Vapor Pressure of Ethanol

  1. A 4.7-L sealed bottle containing 0.33 g of liquid ethanol, C2H6O, is placed in a refrigerator and reaches equilibrium with its vapor at -11 degrees C.

    a) What is the mass of ethanol present in the vapor?
    b) When the container is removed and warmed to room temperature, 20 degrees C, will all the ethanol vaporize?
    c) How much liquid ethanol would be present at 0 degrees C?

    The vapor pressure of ethanol is 10 torr at -2.3 degrees C and 40 torr at 19 degrees C.

    My attempt:

    a) I have no idea how to approach this.
    b) No, because there will be condensation occurring at the same rate as vaporization, once it reaches equilibrium again. (?)
    c) I think I need to use PV=nRT for this... but I don't know how to relate the 2 different temperatures I'm getting (0 C and 19 C)

    Any suggestions would be appreciated, thank you!!
  2. jcsd
  3. A sketch of the solution.

    To attack all three problems here, you need to calculate the vapour
    pressure of ethanol as a function of the absolute temperature T.
    Use the Clausius Clapyron equation for this:
    log p = A/T +B,
    where p is the vapour pressure, T the absolute temperature and A and B are
    constant. You can easily get A and B from the given vapour pressures at
    T=270.85 K and T=292.15 K, which will enable you to calculate the vapour pressure
    at 262.15 K, 273.15 K and 293.15 K, or indeed any other temperature.

    To find the mass of ethanol in the vapour phase, use the ideal gas
    equation to get the number of moles, n:
    n = pV/(RT), since V is given and p is the vapour pressure.
    From n and the molecular weight of ethanol (C2H5OH), you can calculate the
    mass of ethanol in the vapour. Since the vessel is sealed, the amount of
    ethanol in the liquid and vapour phases together remains constant.

    The answers to (a) and (c) should be easy. The answer to (b) is a bit
    trickier because it could be that there is not enough ethanol to produce
    the required partial pressure in the vapour. In this case, all of the
    liquid will evaporate and your answer to (b) is incorrect.
    You will have to do the calculation to see what comes out.

    I assume you can take it from here, being careful to choose a consistent
    set of units.
  4. pkleinod gave a excellent answer

    there is another form of the Clausius Clapeyron equation


    which might be easier for your specific question as you can plug in all the variables and not have to solve for C

    the R value is a constant, 8.31457 J/K
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?
Similar discussions for: Vapor Pressure of Ethanol