Paradox in isentropic expansion? (statistical mechanics)

In summary, the conversation discusses a scenario where a liquid is mixed with a substance with a higher boiling point, resulting in equilibrium between the liquid and its vapor. The vapor is then expanded adiabatically, leading to a decrease in the concentration of the substance and an increase in the entropy of the water. However, when the vapor is compressed again, the substance will dissolve and the entropy of the system will return to its original state. The question arises about what happens to the entropy when the substance is removed from the container, and it is determined that this would require work on the system and potentially violate the second law of thermodynamics.
  • #1
Curl
758
0
If I have a container containing a liquid mixed with some other substance that has much a higher boiling point (i.e. water and salt). This liquid will be in equilibrium with its vapor (the salt vapor pressure is negligible).

Now I quasi-statically adiabatically expand this vapor. Isentropic process right?

But after some time, I've essentially distilled some salt. I can cover up the salt rocks, then begin to compress the vapor, again slowly. This compression is Isentropic right?

What's going on, what happened to the entropy of this system?

Sorry if this is a bad question with a trivial answer, it is late and I'm tired. Thanks for the help.
 
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  • #2
The entropy stays constant. Where is the problem?
 
  • #3
If you separate the salt out of the water the entropy decreases.

It is called "entropy of mixing" http://en.wikipedia.org/wiki/Entropy_of_mixing

If my process did not add entropy to the gas, yet decreased the entropy of mixing by decreasing the concentration of salt in water, then there is a problem.

The question here is to explain how the system has constant entropy before and after the process. The entropy decreased by distilling the salt must be made up somewhere else somehow. I can't figure out where, which is why I ask this.
 
  • #4
I think I understand your question now, and it's a good one. Basically, you seem to be asking something similar to, "what happens when water evaporates, leaving behind salt crystals?"

The entropy of the salt has decreased by phase separating, it's true. But the entropy of the water increased- the (vapor) water molecules have more available volume, because the density is lower than the liquid phase. So, the total entropy change can be zero (or increase) as the water evaporates away, leaving behind the salt.

Now, as you compress the vapor phase, driving water back into the liquid phase,the salt will once again dissolve, returning your system to the original configuration and completing a (possibly) reversible cycle.

But by 'covering up' or otherwise removing the salt from the container, you have changed the system- among other things you changed the available volume for the water, changing the entropy of the water phase. But that can only happen by doing work on the system to partition off the salt. I suspect a detailed calculation would show that that work is greater than or equal to the decrease in entropy in order to satisfy the second law, but I haven't found an analysis online... I didn't look too hard, tho.
 
  • #5
Andy Resnick said:
I think I understand your question now, and it's a good one. Basically, you seem to be asking something similar to, "what happens when water evaporates, leaving behind salt crystals?"

The entropy of the salt has decreased by phase separating, it's true. But the entropy of the water increased- the (vapor) water molecules have more available volume, because the density is lower than the liquid phase. So, the total entropy change can be zero (or increase) as the water evaporates away, leaving behind the salt.

Now, as you compress the vapor phase, driving water back into the liquid phase,the salt will once again dissolve, returning your system to the original configuration and completing a (possibly) reversible cycle.

But by 'covering up' or otherwise removing the salt from the container, you have changed the system- among other things you changed the available volume for the water, changing the entropy of the water phase. But that can only happen by doing work on the system to partition off the salt. I suspect a detailed calculation would show that that work is greater than or equal to the decrease in entropy in order to satisfy the second law, but I haven't found an analysis online... I didn't look too hard, tho.

That'd be true if you are expanding into a vacuum (or into a different ideal gas which acts like it is not present). Here, the vapor is constantly in equilibrium with the liquid since it is quasi-static expansion. True that the volume increases, but remember that the system is losing energy by doing work on the boundary (and the overall temperature decreases also because the "latent heat" is absorbed upon vaporization).

So let's forget about the the compression part for one moment, how is the entropy just after expansion unchanged if some salt rocks leaked out?
 
  • #6
Curl said:
That'd be true if you are expanding into a vacuum (or into a different ideal gas which acts like it is not present). Here, the vapor is constantly in equilibrium with the liquid since it is quasi-static expansion. True that the volume increases, but remember that the system is losing energy by doing work on the boundary (and the overall temperature decreases also because the "latent heat" is absorbed upon vaporization).

So let's forget about the the compression part for one moment, how is the entropy just after expansion unchanged if some salt rocks leaked out?

I guess I didn't understand your system. Can you explain a little better about what you mean by "Now I quasi-statically adiabatically expand this vapor."?
 
  • #7
I apologize for not being clear. I drew a picture.

[PLAIN]http://img521.imageshack.us/img521/6024/salt.gif
 

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  • #8
This is a good question. I think the resolution becomes clearer if we look at a related problem. Suppose we have a mixture of 2 ideal gases, A and B in a piston. The piston is pressed down to start with, and right below the plunger of the piston there is a stationary semi-permeable membrane that let's gas B through but not gas A.

Suppose we now expand the piston adiabatically and reversibly. In fact, let's take the limit where we expand the piston infinitely far. Now comes the reason I chose ideal gases, because we can calculate the entropy exactly. The entropy change for a pure gas changing from a state (V1,T1) to a state (V2,T2) is
[tex]\Delta S = nc_{V}\ln{(\frac{T_{2}}{T_{1}})} \ + nR\ln{(\frac{V_{2}}{V_{1}})}[/tex]
For an ideal gas, another ideal gas is just like a vacuum to it, so we can consider the gases separately by considering the volume occupied by both gases and the temperature which they share. This treats the entropy of mixing implicitly. Gas B gets expanded to final volume V2, which we can take as 'given' and calculate the final temperature from constancy of entropy:
[tex]\Delta S_{B} = n_{B}c_{V}\ln{(\frac{T_{2}}{T_{1}})} \ + n_{B}R\ln{(\frac{V_{2}}{V_{1}})}[/tex]
Meanwhile, gas A changes its temperature to T2, but keeps its volume at V1:
[tex]\Delta S_{A} = n_{A}c_{V}\ln{(\frac{T_{2}}{T_{1}})}[/tex]

For simplicity let's take nA=nB=n. The total entropy change needs to be 0 so:
[tex]\Delta S_{TOT} = 0 = 2nc_{V}\ln{(\frac{T_{2}}{T_{1}})} \ + nR\ln{(\frac{V_{2}}{V_{1}})}[/tex]
[tex]\rightarrow\ T_{2} = T_{1}(\frac{V_{1}}{V_{2}})^{\frac{R}{2c_{V}}}[/tex]

Now, just like in your example, let's block off the semi-permeable membrane and put some solid material there to block both gases before we compress. This way we will separate the gases isentropically. Let's assume V2>>V1, so that the amount of gas B in the same part of the container with A is negligible. There are two options for how we compress adiabatically. We could make the blocking material insulating or conducting. Let's make it insulating so we can use the normal formula for an adiabatic reversible compression, and say the gas finally reaches temperature T3. Let's compress it until the volume of B alone is back to V1 (but of course the whole container now has volume 2*V1 because the gases are separated):
[tex]T_{3}=T_{2}(\frac{V_{2}}{V_{1}})^{\frac{R}{c_{V}}}[/tex]
Inserting the expression for T2:
[tex]T_{3}=T_{1}(\frac{V_{1}}{V_{2}})^{\frac{R}{2c_{V}}}(\frac{V_{2}}{V_{1}})^{\frac{R}{c_{V}}}[/tex]
[tex]\ \ = T_{1}(\frac{V_{2}}{V_{1}})^{\frac{R}{2c_{V}}}[/tex]

Now for the moment of truth. Gas A has gone from (V1,T1) to (V1,T2). Gas B has gone from (V1,T1) to (V1,T3). The total entropy change is:
[tex]\Delta S_{TOT} = \Delta S_{A} + \Delta S_{B} = nc_{V}\ln{(\frac{T_{2}}{T_{1}}}) + nc_{V}\ln{(\frac{T_{3}}{T_{1}}})[/tex]
[tex]\ \ = nc_{V}\ln{(\frac{V_{1}}{V_{2}})^{\frac{R}{2c_{V}}} + nc_{V}\ln{(\frac{V_{2}}{V_{1}})^{\frac{R}{2c_{V}}} = 0[/tex] (!)What has happened is that before the two components were mixed and occupied a total volume V1. Now they are separated and occupy a total volume 2V1. These two effects cancel out because expanding the total system volume increases entropy, while separation of components decreases it.

So what if you now allow the solid separator to be a piston that equalizes pressure between the two chambers and compress the entire thing adiabatically so that the total volume is now V1? You can do the calculation and you will find that the temperature has actually increased! The work in compressing the gas back to its original volume will actually be higher when the gases are separated. The temperature increase increases entropy and now the temperature increase balances out the entropy of mixing term.

Keep in mind about the salt example: there will now be an enthalpy of mixing term that will make things more complicated. But the end result will be similar: The combined effect of changed overall system volume and the temperatures of the components will balance out the lost entropy of mixing.
 
  • #9
Thanks LeonhardEuler.
I've actually thought of this example before. Mine is a little simpler though, it goes like this:
Take a box with 2 gasses, A and B. On one end of the box, you have a lid with asemi-permeable membrane invisible to A (only blocks B) and on the other end, you have the opposite (a membrane invisible to B, blocking A).
Now you bring these two membranes together (from the ends of the box, to the middle). What you have done is separate gas A to the left, and gas B to the right.
Now seal up the 2 chambers, saw your box in half, and expand each one adiabatically to the volume of the original box. Now each gas is in its exact initial state. Boom, no calculation necessary.

The reason I was asking about the salt problem is that the salt isn't really an ideal gas while dissolved (are ions free to move, or are they bonded?) and when it precipitates it is a solid with a small volume. I know that volume of water + volume of solid salt is not equal to volume of water with salt dissolved in it.

Could it be that the enthalpy of evaporation is different when salt concentration is higher? That could have something to do with it.
 
  • #10
It will be very similar in both cases. Remember that the vapor pressure of water depends on the salt concentration. In an ideal solution, Raoult's Law will be followed so that the vapor pressure will be proportional to the fraction of water. Ionic salts obviously deviate from from Raoult's Law, but the qualitative behavior is similar.

Just like in the ideal gas case where if in the last step you want to bring everything back to volume V1, you end up doing more work in the compression stage than is lost in the expansion phase, the compression phase with no salt will require more work because there will be no salt to lower the vapor pressure, so this will result in a higher temperature and more entropy to compensate for the entropy of mixing.
 
  • #11
LeonhardEuler said:
It will be very similar in both cases. Remember that the vapor pressure of water depends on the salt concentration. In an ideal solution, Raoult's Law will be followed so that the vapor pressure will be proportional to the fraction of water. Ionic salts obviously deviate from from Raoult's Law, but the qualitative behavior is similar.

Just like in the ideal gas case where if in the last step you want to bring everything back to volume V1, you end up doing more work in the compression stage than is lost in the expansion phase, the compression phase with no salt will require more work because there will be no salt to lower the vapor pressure, so this will result in a higher temperature and more entropy to compensate for the entropy of mixing.

Nailed. This is the answer, for future reference.

The confusion comes from looking just at the piston, since from its perspective it's only doing reversible work. But on the way down, it's a "different gas" (meaning it behaves differently) so you're comparing apples to oranges.

Thanks.
 

1. What is the definition of a paradox in isentropic expansion?

A paradox in isentropic expansion refers to a situation where the expansion of a gas appears to violate the laws of thermodynamics. Specifically, it refers to the paradoxical increase in entropy during an isentropic (or reversible and adiabatic) process, where entropy is expected to remain constant.

2. How does statistical mechanics explain the paradox in isentropic expansion?

Statistical mechanics explains the paradox in isentropic expansion by considering the microstates of the gas particles. During an isentropic process, the particles are allowed to move freely and their positions and velocities are constantly changing. This results in a distribution of microstates with varying entropy, leading to an overall increase in entropy at the macroscopic level.

3. Can the paradox in isentropic expansion be resolved?

Yes, the paradox in isentropic expansion can be resolved by considering the assumptions made in the thermodynamic model. The paradox arises due to the assumption of an ideal gas with no intermolecular forces. In reality, all gases have some degree of intermolecular interactions, which can decrease the entropy of the system and resolve the paradox.

4. What are the implications of the paradox in isentropic expansion?

The paradox in isentropic expansion has important implications for our understanding of thermodynamics and the limitations of the ideal gas model. It also highlights the importance of considering microscopic behaviors in explaining macroscopic phenomena.

5. Are there real-world examples of the paradox in isentropic expansion?

Yes, there are several real-world examples of the paradox in isentropic expansion, such as the Joule-Thomson effect, where the expansion of a gas through a porous plug results in a decrease in temperature. This is contrary to the expected increase in temperature during an isentropic process.

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