1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Temperature and entropy for two gases mixing

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A system is made up of two halves. In one there's 10kg neon gas with the temperature ##20 \circ##C, in the other 10kg nitrogen gas with the temperature ##100 \circ## C. Suppose the septum is removed so that thermodynamic equilibrium may appear and the gases mix.

    Calculate the final temperature ##T_f## and the total change in entropy.

    2. Relevant equations
    ##pV=nRT##
    Internal energy (ideal gas)
    ##U = nC_VT##

    3. The attempt at a solution
    I got two solutions (for the first question), one done by me and one by our lecturer (who I unfortunaly can't ask) and what I'm wondering why mine is wrong (if it is).
    My solution:
    ##n_1 = \frac{10\cdot 10^3}{20.17} = 495.8 mol##
    ##n_2 = \frac{10 \cdot 10^3}{2*14} = 357 mol##
    Assuming the gas is ideal we get
    ##p_1=n_1RT_1/V_1## and ##p_2 = n_2RT_2/V_1##
    Now assuming the gases where alone one at a time in the container double the volume we would get
    ##p_{f1} = n_1RT_f/(2V_1)## and ##p_(f2) = n_2RT_f/(2V_1)##
    Assuming the mixing is done first by an isothermal and then and isokor process (our teacher made the exact same assumption as well)
    ##p_f = p_{f1} +p_{f2} = \frac{n_1RT_1}{2V_1} + \frac{n_2RT_2}{2V_2} = \frac{R}{2V_2}(n_1T_1+n_2T_2)##
    I now make the assumption that pressure from both gases is equal to the pressure from each of the gases added together (I don't know if this is correct but I think there is some thermodynamic principle that allow me to use the superposition principle here).
    ##p_f\cdot 2V_1 = (n_1+n_2)RT_f##
    Inserting ##p_f## into the equation gives us
    ##T_f = \frac{n_1T_1+n_2T_2}{n_1+n_2}##

    While the one done by teacher uses internal energy and instead end up at the answer of
    ##T_f = \frac{3/2n_1T_1+5/2n_2T_2}{3/2n_1+5/2n_2}##


    I could add the solution from our teacher too if it's needed (it's rather lengthy) but I'm mostly wondering if anyone could see what I do wrong. Both solutions give numeric values well within the solution at the back of the book but I would assume I'm wrong, I just don't know why.
     
  2. jcsd
  3. Mar 26, 2015 #2
    Your assumption that, after mixing, the partial pressures of the two gases is the same is incorrect.

    The key to this problem is recognizing that there is no change in internal energy. That's what your teacher did. You can see that from the 5/2 and 3/2 factors.

    Chet
     
  4. Mar 26, 2015 #3
    Thanks again Chet, always helping me out!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Temperature and entropy for two gases mixing
  1. Mixing of gases (Replies: 2)

Loading...