Question

Mixing Gases Consider two containers, . Both have volume 0.1 m3, and pressure 106 pa One contains monatomic (3 degrees of freedom) He at T 128 K and One contains diatomic (5 degrees of freedom) N2 at T- 258 K. A valve is opened allowing these two gases to mix. They are kept thermally isolated from the outside You can treat them as i deal gases. 1) What is the change in internal energy under this process? Submit Help 2) What is the final temperature of the mixed gas? 186.84 K Submit Help 3) 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 gaseshave more volume accessible? Submit Help 4) What is the change of dimensionless entropy solely due to the fact that the gasses' temperatures have changed? AStemp/k-botemp Submit Hide Help Helpi You should ute fact tht(Nurn that into an integral :AS, ap for constant vo You should also use the fact that Cv o change variables in the integral. You know Cv from equipartition. NOTE: If you are sure you are doing the problem correctly, but the answer is still wrong, make sure that the temperature you calculated in part 3 is accurate. A small error there will give you the wrong answer here 5) What is then the total change of dimensionless entropy? Submit 6) What is the change of the standard entropy in this process? J/KI Submit

Asking for Q3,4,5,6.

Answer 1

(3)

$dimensionless \ entropy \ (\sigma _{vol}) = \frac{\Delta S_{vol}}{K} = Nln(\frac{V_{final}}{V_{initial}})$

where N is the total number of particles. since both the containers have identical volume (0.1 m³), final volume will be 0.2 m³.

total number of particles = 6.83 * 10²⁵

entropy = 8.83 * 10²⁵ * ln (2)

= 6.12 * 10²⁵