How Does Gas Expansion Relate to Changes in Molar Internal Energy and Enthalpy?

[CAF123]

Homework Statement

A rigid thick walled insulating chamber containing a gas at a high pressure $P_i$ is connected to a large insulating empty gas holder where the pressure is held constant at $P_A$ with a piston. A small valve between the two chambers is opened and the gas flows adiabatically into the cylinder. Prove $n_i u_i - n_f u_f = h' (n_i - n_f)$, where $n_i$= no. of moles of gas initially in chamber, $n_f$ = no. of moles of gas left in chamber, $u_i$ = molar internal energy of gas initially in chamber, $u_f$ = molar internal energy of gas left in chamber and $h'$ = final molar enthalpy of gas in cylinder.

Homework Equations

First Law of thermodynamics,
State function H = U + PV

The Attempt at a Solution

The LHS side of the show that is the change in internal energy of the gas left in the chamber. It decreases because the gas does work expanding against the constant pressure $P_A$. Hence $$\Delta U = W = P_A \int_{V_i}^{V_f} dV,$$ where $V_i$ is the volume occupied by gas after expansion and $V_f$ = volume occupied afterwards.

I can write this as $P_AV_f - P_A V_i = P_A(V_i + V') - P_AV_i$ since $V_f$ is composed of both chamber and cylinder. Terms cancel to give $$ W = P_AV' = P_A \left(\frac{(n_i - n_f)RT}{P_A}\right)$$, but this does not seem close to the show that.

Many thanks.

 

[Chestermiller]

Let u' be the final molar internal energy of the gas in the cylinder. In terms of u_i and n_i, what is the initial internal energy of the gas in the combined system? In terms n_i, u_f, n_f, and u', what is the final internal energy of the combined system? In terms of P_A and V (where V is the final volume of gas within the cylinder), how much work was done? Write the first law expression involving the change in internal energy of the combined system and the work which was done.

Chet

 

[CAF123]

Let u' be the final molar internal energy of the gas in the cylinder. In terms of u_i and n_i, what is the initial internal energy of the gas in the combined system?

The initial state is just composed of the chamber so that is $n_i u_i$.

In terms n_i, u_f, n_f, and u', what is the final internal energy of the combined system?

That is $u_f n_f + (n_i -n_f)u'$

In terms of P_A and V (where V is the final volume of gas within the cylinder), how much work was done?

I think it is what I derived above: $P_A V$

Write the first law expression involving the change in internal energy of the combined system and the work which was done.

So, $$u_f n_f + (n_i - n_f)u' - n_iu_i = P_AV \Rightarrow u_f n_f - u_i n_i = P_A V - (n_i - n_f)u'$$
I can relate $h' = u' + P_AV$ and sub for $P_AV$ but this is not quite the result either.

Thanks.

 

[Chestermiller]

The initial state is just composed of the chamber so that is $n_i u_i$.


That is $u_f n_f + (n_i -n_f)u'$


I think it is what I derived above: $P_A V$



So, $$u_f n_f + (n_i - n_f)u' - n_iu_i = P_AV \Rightarrow u_f n_f - u_i n_i = P_A V - (n_i - n_f)u'$$
I can relate $h' = u' + P_AV$ and sub for $P_AV$ but this is not quite the result either.

Thanks.

Very nice job. Now, only one more step. Recognize that V is the total volume of the cylinder, not the final volume per mole. Call the final volume per mole v'. What is v' in terms of the n's and V?

Chet

 

[Nickie]

Your approach is on the right track, but there are a few things that need to be clarified. First, the equation you are trying to prove is actually the definition of enthalpy, not a consequence of the first law of thermodynamics. So instead of trying to prove it, you should use it as a starting point and then apply the first law to derive the desired equation.

Secondly, the equation you have written for the work done by the gas is not correct. The gas is expanding into a larger volume, so the work done by the gas should be positive, not negative. Also, the integration limits should be from the initial volume to the final volume, not the other way around.

With those corrections, the equation for the work done by the gas should be:

$$W = P_A \int_{V_i}^{V_f} dV = P_A (V_f - V_i)$$

Now, using the definition of enthalpy, we can write:

$$h' = u' + P_A V'$$

where $u'$ is the molar internal energy of the gas in the cylinder and $V'$ is the volume of the gas in the cylinder. We can also write the initial and final molar internal energies of the gas in the chamber as:

$$u_i = u' + u_{ch}$$

$$u_f = u' + u_{ch}'$$

where $u_{ch}$ and $u_{ch}'$ are the molar internal energies of the gas in the chamber before and after the expansion, respectively. Substituting these equations into the desired equation, we get:

$$n_i u_i - n_f u_f = n_i (u' + u_{ch}) - n_f (u' + u_{ch}') = n_i u' + n_i u_{ch} - n_f u' - n_f u_{ch}' = (n_i - n_f) u' + n_i u_{ch} - n_f u_{ch}'$$

Now, using the first law of thermodynamics, we know that the change in internal energy of the gas is equal to the work done by the gas, so we can write:

$$n_i u_{ch} - n_f u_{ch}' = W = P_A(V_i + V') - P_AV_i = P_A V'$$

Substituting this into the previous equation