Maximum work obtained by mixing 2 gases

In summary, the maximum mechanical work that can be obtained is the difference in pressure between the two chambers.
  • #1
superduper
2
0

Homework Statement


2 boxes containing the same number of moles of 2 ideal identical gases with the same adiabatic index (this is given as gamma), at the same initial temperature Ti but with different volumes, V1 and V2 are brought together. Find the maximum mechanical work that can be obtained.

Homework Equations

The Attempt at a Solution


If the gases have all the parameters identical but the volumes, that means that they also have different pressures, so when we are mixing them, the gas with the higher pressure will do work on the gas with lower pressure. However, I have no idea how to calculate this work and the answer given is a big messy expression. It dosent say anything about the recipients being adibatically isolated, but I guess I have to assume that? The temperature will be constant? I think I should calculate the variation of entropy for the system and then relate this to the first principle to get the work done, but I have no idea how to do that.
 
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  • #2
Hi again. Welcome to Physics Forums.

I don't think the problem intends for you to assume that the gases are intimately mixed. I think it intends for you to assume that there is an insulated partition separating the two gases, and to find the maximum amount of work that they can do on the partition. If I'm right, my solution to this problem should match the given answer. Do you want me to reveal my solution so that you can compare it with the given answer?

If my interpretation is correct, then in both chambers, an adiabatic reversible expansion occurs. So, in each chamber, PVν=constant. Let's focus on chamber 1. Let ##P_0## represent the initial pressure nRT/V1, let P represent the final pressure when the pressures on both sides of the partition have equilibrated, and let ##V_{1f}## represent the final volume. In terms of ##P_0##, V1, and ##V_{1f}##, what is the final pressure P in chamber 1?

Chet
 
Last edited:
  • #3
Maybe you can provide the expression for the given answer, and we can try to determine how it was derived.

Did the expression have a bunch of gamma's and volume ratios in it?
 
  • #4
Does it look anything like this:

$$W=nRT_i\left[2-\left(\frac{1+(V_2/V_1)^{\frac{\gamma-1}{\gamma}}}{1+(V2/V_1)}\right)^{\gamma-1}-\left(\frac{1+(V_1/V_2)^{\frac{\gamma-1}{\gamma}}}{1+(V1/V_2)}\right)^{\gamma-1}\right]$$
 
  • #5
Sorry for the late response, but I'm away from uni and don't have the book at my disposal to see if that's the exact answer, but, yes, I remeber it has a bunch of gammas and volume ratios involved
 
  • #6
superduper said:
Sorry for the late response, but I'm away from uni and don't have the book at my disposal to see if that's the exact answer, but, yes, I remeber it has a bunch of gammas and volume ratios involved
So, where do you want to go from here? How would you like me to help you?
 

1. What is the concept of maximum work obtained by mixing 2 gases?

The concept of maximum work obtained by mixing 2 gases is related to the thermodynamic process of mixing two different gases at different temperatures and pressures. When these gases are mixed, there is a potential for work to be done, which can be calculated using thermodynamic principles.

2. How is maximum work obtained by mixing 2 gases calculated?

The maximum work obtained by mixing 2 gases is calculated using the Gibbs free energy equation, which takes into account the initial and final temperatures, pressures, and molar fractions of the gases. This calculation can be complex and is typically done using computer simulations or specialized software.

3. What factors affect the maximum work obtained by mixing 2 gases?

The maximum work obtained by mixing 2 gases is affected by several factors, including the initial and final temperatures and pressures of the gases, the types of gases being mixed, and the molar fractions of each gas. The efficiency of the mixing process and any external energy inputs can also impact the maximum work obtained.

4. Can the maximum work obtained by mixing 2 gases ever be negative?

No, the maximum work obtained by mixing 2 gases cannot be negative. This is because the Gibbs free energy equation only produces positive values for work. However, it is possible for the actual work obtained by mixing 2 gases to be less than the maximum calculated value due to inefficiencies or external factors.

5. What are some real-world applications of maximum work obtained by mixing 2 gases?

The concept of maximum work obtained by mixing 2 gases has applications in various industries, such as power generation, refrigeration, and chemical processes. For example, in power plants, the mixing of hot and cold gases can be used to drive turbines and generate electricity. In refrigeration systems, the mixing of gases at different pressures and temperatures can be used to produce cooling effects. Additionally, the concept is important in understanding and optimizing chemical reactions and processes.

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