy. And in a reversible process that can't happen.

Think of gas in a piston, where the piston is held down by a pile of sand. The reversible expansion involves slowly removing sand, grain by grain, from the piston. As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.

### If irreversible:

...then we still don't have enough information because there are many ways the process can be irreversible. But one particular, extreme case of irreversibility is if the adiabatic expansion occurs without transfer of any energy, then we can solve the problem. In this extreme scenario, neither heat nor work are not transferred. You can thus infer that $$dU=0$$ for the gas, so

$$dU=-TdS + p dV=0$$

$$TdS = p dV$$

Using your definition for $$dS$$ for an ideal gas,

$$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$$

Keep in mind that $$\frac{RT}{V}$$ for an ideal gas is $$p$$, so you should get that

$$C_V dT = 0$$

Which is also a differential equation that you can integrate to get the temperature change. (The obvious solution is that the temperature does not change in this case.)

In this second case, now that we know that $$T$$ is constant, we know $$dT$$ is zero, and thus using your equation for $$dS$$, we get that $$dS = \frac{R}{V}dV > 0$$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible. The potential to do work has been lost.

Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum). If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.

In calculating the change in entropy of a closed system that has undergone an irreversible process, you need to first focus exclusively on the initial and final equilibrium states. You then need to identify a reversible process path between the exact same pair of equilibrium states. The reversible process path does not necessarily have to bear any resemblance to the actual irreversible process path. Therefore, even if the irreversible process path is adiabatic, the reversible path does not have to be. You then calculate the integral of dq/T for the reversible process path. This is equal to the entropy change.

Pretty simple, huh. I don't know why it isn't taught properly to students.

• This doesn't seem to answer the question. – Curt F. Jul 20 '15 at 5:43
• Ok, don't exactly answer the op's question but is enough to solve the problem; +1. – user5764 Jul 20 '15 at 16:33
• Regarding the original question of why the equation for dS in terms of dT and dV for an ideal gas gives the correct answer, it is because the equation applies to the changes in dS, dT, and dV over a continuous sequence of thermodynamic equilibrium states. This is the same thing as a reversible path. So, all one needs to do is to integrate this equation from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state determined obtained with the irreversible process. Note that the two terms on the right hand side of the equation are exact differentials. – Chet Miller Jul 21 '15 at 1:53

An important thing to know is that on the adiabatic processes, the system is isolated, due to that no heat can be given or taken, but that doesn't mean that the volume can't change. Because we can't take heat, we can lower/rise the temperature from T1 to T2 . We don't have any work done only when the volume is constant. Entropy will be at a constant volume or pressure only when you see the index v ** or **p in the formula. If the process would be reversible ( which i think it is) only the change of temperature should be in infinitezimal amounts. Hope I understood the question right

• So does entropy change in this situation? – RobChem May 8 '15 at 13:26
• Yes it does, everytime there is work done, entropy will also change. – Ndrina Limani May 8 '15 at 13:27
• But there is no work done? – RobChem May 8 '15 at 13:29
• The gass expand 50% more, of course there is work done, but probably not in the sorroundings ! – Ndrina Limani May 8 '15 at 13:31
• If it is adiabatic, surely it's expanding into a vacuum. Therefore, no work is done. Or is this not the case? – RobChem May 8 '15 at 13:32