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

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SUMMARY

The discussion focuses on calculating the volume of liquid carbon tetrachloride (CCl4) remaining in a sealed 5.00 L flask at 30.0°C after equilibrium is reached. The vapor pressure of CCl4 at this temperature is 143.0 mmHg, which converts to 0.188 atm. The calculations reveal that 0.014 moles of CCl4 remain as liquid after accounting for the vapor phase, indicating that some liquid CCl4 persists in the flask. The key takeaway is that the volume used in the vapor pressure equation should be the total flask volume minus the volume of liquid CCl4 remaining.

PREREQUISITES
  • Understanding of vapor pressure and its relationship to liquid equilibrium.
  • Familiarity with the Ideal Gas Law (PV=nRT).
  • Knowledge of converting units between mmHg and atm.
  • Basic skills in stoichiometry for calculating moles and mass from volume and density.
NEXT STEPS
  • Learn about the concept of saturated vapor pressure and its implications in closed systems.
  • Study the Ideal Gas Law in more depth, particularly its applications in equilibrium problems.
  • Explore the calculation of moles and conversions between mass, volume, and density in liquid-gas systems.
  • Investigate the effects of temperature on vapor pressure and phase changes in substances like CCl4.
USEFUL FOR

Chemistry students, particularly those studying thermodynamics and phase equilibria, as well as educators looking for practical examples of vapor pressure calculations in closed systems.

Qube
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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.
 
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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.
 
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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.
 
Qube said:
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.
 
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