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Homework Statement
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.
Homework Equations
##pV=nRT##
Internal energy (ideal gas)
##U = nC_VT##
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.