Thermodynamics: Two gases in a container

[heyhey281]

Homework Statement

Consider
a) two gases, each in a container with volume V
b) two gases in a container with volume V
Which system is more disordered, where is the entropy higher?

Relevant Equations

Ideal gas equation
(Van der Waals equation)

Ideal gas:
If the gases are of different type, I would say the entropy stays the same. The total entropy is in both cases just the sum S = S1 + S2, where S1 is the entropy of the first gas and S2 the entropy of the second gas.
If the gases are of the same type, I think the entropy change is also 0.

(I am not too sure if you can even distinguish ideal gases / call them different types?)

Real gas:
Due to the intermolecular forces, I think the molecules might "cluster" a little bit more which would lower the entropy? But on the other hand, I could first put both gases into the common volume V and hypothetically assume that there is absolutely no interaction between them ("switch off" the interaction between them before I mix them). Then I could “switch on” the interaction and wait until an equilibrium is reached. Since the system is isolated, the entropy cannot have possibly decreased?

 

[kuruman]

I am not too sure if you can even distinguish ideal gases - - -

In what sense? Your body does it all the time. You keep some of the oxygen that you breathe in and you breathe out carbon dioxide.

More to the point, I think you need to consider entropy of mixing.