[gingi85]

How would I go about calculating Q for isothermal free expansion in a van der Waals gas?

[Hunt]

T is const and the expansion W = 0 , so dQ = dU and the expression of U becomes dependent only on the volume. You can write that dU= Pi dV where Pi = internal pressure of the gas. You can derive that Pi = a n^2/V^2 for a van der waal gas and then integrate to get the expression for Q.

[gingi85]

. and the expression of U becomes dependent only on the volume. You can write that dU= Pi dV where Pi = internal pressure of the gas .

How do you get that U is dependent only on volume if dU = dQ? And how did yu get the expression dU = Pi dV?

Thanks. Please excuse my ignorance.

[Yggdrasil]

In thermodynamics problems, it is always helpful to start by choosing the thermodynamic potential you'd like to work with, and your set of thermodynamic variables.

In this case, we obviously want to look at internal energy.  Now, when you look at internal energy, the natural choice of variables is S and V because then the total differential reduces to:

dU = T dS - p dV

However, because T is fixed, it makes sense to choose T as one of our variable of choice instead.  So, to begin we write U as a function of T and V:

U = U(T,V)

and then write convert this to a differential equation:

dU = (dU/dT)_v dT + (dU/dV)_T dV

Since (dU/dT)_v = C_v by definition, this simplifies to:

dU = C_v dT + (dU/dV)_T dV

As a side note, let's consider what happens in the case of an ideal gas.  For an ideal gas, U depends only on temperature, so (dU/dV)_T = 0.  This makes our differential equation very simple:

dU = C_v dT

upon integration, we get

ΔU = C_v ΔT (more commonly written as ΔU = n C_v ΔT)

which should be a very familiar equation.


Anyway, back to the free isothermal expansion of a van der waals gas.  Here dT = 0 (because the process is isothermal), so our differential simplifies to:

dU = (dU/dV)_T dV

We know that:

(dU/dV)_T = T (dP/dT)_v - P

(I can write out the full derivation of this equation if you want, I just need to check my notes to make sure it's all correct), and you can calculate (dP/dT)_v for a van der Waals gas.  So, you should be able to take it from this point on.

[Hunt]

Pi = internal pressure of a gas = (dU/dV)_T
From dimensional analysis , energy over volume = pressure. The internal pressure is used to account for attractive/repulsive forces in a gas.

For an ideal gas, Pi = 0 .
For a van der waal gas, Pi = a n^2/V^2

Inorder to get the expression for Pi , you use the equation Yggdrasil already posted :

(dU/dV)T = T (dP/dT)v - P

How to get this expression? You need to know the maxwell relations. ( Review any pchem/thermodynamics book for that )

dU = TdS - PdV ==> Pi=(dU/dV)_T= T(dS/dV)_T - P

dA = -SdT - PdV
Maxwell relation : (dS/dV)_T =  (dP/dT)_V

the best way to understand this is to get yourself a good thermodynamics book and go over the maxwell relations, see the math behind them, how they're used , etc.

As for the rest...Yggdrasil already explained thoroughly.