Compression of gases that liquefies

[Chestermiller]

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.

 

[sgstudent]

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]

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.