How Much CCl4 Remains Liquid at Equilibrium in a Sealed Flask?

[Qube]

Homework Statement

If 5.00 mL of liquid carbon tetrachloride (CCl4, density = 1.587 g/mL) was injected into a sealed 5.00 L flask at 30.0°C, what volume (if any) of the CCl4 would remain as liquid after equilibrium is reached? (the vapor pressure of CCl4 at 30.0°C is 143.0 mmHg)

Homework Equations

Divide mm Hg by 760 to yield pressure in atm.

PV=nRT

Vapor pressure = pressure of liquid solution at equilibrium.

The Attempt at a Solution

VP = 0.188 atm = 143 / 760. I know this.

P(solution) = nRT/V.

Do I set this equal to the VP? This seems rather unlikely as I'd have to solve for two variables at once - the volume of the liquid solution - and the moles of the liquid left.

 

[Borek]

Calculate how much gaseous CCl₄ would be present assuming saturated vapor. If it is lower than the amount present - there will be some liquid left. If it is higher than the amount present - all CCl₄ will evaporate and there will be no saturated vapor.

 

[Qube]

Using P=nRT/V I found the moles of gaseous CCl4 present assuming saturated vapor. I found there was 0.0377 moles of CCl4(g).

I then found the grams of CCl4 added to the container by multiplying milliliters by density. I then found the moles of CCl4 present by dividing grams by molar mass. Subtracting the gaseous number of moles from the total number of moles injected we can find the number of moles of CCl4(l). That number is 0.014 moles.

We can then convert moles back into grams and back into volume given density.

Question:

1) Vapor pressure of gaseous CCl4 = nRT/V. What should my value for V be? Container volume, or free volume - 5 milliliters subtracted from 5 liters?

I'm guessing the amount of CCl4 injected is purposely small so it doesn't matter on a multiple-choice test, and that it's sort of hard to guess the free volume given that we're supposed to be finding the amount of volume of liquid CCl4 remaining.

 

[Borek]

1) Vapor pressure of gaseous CCl4 = nRT/V. What should my value for V be? Container volume, or free volume - 5 milliliters subtracted from 5 liters?

To be exact - neither. It should be 5L minus the volume of the liquid left (which, as you already know, is lower than 5 mL). But error you are making using 5L is negligible.

 

[DeeNos]

As a scientist, it is important to approach problems systematically and logically. In this case, we can use the ideal gas law (PV = nRT) to determine the amount of liquid carbon tetrachloride that remains after equilibrium is reached.

First, we need to convert the given vapor pressure of CCl4 (143 mmHg) to atm. This can be done by dividing it by 760, as stated in the problem. This gives us a vapor pressure of 0.188 atm.

Next, we can use the ideal gas law to calculate the pressure of the liquid solution at equilibrium. Since the flask is sealed, the total pressure inside the flask will be equal to the vapor pressure of CCl4 plus the pressure of the remaining liquid.

Therefore, we can set up the equation as follows:

P(solution) = P(vapor) + P(liquid)

Substituting in the values we know, we get:

P(solution) = 0.188 atm + P(liquid)

We also know that the volume of the flask is 5.00 L and the temperature is 30.0°C, which can be converted to 303 K.

Now, we can rearrange the ideal gas law to solve for the pressure of the remaining liquid:

P(liquid) = P(solution) - 0.188 atm

P(liquid) = (nRT/V) - 0.188 atm

Since we are looking for the volume of liquid remaining, we can rearrange the equation again to solve for V:

V = (nRT) / (P(liquid) + 0.188 atm)

Now, we need to determine the number of moles of liquid carbon tetrachloride present in the flask. This can be done by using the given density of CCl4 (1.587 g/mL) and the initial volume of 5.00 mL. This gives us a mass of 7.935 g.

Next, we can use the molar mass of CCl4 (153.82 g/mol) to convert the mass to moles. This gives us a value of 0.0516 moles.

Now, we can substitute this value for n in our equation for V:

V = (0.0516 moles x 0.08206 L atm/mol K x 303 K) / (P(liquid) + 0.188 atm)

Finally, we can solve for