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

Click For Summary

Discussion Overview

The discussion revolves around determining the volume of liquid carbon tetrachloride (CCl4) that remains in a sealed flask at equilibrium after injecting a specific amount of the liquid. The context includes calculations involving vapor pressure, moles of gas, and the relationship between liquid and gaseous phases at a given temperature.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving 5.00 mL of CCl4 in a 5.00 L flask at 30.0°C, asking how much remains liquid at equilibrium, referencing the vapor pressure of CCl4.
  • Another participant suggests calculating the amount of gaseous CCl4 assuming saturated vapor to determine if any liquid remains, indicating that if the gaseous amount is lower than the injected amount, some liquid will remain.
  • A participant calculates the moles of gaseous CCl4 at saturation and finds there are 0.0377 moles, then calculates the total moles of CCl4 injected and determines that 0.014 moles remain as liquid.
  • There is a question regarding the appropriate volume to use in the vapor pressure equation, with one participant suggesting it should be the free volume after accounting for the liquid, while another notes that using the total container volume is a negligible error.

Areas of Agreement / Disagreement

Participants express uncertainty about the correct volume to use in calculations, with differing opinions on whether to use the total container volume or the free volume. There is no consensus on the best approach to determine the volume of liquid CCl4 remaining.

Contextual Notes

Participants highlight the importance of considering the volume of liquid remaining when calculating vapor pressure, and the discussion includes assumptions about the negligible impact of certain approximations.

Qube
Gold Member
Messages
461
Reaction score
1

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.
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person
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.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
3K
Replies
1
Views
12K
Replies
3
Views
6K
  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 4 ·
Replies
4
Views
5K