Change in temperature in the Gay-Loussac-Joule experiment

[Emspak]

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 n_A and n_B 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.

 

[BvU]

Initial total energies left and right are not equal: na must be different from nb (otherwise opening the valve doesn't do anything!). So starting with 2u1 is not a good idea.

Write an energy balance in full, using symbols for what you don't know. Then go looking for things that are the same, things that are related by an equation, etc.

 

[Emspak]

OK, looked at that way,

gas 1: $u_1 = c_vT_1 - \frac{a}{n_Av_1} + u_0$
gas 2: $u_2 = c_vT_1 - \frac{a'}{n_Bv_2} + u_0'$

T will be the same, I made $a$ and $a'$ differ because they are two different gases.

$u_1 + u_2 = u_f$ because the energy is conserved. $V = n_Av_1$ and $V = n_Bv_2$ because the $v$ is the volume of 1 mole and $V$ is the same (they are in equal-sized vessels). So $n_Av_1 = n_Bv_2$.

If I add the two I end up with $u_1 + u_2 = c_vT_1 - \frac{a}{n_Av_1} + u_0+ c_vT_1 - \frac{a'}{n_Bv_2} + u_0'$ which gets me
$2c_vT_1 - \frac{a}{n_Av_1} - \frac{a'}{n_Bv_2} + u_0' + u_0$

which should equal the energy when I am done letting the gases mix, which is

$u_f = c_vT_2 - \frac{a''}{(n_Av_1+n_bv_2)}+ u_0' + u_0$

But I wasnt sure if I was constructing the $u_f$ correctly here.

EDIT:
Putting it all together: $c_vT_2 - \frac{a''}{(n_Av_1+n_bv_2)}+ u_0' + u_0 = 2c_vT_1 - \frac{a}{n_Av_1} - \frac{a'}{n_Bv_2} + u_0' + u_0$

$c_vT_2 - \frac{a''}{(n_Av_1+n_bv_2)} = 2c_vT_1 - \frac{a}{n_Av_1} - \frac{a'}{n_Bv_2}$
$c_vT_2 - 2c_vT_1 = \frac{a''}{(n_Av_1+n_bv_2)} - \frac{a}{n_Av_1} - \frac{a'}{n_Bv_2}$

am I getting there..?

 

[BvU]

I see, you interpret "Each vessel contains n_A and n_B moles of gas" as n_A moles of gas A on the left and n_B moles of gas B on the right. This means you are after an internal energy of mixing or something. .

Sorry to have confused you in that direction with my total energy vs energy per mole.

Gay-Lusssac and Joule were just letting one gas expand adiabatically (against near-vacuum) to study if the internal energy of such a gas depends on the temperature AND the volume, or on the temperature only (ideal gas: temperature only). Within their error margins they found ideal gas behaviour, an indication that the Joule coefficient $\eta \equiv \left ( {dT\over dv} \right )_u$ is very small.

You'd better stick to different numbers of moles of the same gas in the two volumes to avoid two a's, two c_v's (!), two u₀ etc.

You also want to decide on the units of a, c_v and u₀ because now you end up with unwanted factors of two and a u₀.

Usually we take upper case for extensive properties (total energy U) and lower case for intensive properties (energy per mole u) if the distinction must be made. For e.g. temperature and pressure that isn't necessary.

Then: you don't need van der Waals any more, because that is already worked out in the expression for u.

With all that, the exercise is a piece of cake: write out $n_A u_A + n_B u_B = (n_A+n_B) u_f$, work around to an expression for $T_1 - T_2$ and there is the Joule coefficient !

 

[Emspak]

According to the exercise, there are two different gases tho. It says specifically that you are allowing them to mix. Does the same ting apply then? The answer we are "supposed" to get looks like this:

$$\Delta T = \frac{(2n_An_B - n^2_A - n^2_B) a}{2c_vV(n_A+n_B) }$$

which would say to me that we are supposed to treat a as equal between the two gases, as well as $c_v$.

 

[BvU]

Ah, there is an answer that we have to reproduce... Funny answer: one a, one c_v, u₀ gone as well. Numerator is $-(n_A - n_B)^2$. Wonder how the difference comes in...
Could it be that there is something like $u = c_VT - \frac{a}{V} + u_0$ should have been $u = c_VT - \frac{a}{v} + u_0$ ?