Calculating Change in Entropy for Gas Mix

[ajbeach2]

Take two containers, each with volume 0.1 m3, containing ideal gases of monatomic He at T= 113 K and diatomic N2 at T = 298 K, respectively. Each gas is at pressure p= 1000000 Pa. A valve is opened allowing these two gases to mix. They are kept thermally isolated from the outside.

1) Now we want to calculate the change in entropy during this process. What is the change of dimensionless entropy solely due to the fact that the gases have more volume accessible?


2). What is the change of dimensionless entropy solely due to the fact that the gases' temperatures have changed?


3) What is then the total change of dimensionless entropy?


4) What is the change of the standard entropy in this process?



Equations: PV = nkt
Entroy = ln(omega)
omega = total number of microstates
k = 1.381e-23

change in entropy = Number of particles * ln(Volume final/volume initial)




i calculated the number of particles for gas A as 6.408079306e25 and for B 2.429909267e25

i used this in the change in entropy equations which relates the change in entropy to the number of particles times the ln of the change in volume. For the first question i got
6.126026862e25 which is right. I don't know how to calculate parts 2,3 or 4

 

[Albert11317]

8 years later, and I'm struggling with the exact same problem.

 

[Chestermiller]

8 years later, and I'm struggling with the exact same problem.

If you let the two gases thermally equilibrate with one another, what would the final temperature be?