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A gas at room temperature must have a vapor pressure greater than atmospheric pressure?

  1. Oct 25, 2014 #1
    This is for "ammonia". The problem says that ammonia is a gas at room temperature. And that this tells us:

    "the fact that ammonia is a gas at room temperature tell us that vapor pressure of ammonia must be greater than atmospheric pressure".

    I know that for water, when the boiling point is reached, vapor pressure = atmospheric pressure and a phase change occurs (is the direction determined by the input or "taking away" of energy?).

    I thought that the pressure in a system could not be higher than 760mmHG though? So how is ammonia have a partial pressure higher than this and how does this tell us that we are dealing with a gas?
     
  2. jcsd
  3. Oct 25, 2014 #2

    NascentOxygen

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    Staff: Mentor

    You seem to be not seeing the distinction between vapour pressure when the gas exists in equilibrium with its liquid phase, and partial pressure of a component in any mixture of gases.
     
  4. Oct 25, 2014 #3
    But in a system where atmospheric pressure is relevant, wouldnt the total pressure of the mixture have to equal 760mm Hg?
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    OH Wow youre completely right. Thank you I think i've almost got it.

    So the vapor pressure is not indicative of the partial pressure of the system. It's more an inherent quality that tells you how much of a pure gas would be in its vapor state or liquid state at a given temperature.

    While partial pressure is just the distribution of pressures for each gas component based upon the mole ratio and the total pressure possible in the system (assuming ideal gases).

    So then what happens when vapor pressure = atmospheric pressure, but youre not adding any extra heat?

    If vapor pressure = atmospheric pressure, and you add heat then bubbles (pockets of vapor) can form inside the solution which can then escape because the pressure of the atmosphere and of the liquid's vapor is the same, right?

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    Last question but, if your vapor pressure is greater than atmospheric pressure, then all of the liquid should be able to turn into gas right?

    But what if there are other gases in the system that contribute to the total pressure?

    If for gas A Ppure=760mm, and its vapor pressure >atm, then if you add another gas into the system, then some of gas A will be in liquid form right? Would that mean the vapor pressure (assuming you kept moles of gas A constant) went down upon addition of a new gas?
     
  5. Oct 26, 2014 #4

    NascentOxygen

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    Yes. Alternatively, if you take a sample of gas and isothermally increase the pressure, at some point it will begin to condense. That pressure is the vapor pressure.

    No. You are not changing the partial pressure of gas A and an ideal gas behaves as though it is the only species present, there is no interaction between different gases.

    Vapor pressure is a physical property determined by temperature only, for any particular gas, so it is not changed by the addition of a different gas. Neither is there any change by this in the partial pressure of A (unchanged number of molecules, unchanged container volume).

    This topic has long fallen into disuse with me, so I sought confirmation on google---see excerpt below, from Fundamentals of Food Process Engineering, by Romeo T Toledo. In particular, note his final sentence.

    shot_000007-1.jpg
     
  6. Oct 26, 2014 #5
    I see okay thank you so much that clears up a lot.

    So if i had a gas with a vapor pressure above atmospheric pressure, and its Pressure was 760mmHg if pure, then if i added another gas to it, it would still maintain the same pressure, it's just that its mole fraction would change (from 1 to 760mmHg/the new total amount of gas in the system) right?
     
  7. Oct 26, 2014 #6

    NascentOxygen

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    The first gas would have the same partial pressure (I think that's what you meant to write?). But addition of a second gas will cause the pressure to increase, pressure being Σ partial pressures.
     
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