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Dalton's law for ideal gases at different temperatures

  1. Mar 7, 2016 #1
    How would Dalton's law be affected when there are two ideal gases in a container at different temperatures?

    Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease.

    Now, we can no longer use P(A) + P(B)= P(T)

    ; P(A) is the partial pressure of gas A,

    P(B) is the partial pressure of gas B

    and P(T) is the total pressure exerted by the mixture of gas A and gas B

    Am I correct with this logic?

    A glass bulb of volume 400 ml is connected to another bulb of volume 200 mL by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 293 K and 1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.

    In the above problem, we cannot simply use PV= nRT to calculate final pressure because we don't know the final common temperature. So, we have to individually calculate the partial pressure of both the gases and then sum them up to get the total pressure. But according to me, as stated above, that should not be correct.

    Is my logic correct or I am missing out on something? Also, how should I solve using my logic and please tell if there are other ways to go about solving this problem.
     
  2. jcsd
  3. Mar 7, 2016 #2
    I think these laws apply to only equilibrium systems only. We can apply the law only when the two attain a common temperature. and becomes a homogeneous one system having a common temperature and also a common total pressure. What is measured is common pressure only. Partial pressures are calculated not measured.
     
  4. Mar 7, 2016 #3

    DrDu

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    In your example, the pressure will be the same in both bulbs but the amount of air in the two bulbs (aka n_1 and n_2) will be different, namely: ##n_1=PV_1/RT_1## and ##n_2=PV_2/RT_2##.
     
  5. Mar 7, 2016 #4
    How does this answer my question?
     
  6. Mar 7, 2016 #5

    DrDu

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    I give you another equation: ##n=n_1+n_2##. Now you have 3 equations for the three unknown quantities P, n_1 and n_2 which you may solve to get P.
     
  7. Mar 7, 2016 #6
    I got the answer this way. Can you tell me if my reasoning about the Dalton's law was correct?
     
  8. Mar 8, 2016 #7
    Can you tell me if my reasoning about the Dalton's law was correct or not??
     
  9. Mar 8, 2016 #8

    DrDu

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    I don't quite understand it. You talk of gas A and B, but in your example, both flasks contain dry air.
     
  10. Mar 8, 2016 #9
    I only know of dry air as air having no content of water vapour. Why should it matter?
     
  11. Mar 8, 2016 #10

    DrDu

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    Yes, but what is gas A and what B in your example?
     
  12. Mar 8, 2016 #11
    The gases A nd B have got to do nothing with that problem. I just used them to explain a concept that I thought would be utilized in the problem.
     
  13. Mar 8, 2016 #12

    DrDu

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    I find this a bit confusing. Why do you introduce A and B if they are not related to your problem? Maybe you could rephrase your question?
     
  14. Mar 8, 2016 #13
    Well, I don't know any other way to explain that thing. You can think of my question as having two parts. Firstly I want to know the flaw in my reasoning in -
    "Let two gases : Gas A and gas B be placed in a container . Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease."


    And then I wanted to know if that logic could be extended in the problem below.
     
  15. Mar 8, 2016 #14

    DrDu

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    Yes, sort of, but I would not talk of different gasses only because they have different temperature. If the gasses of different temperature come into contact, one will lose kinetic energy and the other one will gain, but there will be hardly a pressure difference as the gasses will start to flow.
    That's basically how wind comes around: Temperature differences generate minute pressure differences which lead to strong currents of air.
     
  16. Mar 8, 2016 #15
    What if the two gases are at different pressures before they are mixed? Then there would be pressure and heat exchange happening at the same time?
     
  17. Mar 8, 2016 #16

    DrDu

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    Yes, you will get a strong turbulent flow and rapid equilibration of pressure and temperature.
     
  18. Mar 8, 2016 #17
    That would be a very chaotic situation? If this is true, then we cannot solve the problem :

    A glass bulb of volume 400 ml is connected to another bulb of volume 200 mL by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 293 K and 1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.


    using Pv=nrt ; p is the total moles in both the bulbs , v is the total volume and T is the "equilibrium temperature" since we cannot find equilibrium temperature. Then maybe we want to use the relation : sum of partial pressures of individual gases = total pressure
    So now we will use Pv =nRt to calculate the partial pressure but then which temperature should be substitute here for both the gases? It cannot be 273 K and 373 K for sure because temperature will be affected when the gases interact?
    Or this approach is wrong altogether?
     
  19. Mar 8, 2016 #18
    DrDu. Elena14 keeps talking about partial pressures. Have you ever, in you broad experience, heard to the term "partial pressure" being used in connection with a single component system (e.g., this one)? I haven't. My interpretation of this problem statement was the same as yours, and I obtained the same set of equations and results as you did. I am just not able to decipher what Elena14 is getting at. As far as I can judge, Dalton's law does not apply to this system.

    Chet
     
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