physical chemistry - 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)

Answer

"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.