# Compression of gases that liquefies

• I
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.

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

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.
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?

Chestermiller
Mentor
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.

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.

Chestermiller
Mentor
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.