# Compression of gases that liquefies

• I
• sgstudent
In summary, the conversation discusses a closed system with water and a gas, A, at different pressures and temperatures. As the system is compressed with 2atm of pressure, the partial pressures of A and water change due to their phase transitions and mole fraction in the liquid. The final pressure is specified to be 2atm, but it is unclear if the energy of compression is adiabatic or isothermal. There is also a discussion on the ideal gas and liquid mixture models used in the calculations.

#### sgstudent

For example we had a closed system with water at 298K, so it's pressure is 0.031atm. To it we add a gas, A such that we get 1atm of total pressure. That gas undergoes a phase transition at 1.5atm and 298K. So initially, the partial pressure of water is a little smaller than 0.031atm due to the dissolution of some of that gas, and the rest of the pressure is from that gas.

As we compress this mixture of gases with 2atm of pressure what would happen? How would the A and water's partial pressure change?

Since A is able to increase its partial pressure further before liquefying, I feel that it should increase to 1.5atm. But that results in a decrease in the water's partial pressure because it's mole fraction in the liquid continues to drop further. So the total pressure will be 1.5atm from A, and less than 0.031atm from water. But then since we're applying 2atm of pressure, how does that equal out?

Something like the uploaded image but instead of air we replace it with a different gas.

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Are you actually describing a stable final situation here? The final pressure in the cylinder has to be 2Atm .

russ_watters
Are you saying that pure A would undergo a phase transition at 1.5 atm and 298K?

Chestermiller said:
Are you saying that pure A would undergo a phase transition at 1.5 atm and 298K?
Yes so in a phase diagram for A it would transition at 1.5atm. Does that matter if it's a mixture?
sophiecentaur said:
Are you actually describing a stable final situation here? The final pressure in the cylinder has to be 2Atm .
I get that the final internal pressure has to be 2atm. But I don't see how that can be achieved with the parameters I've set because A would liquefy once it's pressure is at 1.5atm, and the water would liquefy at pressure less than 0.031atm ad well.

sgstudent said:
Yes so in a phase diagram for A it would transition at 1.5atm. Does that matter if it's a mixture?

I get that the final internal pressure has to be 2atm. But I don't see how that can be achieved with the parameters I've set because A would liquefy once it's pressure is at 1.5atm, and the water would liquefy at pressure less than 0.031atm ad well.
Are the liquid phases of A and water miscible?

Chestermiller said:
Are the liquid phases of A and water miscible?
Yes I would assume so for this scenario. So I guess I'd use Raoult's law for water and Henry's law for A to calculate their partial pressures at each point? So initially, when I first introduced A to the pure water system, how would I know how much it dissolves? We would just know that the total P is 1atm and initial P is 0.031atm for pure water vapour only.

sgstudent said:
For example we had a closed system with water at 298K, so it's pressure is 0.031atm. To it we add a gas, A such that we get 1atm of total pressure. That gas undergoes a phase transition at 1.5atm and 298K. So initially, the partial pressure of water is a little smaller than 0.031atm due to the dissolution of some of that gas, and the rest of the pressure is from that gas.

As we compress this mixture of gases with 2atm of pressure what would happen? How would the A and water's partial pressure change?

Since A is able to increase its partial pressure further before liquefying, I feel that it should increase to 1.5atm. But that results in a decrease in the water's partial pressure because it's mole fraction in the liquid continues to drop further. So the total pressure will be 1.5atm from A, and less than 0.031atm from water. But then since we're applying 2atm of pressure, how does that equal out?

Something like the uploaded image but instead of air we replace it with a different gas.
It really sounds to me like you are just arguing with your own constraints. If the final pressure is 2atm because you specified it to be 2atm, then the final pressure is 2atm.

Perhaps what you really mean to specify (or find?) is a compression ratio...? I see your title says "compression", but then seemed to misuse the word in the description of the problem. Compression and pressurization are different things: compression is decreasing the volume by a specified amount whereas pressurization is increasing the pressure to a specified quantity.

Also, you didn't specify what happens, if anything, to the energy of compression. Should we assume adiabatic or isothermal?

sgstudent said:
Yes I would assume so for this scenario. So I guess I'd use Raoult's law for water and Henry's law for A to calculate their partial pressures at each point? So initially, when I first introduced A to the pure water system, how would I know how much it dissolves? We would just know that the total P is 1atm and initial P is 0.031atm for pure water vapour only.
You are assuming that the gas phase is an ideal gas mixture. That's fine.

But, for the liquid phase, if you use Raoult's law, you are assuming an ideal liquid mixture (that applies to both species), and, if you assume A obeys Henry's law for the liquid, you are assuming a non-ideal liquid mixture model with a known activity coefficient. What liquid phase mixture model are you assuming (a) Raoult's law, (b) Henry's law, or (c) a different non-ideal fluid mixture model?

russ_watters said:
It really sounds to me like you are just arguing with your own constraints. If the final pressure is 2atm because you specified it to be 2atm, then the final pressure is 2atm.

Perhaps what you really mean to specify (or find?) is a compression ratio...? I see your title says "compression", but then seemed to misuse the word in the description of the problem. Compression and pressurization are different things: compression is decreasing the volume by a specified amount whereas pressurization is increasing the pressure to a specified quantity.

Also, you didn't specify what happens, if anything, to the energy of compression. Should we assume adiabatic or isothermal?
Sorry for the wrong terminology used, I meant that the mixture is compressed with 2atm of pressure. But since gas A undergoes a phase transition at 1.5atm, and water undergoes a phase transition at 0.031atm in their pure states how can their gaseous pressures ever be equal to 2atm?

Edit: it should be a isothermal process where temperature stays at 298K

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Chestermiller said:
You are assuming that the gas phase is an ideal gas mixture. That's fine.

But, for the liquid phase, if you use Raoult's law, you are assuming an ideal liquid mixture (that applies to both species), and, if you assume A obeys Henry's law for the liquid, you are assuming a non-ideal liquid mixture model with a known activity coefficient. What liquid phase mixture model are you assuming (a) Raoult's law, (b) Henry's law, or (c) a different non-ideal fluid mixture model?
I was taught that an ideal dilute mixture would have the solvent follow Raoult's law, and the solute follow Henry's law. So I presumed that A being the added gas should follow Henry's law.

But even if I were to use assume a Raoult's law for both, when I first introduced the gas water's mole fraction in the liquid phase should decrease so the pressure of 0.031atm should decrease. But I don't really know how to tell how much it decreases by and likewise for A, I'm unsure how much of it dissolves as well. So I know that P(H2O(g))<0.031atm and P(A(g))>0.969atm but is there a way to know for sure their actual pressures?

And even if we do know it, when we compress this mixture with 2atm of pressure how would the system change with this pressure? I presumed that A could continue to increase in pressure up til it hit 1.5atm, because that's it's vapor pressure at 298K. But for water it's vapour pressure is 0.031atm at 298K, but since we increased the pressure of A to 1.5atm, by Raoult's law, it's pressure should decrease even more because more A would have dissolved right? I'm quite clueless as to how the system changes when such a scenario happens to be honest.

sgstudent said:
Sorry for the wrong terminology used, I meant that the mixture is compressed with 2atm of pressure. But since gas A undergoes a phase transition at 1.5atm, and water undergoes a phase transition at 0.031atm in their pure states how can their gaseous pressures ever be equal to 2atm?

Edit: it should be a isothermal process where temperature stays at 298K
Well, what does Raoult's law predict for the mole fractions of water and A in the liquid at 1 atm pressure?

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sgstudent said:
how can their gaseous pressures ever be equal to 2atm?
Because you have said it is in your initial description. Something may need to change state or dissolve . . . . . It's not an ideal situation.

Chestermiller said:
Well, what does Raoult's law predict for the mole fractions of water and A in the liquid at 1 atm pressure?
P(H2O(g))=P*(H2O(g))X(H2O) and P(A)=P*(A)X(A), but we only know the values of P*(H2O) and P*(A) so I don't get how we can predict their mole fractions though.

Chestermiller said:
Well, what does Raoult's law predict for the mole fractions of water and A in the liquid at 1 atm pressure?
Oh, wait we can tell it from the graph right? So we would know that the exact pressures of water and A at a total pressure of 1 atm?

So it will look something like this right? So we can get the mole fractions of either component?

But continuing on, when we compress the gas with 2atm of pressure the mole fraction of water goes to 0? I don't really understand how the process continues on from here though.

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We first have to specify how much water and how much A are present total. So, please specify. That way we can determine how much of each species is present in each phase.

Chestermiller said:
We first have to specify how much water and how much A are present total. So, please specify. That way we can determine how much of each species is present in each phase.
Let's say we have 20 moles of H2O in the liquid phase and 5 moles of A in the liquid phase initially. And from the graph, the partial pressure of water would be 0.0248atm and A would be 0.9752atm. Would this be alright?

sgstudent said:
Sorry for the wrong terminology used, I meant that the mixture is compressed with 2atm of pressure.
That's still the wrong terminology. Again, 2atm is a pressure, not a compression amount.
But since gas A undergoes a phase transition at 1.5atm, and water undergoes a phase transition at 0.031atm in their pure states how can their gaseous pressures ever be equal to 2atm?
They can't, but you could have all compressed liquid.

sgstudent said:
Let's say we have 20 moles of H2O in the liquid phase and 5 moles of A in the liquid phase initially. And from the graph, the partial pressure of water would be 0.0248atm and A would be 0.9752atm. Would this be alright?
No. The partial pressure of A would be (0.2)(1.5)=0.3 atm, and the total pressure would be 0.3248 atm. What number of moles of A would you need in the liquid for the total pressure to be 1 atm? Also, you would need to specify the total number of moles of vapor at 1 atm if we are going to compress the system and quantify what's happening next.

Let me guess. You've been studying multicomponent phase equilibrium, but you haven't actually worked any practice problems yet.

Chestermiller said:
No. The partial pressure of A would be (0.2)(1.5)=0.3 atm, and the total pressure would be 0.3248 atm. What number of moles of A would you need in the liquid for the total pressure to be 1 atm? Also, you would need to specify the total number of moles of vapor at 1 atm if we are going to compress the system and quantify what's happening next.

Let me guess. You've been studying multicomponent phase equilibrium, but you haven't actually worked any practice problems yet.
Haven't even gone there yet, we just got introduced to the topic of mixing but I've been thinking a lot to make sure I understood what's going on in the lectures. Sometimes I feel like I overthink too much but I'm just unsure if what I'm overthinking will be looked at further in the module or no, which leads me to these types of situations.

But I'd need 37 moles of A to get 1 atm total pressure. 37/57.14x1.5=0.9752atm.

russ_watters said:
That's still the wrong terminology. Again, 2atm is a pressure, not a compression amount.

They can't, but you could have all compressed liquid.
Hmm, so everything would just liquefy?

sgstudent said:
Hmm, so everything would just liquefy?
Yes.
An actual close example might be neopentane.
Boils at 101 kPa at 9,5 Celsius. Vapour pressure at 20 Celsius is 146 kPa.
So if you compress a cylinder of neopentane at isothermal conditions (compression heat freely conducted away) of just over 20 Celsius, it would condense at 1,5 bar till all of it is liquid, and then you can compress the liquid to 2 bar.
Neopentane is a low polarity alkane, so not miscible with water, though.

sgstudent said:
Haven't even gone there yet, we just got introduced to the topic of mixing but I've been thinking a lot to make sure I understood what's going on in the lectures. Sometimes I feel like I overthink too much but I'm just unsure if what I'm overthinking will be looked at further in the module or no, which leads me to these types of situations.

But I'd need 37 moles of A to get 1 atm total pressure. 37/57.14x1.5=0.9752atm.
37 moles of A is incorrect. I calculate 38.76 moles of A. Please re-analyze the problem and show how you can get the right answer.

sgstudent said:
Hmm, so everything would just liquefy?
That's what we are going to prove by doing some calculations. But, first you need to show that you can get the right answer for the case of 1 atm pressure.

Chestermiller said:
37 moles of A is incorrect. I calculate 38.76 moles of A. Please re-analyze the problem and show how you can get the right answer.

If we had 20 moles of water, (0.031)(20/20+b)+(1.5)(b/20+b)=1atm so solving for B gives us 38.76 moles. Would be this correct?

sgstudent said:
If we had 20 moles of water, (0.031)(20/20+b)+(1.5)(b/20+b)=1atm so solving for B gives us 38.76 moles. Would be this correct?
Excellent. Now you're starting to understand.

So the total number of moles of liquid you have is 58.76. Now, to do the calculation we wish to carry out, you are going to need to specify the number of moles of vapor present in the cylinder at 1 atm total pressure. This is because, when we compress the system, the total number of moles of liquid plus vapor present in the system doesn't change. So please specify the number of moles of vapor present at 1 atm. Then, please determine (from Raoult's law) the number of moles of water and of species A present in the vapor at 1 atm. Then determine the total number of moles of water and the total number of moles of species A present in the cylinder (vapor plus liquid).

Chestermiller said:
Excellent. Now you're starting to understand.

So the total number of moles of liquid you have is 58.76. Now, to do the calculation we wish to carry out, you are going to need to specify the number of moles of vapor present in the cylinder at 1 atm total pressure. This is because, when we compress the system, the total number of moles of liquid plus vapor present in the system doesn't change. So please specify the number of moles of vapor present at 1 atm. Then, please determine (from Raoult's law) the number of moles of water and of species A present in the vapor at 1 atm. Then determine the total number of moles of water and the total number of moles of species A present in the cylinder (vapor plus liquid).

Let's say we have 10 moles of water vapour. So by Raoult's law, P(A)=38.76/58.76*1.5=0.989atm and P(water vapour)=0.011atm. So by Avogadro's principle we would have 89.9 moles of A. Would this be correct?

Water is pretty simple and unusual compound. There is a modest choice of fluids which exhibit near-Raoult behaviour with water, and most of them are high boiling.

What could be near-Raoult behaving gas in the given vapour pressure range is ethylamine.

sgstudent said:
Let's say we have 10 moles of water vapour. So by Raoult's law, P(A)=38.76/58.76*1.5=0.989atm and P(water vapour)=0.011atm. So by Avogadro's principle we would have 89.9 moles of A. Would this be correct?
No. This would be correct if there were 1 mole of water vapor. For 10 moles of water in the vapor phase, the total number of moles in the vapor phase would be 10/0.01055 = 947.7

Let's use 1 mole of water in the vapor phase and 94.77 moles of A in the vapor. So we have 21.0 moles of water in the cylinder and 133.53 moles of A in the cylinder. And at 1 atmosphere pressure, the number of moles of liquid is 58.76 and the number of moles of vapor is 95.77.

Now let's consider the case where we increase the pressure to 1.3 atm. Let L equal the new number of moles of liquid in the system and let V equal the new number of moles of vapor in the system. Using Raoult's law and appropriate material balances, please write down the equations for calculating L, V, and the mole fraction of water in the liquid phase.

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In this system, there is a certain pressure (<1.5 atm) above which you can have only liquid. What is that pressure?

Chestermiller said:
In this system, there is a certain pressure (<1.5 atm) above which you can have only liquid. What is that pressure?
I don't see where there will be a point like that though.
Chestermiller said:
No. This would be correct if there were 1 mole of water vapor. For 10 moles of water in the vapor phase, the total number of moles in the vapor phase would be 10/0.01055 = 947.7

Let's use 1 mole of water in the vapor phase and 94.77 moles of A in the vapor. So we have 21.0 moles of water in the cylinder and 133.53 moles of A in the cylinder. And at 1 atmosphere pressure, the number of moles of liquid is 58.76 and the number of moles of vapor is 95.77.

Now let's consider the case where we increase the pressure to 1.3 atm. Let L equal the new number of moles of liquid in the system and let V equal the new number of moles of vapor in the system. Using Raoult's law and appropriate material balances, please write down the equations for calculating L, V, and the mole fraction of water in the liquid phase.
I thought of this nA/L*1.5+(1-nA/L)*0.031=1.3 but I can't seem to think of the other equation to solve the simultaneous equations though. Could I get a nudge in the right direction?

sgstudent said:
I don't see where there will be a point like that though.

I thought of this nA/L*1.5+(1-nA/L)*0.031=1.3 but I can't seem to think of the other equation to solve the simultaneous equations though. Could I get a nudge in the right direction?
Your Roault's law equation is correct, but I'm going to write it in terms of the liquid mole fraction of water x, rather than the liquid mole fraction of A. I hope that's OK with you:
$$0.031x+1.5(1-x)=P\tag{1}$$where P is the total pressure.

OVERALL MASS BALANCE:
The number of moles of liquid L plus the number of moles of vapor V must be equal to the total number of moles of water in the container (21) plus the total number of moles of A in the container (133.53). So,
$$L+V=154.53\tag{2}$$

MASS BALANCE ON WATER:
The number of moles of water in the liquid phase (Lx) plus the number of moles of water in the vapor phase (##\frac{0.031x}{P}V##) must be equal to the total number of moles of water in the container (21):
$$Lx+\frac{0.031x}{P}V=21\tag{3}$$

We have liquid in the cylinder with vapor in the head space, and we increase the pressure P in the cylinder by forcing a piston down on the top of the vapor. We would like to determine whether there is a certain pressure beyond which all the vapor has been squeezed into the liquid, and only liquid remains below the piston. Our game plan is to choose a sequence of increasing values for P and solve for the mole fraction of water in the liquid x, and the number of moles of liquid L and vapor V. The calculation goes like this:

1. Choose a value for P
2. Solve Eqn. 1 for the corresponding value of x
3. Substitute the values for P and x into equation 3
4. Solve Eqns. 2 and 3 simultaneously for L and V

I would like you to carry out this procedure for values of P equal to 1.0, 1.1, 1.2, 1.25, and 1.3 atm. Then make a graph of x vs P and a second graph of L and V vs P.

Chestermiller said:
Your Roault's law equation is correct, but I'm going to write it in terms of the liquid mole fraction of water x, rather than the liquid mole fraction of A. I hope that's OK with you:
$$0.031x+1.5(1-x)=P\tag{1}$$where P is the total pressure.

OVERALL MASS BALANCE:
The number of moles of liquid L plus the number of moles of vapor V must be equal to the total number of moles of water in the container (21) plus the total number of moles of A in the container (133.53). So,
$$L+V=154.53\tag{2}$$

MASS BALANCE ON WATER:
The number of moles of water in the liquid phase (Lx) plus the number of moles of water in the vapor phase (##\frac{0.031x}{P}V##) must be equal to the total number of moles of water in the container (21):
$$Lx+\frac{0.031x}{P}V=21\tag{3}$$

We have liquid in the cylinder with vapor in the head space, and we increase the pressure P in the cylinder by forcing a piston down on the top of the vapor. We would like to determine whether there is a certain pressure beyond which all the vapor has been squeezed into the liquid, and only liquid remains below the piston. Our game plan is to choose a sequence of increasing values for P and solve for the mole fraction of water in the liquid x, and the number of moles of liquid L and vapor V. The calculation goes like this:

1. Choose a value for P
2. Solve Eqn. 1 for the corresponding value of x
3. Substitute the values for P and x into equation 3
4. Solve Eqns. 2 and 3 simultaneously for L and V

I would like you to carry out this procedure for values of P equal to 1.0, 1.1, 1.2, 1.25, and 1.3 atm. Then make a graph of x vs P and a second graph of L and V vs P.
Prof chestermiller, could I continue on with this on the next weekend as my midterms are next week and this chapter isn't tested. Just wanted to let you know that I haven't ghosted you for this problem.

sgstudent said:
Prof chestermiller, could I continue on with this on the next weekend as my midterms are next week and this chapter isn't tested. Just wanted to let you know that I haven't ghosted you for this problem.
Certainly you can continue at your convenience.