Calculating Change in Entropy for Gas Mix

In summary, the conversation discusses mixing two containers of ideal gases with different temperatures and volumes, and calculating the change in entropy during the process. The change in dimensionless entropy is calculated by using the number of particles and the change in volume, and for the first question, the value obtained was 6.126026862e25. The second and third questions involve calculating the change in dimensionless entropy solely due to the change in temperature and the total change of dimensionless entropy, respectively. The fourth question asks for the change in standard entropy during the process. The equations used are PV = nkt, where k = 1.381e-23, and entropy = ln(omega), where omega is the total number of microstates.
  • #1
ajbeach2
1
0
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
 
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  • #2
8 years later, and I'm struggling with the exact same problem.
 
  • #3
Albert11317 said:
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?
 

1. What is entropy and why is it important in gas mix calculations?

Entropy is a measure of the disorder or randomness in a system. In gas mix calculations, it is important because it helps us understand the behavior of gases under different conditions and allows us to predict how they will change over time.

2. How is entropy calculated for gas mixtures?

Entropy for gas mixtures can be calculated using the formula S = -R∑(n_i ln(n_i)), where S is the entropy, R is the gas constant, and n_i is the mole fraction of each gas component. This formula takes into account the number of gas molecules and their distribution in the mixture.

3. What factors affect the change in entropy for gas mixtures?

The change in entropy for gas mixtures is affected by temperature, pressure, and the types and amounts of gases present. Generally, an increase in temperature or pressure leads to an increase in entropy, while a decrease in temperature or pressure leads to a decrease in entropy.

4. Can entropy be negative for gas mixtures?

Yes, entropy can be negative for gas mixtures. This occurs when the system becomes more ordered or less random, which is often the case in processes such as compression or mixing of gases with different molecular sizes.

5. How can the change in entropy for gas mixtures be used in practical applications?

The change in entropy for gas mixtures can be used to predict the direction and extent of chemical reactions, as well as the behavior of gases in industrial processes such as combustion and refrigeration. It is also useful in understanding the thermodynamic properties of gas mixtures and developing efficient energy systems.

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