Entropy variation during free expansion

[becko]

Homework Statement

A system that consists of $$n$$ moles of ideal gas does a free expansion (to the vacuum) from a volume $$V$$ to a volume $$2V$$. a) What is the variation of entropy of the gas?; b) of the universe?; c) if the expansion was reversible and isothermal, what would be the variation of the entropy of the gas?; d) of the universe?

Homework Equations

Ideal Gas Law: $$PV=nRT$$
Conservation of Energy: $$\texttt{d}U=T\texttt{d}S-p\texttt{d}V$$

The Attempt at a Solution

a)
not sure how to do this. I think it should be zero.

b)
I know that the entropy is additive, so that the change of entropy in the universe is just the change of entropy in the system, plus the change outside the system. Supposedly there is no change of entropy outside the system, so that the answer to this question should be the same as a).

c)
Since the temperature doesn't change and this is an ideal gas, $$\texttt{d}U=0$$. Therefore
$$\texttt{d}S=\frac{p\texttt{d}V}{T}$$
Using the ideal gas law:
$$\texttt{d}S=\frac{nR\texttt{d}V}{V}$$.
Integrating gives the result:
$$\Delta S=nR\texttt{ln}2$$

d)
The same as in c).


Please, help me with a) and b). I believe the answer should be zero, but then I think that I can travel from the initial state to the final state through the reversible and isothermal line, just as in c) and d), in which case the answer to a) and b) should be the same as in c) and d) (since entropy is a function of state). Can someone clear this up for me?

 

[anathema]

Hello!

For part a), you are correct that the change in entropy for the gas itself is zero. This is because the process is a free expansion to vacuum, meaning that there is no external pressure acting on the gas and thus no work done. Since the change in internal energy is also zero (no change in temperature or number of moles), the change in entropy for the gas is zero according to the equation \Delta S = \frac{Q}{T}.

For part b), you are also correct that the change in entropy for the universe is also zero. This is because the process is adiabatic (no heat transfer) and reversible (no increase in entropy due to irreversibilities). Therefore, the entropy of the universe remains constant.

For part c), you are correct in your approach. Since the process is isothermal and reversible, the change in internal energy is zero and the change in entropy is given by \Delta S = \frac{Q_{rev}}{T}. Since the process is reversible, the heat transfer is equal to the work done, which can be calculated using the ideal gas law. Therefore, the change in entropy for the gas is \Delta S = nRln2.

For part d), the change in entropy for the universe is also \Delta S = nRln2. This is because the process is isothermal and reversible, so the change in entropy for the gas is the same as the change in entropy for the universe.

I hope this helps clarify things for you. Keep up the good work!

 

[nlightner]

I would like to clarify that the variation of entropy during free expansion is not zero. In fact, it is an irreversible process and the entropy of the gas will increase during this expansion. This is because the gas is expanding into a larger volume, leading to a decrease in its density. As a result, there is an increase in the number of accessible microstates for the gas molecules, leading to an increase in entropy. The exact value of the entropy change can be calculated using the formula \Delta S = nR\texttt{ln}(V_f/V_i), where V_f and V_i are the final and initial volumes respectively.

As for the variation of entropy in the universe, it is important to note that the concept of entropy applies to the entire universe, not just to the gas system. While the entropy of the gas increases during free expansion, the entropy of the surroundings (or the vacuum in this case) remains constant. This means that the overall entropy change in the universe is positive, as the gas system has contributed to an increase in entropy. This is in line with the second law of thermodynamics, which states that the total entropy of a closed system can never decrease.

In the case of a reversible and isothermal expansion, the entropy of the gas will remain constant as the temperature and pressure are constant. However, there will still be an increase in the entropy of the universe, as the gas is expanding into a larger volume and contributing to an increase in the number of accessible microstates. Therefore, the answers to c) and d) should be the same as the answers to a) and b), respectively.

I hope this clarifies the concept of entropy variation during free expansion for you. If you have any further questions, please feel free to ask.