Entropy change when mixing two gases

[toothpaste666]

Homework Statement

1.00mole of nitrogen (N2) gas and 1.00mole of argon (Ar) gas are in separate, equal-sized, insulated containers at the same temperature. The containers are then connected and the gases (assumed ideal) allowed to mix.
A) What is the change in entropy of the system?
B) What is the change in entropy of the environment?
C)Repeat part A but assume one container is twice as large as the other.

Homework Equations

dS = dQ/T
PV=nRT
dE = dQ - dW

The Attempt at a Solution

This problem is stumping me. This is my attempt. since the T is constant, the internal energy won't change

so dQ = dW
dQ = PdV

Q = \int_{Va}^{Vb} PdV = nRT \int_{Va}^{Vb} \frac{dV}{V}

Q = nRTln\frac{Vb}{Va}

s = Q/T so

s = nRln\frac{Vb}{Va}

the final volume is twice the initial so

s = nRln\frac{Vb}{Va} = nRln\frac{2Va}{Va} = nRln(2)

where n is 2 mols because you combine the mols of each gas.
and this would be the entropy change of the system? for the environment i think it would be 0 because the system is isolated. Am I on the right track?

 

[Chestermiller]

Yes. Since they are in the ideal gas region, you knew to treat each gas separately. Nice analysis.

Chet

 

[toothpaste666]

thank you!