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Thermodynamics final temperature of gas

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed. One vessel of volume 16.8 L contains oxygen at a temperature of 300 K and a pressure of 1.75 atm. The other vessel of volume 22.4 L contains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is opened, the gases in the two vessels mix, and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure?
    (Ans. (a) 380 K , (b) 2.06 x 105 Pa or 2.04 atm)


    2. Relevant equations



    3. The attempt at a solution
    The teacher used the method of
    heat in=heat out
    (m1)c(T-300)=(m2)c(450-T)
    and obtain T=380K
    I did it by
    Total Energy of system is conserved.
    For an ideal gas, U=KE
    3/2(n1)R300+3/2(n2)R450=3/2(n1+n2)RT
    T=380K
    Is the method i used correct?
    And how come the teachers method is correct? The pressure is different in the vessels. So shouldn't that be taken into account?
     
  2. jcsd
  3. Oct 11, 2011 #2
    well both the final temperature and final pressure of the two systems are the same at the end, which is the time you are calculating for.

    you used conservation of energy, and the teacher used conservation of heat. There are two energy components to U (the total internal energy). Those two components are heat, and work. Work is only done if there is change in volume, but in this case the vessels do not grow or shrink. So the change in heat will be equal to the change in total internal energy will be equal to 0 for this case(thermally insulated). This change being 0 allows you to easily relate the final and initial energy values.

    EDIT: To be more specific as to why your equations are the same, let me try to put it this way.
    He used mass, you used moles. Those have a linear relationship with each other since it is oxygen thats in both tanks. You used R, he used c, both constants that will give the units of energy. Then you used temperature, and he used temperature. So as you can see, both formulas are quite similar.
     
    Last edited: Oct 11, 2011
  4. Oct 11, 2011 #3

    Andrew Mason

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    Your method is correct but you are using the incorrect heat capacity. [itex]O_{2}[/itex] is a diatomic gas and its heat capacity at constant volume is 5R/2 in this temperature range. But, since heat capacities cancel out it doesn't matter.

    Your method is equivalent to your teacher's method. Your teacher is using the heat capacity per unit mass and you are using the molar heat capacity. Since the two gases are the same they all cancel out:

    m1c(T-300) = m2c(450-T)
    m1cT - m1c(300) = m2c(450) - m2cT
    (m1+m2)T = m2(450)+m1(300)

    AM
     
  5. Oct 11, 2011 #4
    I am not using molar heat capacity? on my notes it says Kinetic energy=3/2nRT??. But i am confused with the teacher's method. Is heat dependent on pressure? the teachers method is i see it as putting solution1 of temp1 and solution2 of temp2 together and finding final temp. But with gas of differernt initial pressures, this method still works without considering the pressures ?
     
  6. Oct 11, 2011 #5
    You're asking a good question. Yes the pressure and the heat are related, which I believe covers the missing link in your understanding. you have constant pressure, constant temperature, constant volume, and constant moles inside the whole system at its final time. You know what all of these values are, therefore you will be able to find the temperature. He didn't need to take into account pressure, just as he didn't take into account Volume because PV = nRT. He has nRT, just in the form of m.c.dT

    Your teacher is saying that all of the heat lost by the hotter gas must be gained by the cooler gas. What he is doing is simply calculating the total heat inside the system, and given all of those constant parameters i mentioned in the first paragraph, he can find the final temperature.

    His equation is very similar to your equation. It is like this:

    (m1)c(T1) + (m2)c(T2) = (m1+m2)c(Tfinal)
     
  7. Oct 11, 2011 #6
    "He didn't need to take into account pressure, just as he didn't take into account Volume because PV = nRT. He has nRT, just in the form of m.c.dT." Thank you.
     
  8. Oct 11, 2011 #7

    Andrew Mason

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    I thought you were using 3R/2 as the heat capacity. I see that you were using a kind of principle of conservation of translational kinetic energy. Unfortunately, there is no such principle. So your method will only work if the two gases have the same heat capacity, Cv.

    For example, it will not work if one gas is He and the other is O2. The internal energy of He is stored as translational KE. However, O2 stores kinetic energy as rotational as well as translational KE. But only translational KE determines temperature. So for a given change in internal energy there will be a greater change in the temperature of He than of O2.

    AM
     
    Last edited: Oct 11, 2011
  9. Oct 12, 2011 #8
    Ok, today the lecturer started on kinetic theory of gasses, i realised jR/2 is the heat capacity. and for oxygen, j=5. Thanks.
     
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