- #1
Emspak
- 243
- 1
Homework Statement
Consider the apparatus from the Gay-Loussac-Joule experiment. You have 2 containers of gas, with a valve between them. The volume of each container is V. Each vessel contains nA and nB moles of gas. Their temperatures (T1) are initially the same.
When the valve is opened and the gases are allowed to mix, find an expression that shows the change in temperature.
Homework Equations
We will assume a Van der Waals gas with an equation of state as follows:
##nRT = \left(P+\frac{a}{V^2}\right)(V-b)##
and the internal energy of both gases can be described by:
##u = c_VT - \frac{a}{V} + u_0##
The Attempt at a Solution
OK, I assumed we need the conservation of energy, so that there will be a ##u_i## (initial energy) and a ##u_f## (final).
I figured at first I should just add the energies of the two gases as follows:
gas 1: ##u = c_vT - \frac{a}{V} + u_0##
gas 2: ##u = c_vT - \frac{a}{V} + u_0##
and the total energy has to be the same as the van der Waals gas in a space 2x the volume, so ##u_f = c_vT - \frac{a}{2V} + u_0##.
I also know that the equation of state for the combined gas should be
##(n_A + n_B)RT = \left(P+\frac{a}{V^2}\right)(V-b)##
and fro there I should get a reasonable expression for the change in temperature.
The problem is I am not quite sure how to make the connection. My first attempt at finding a delta T was this:
##2u_i = 2c_vT_1 - 2\frac{a}{V} + 2u_0 = u_f = c_vT_2 - \frac{a}{2V} + u_0##
##2c_vT_1 - c_vT_2 = 2\frac{a}{V} - \frac{a}{2V} - u_0##
##(2T_1 - T_2)c_v = \frac{a}{2V} - u_0##
##(2T_1 - T_2) = \frac{a}{2Vc_v} - u_0##
But i am sensing that isn't right because I am not incorporating ##n_A## and ##n_B##.
Anyhow, I feel like I am almost getting i but there is some crucial step I am missing.
EDIT: I used twice the initial energy to account for there being two volumes of gas, and wasn't sure if I should change the a constants (it didn't seem like it given the parameters of the problem).
ANy help is much appreciated. Thanks.