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Year 12: Cambridge Physics Problem (Pressure inside a vessel)

  1. Jun 19, 2012 #1
    A vessel is divided into two parts of equal volume by a partition in which there is a very small hole. Initially, each part contains gas at 300K and a low pressure, p. One part of the vessel is now heated to 600K while the other is maintained at 300K. If a steady state is established when the rate at which molecules pass through the hole from each side is the same, find the resulting pressure difference between the two parts.

    Attempt:

    I'm assuming that the number and mass of molecules inside remain the same and that the temperature of the two parts during the steady state is the same.

    So we have
    [tex]N_1+N_2=2N[/tex], where N is the number of molecules inside each part before heating and N1 and N2 denote the number of molecules inside each part after heating.

    Also pressure is proportional to <c>2, implying T is proportional to <c>2, and that p is proportional to T.

    We also have [tex]N_1<c>_1=N_2<c>_2[/tex], where N_i<c>_i denotes the rate at which molecules pass through the hole from one side to another.

    So now [itex]<c>^2 \propto T

    \text{and }N_1<c>_1=N_2<c>_2

    \Rightarrow \dfrac{N_1}{N_2}= \dfrac{<c>_2}{<c>_1}

    \Rightarrow \dfrac{N_1^2}{N_2^2}= \dfrac{<c>_2^2}{<c>_1^2}

    \Rightarrow \dfrac{N_1^2}{N_2^2}= \dfrac{T_2}{T_1}

    \Rightarrow \dfrac{N_1}{N_2}= \(\dfrac{T_2}{T_1})^{1/2}[/itex]

    And I'm stuck. I'm supposed to find the difference of pressure in terms of p but how do I do that when the terms which I have introduced are nowhere close to p? The closest one I could get are p1 and p2. Help?
     
  2. jcsd
  3. Jun 19, 2012 #2
    I believe this is for ideal gases. Let the volume of each compartment be V. Write down the ideal gas equations for the initial and final conditions(separately). The initial condition will give you an equation in p, which you can use to find out difference in pressures.

    [tex](2V)p = 2N R T_i[/tex]
     
  4. Jun 19, 2012 #3
    This question is kind of stupid because if there is any kind of mass transfer between the two sections of the vessel, there will also be heat transfer making the problem quite difficult. If there is no heat transfer then as a first approximation you could apply the ideal gas law to each section of the vessel:

    PV = nRT;
    P = pressure
    V = volume
    n = number of molecules
    R = universal gas constant
    T = temperature of the section
     
  5. Jun 19, 2012 #4
    At the initial situation of the problem, the ideal gas law, as I suggested, can obviously be applied for the whole vessel. For the final situation, at equilibrium, meaning net transfer of heat being zero, the ideal gas law is applicable.
     
  6. Jun 19, 2012 #5
    You can have equilibrium with a temperature gradient inside both the chambers, which again would make the problem much more difficult. If you want to solve the problem at "steady state" you need to solve the heat equation and determine the temperature distribution inside both vessels. It may not vary with time but it certainly wont be an abrupt change at the interface. of the wall.
     
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