How Do You Calculate Equilibrium Conditions After Mixing Two Different Gases?

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Discussion Overview

The discussion revolves around calculating the equilibrium conditions after mixing two different gases in adiabatic containers. Participants explore the necessary equations and principles to determine the final pressure and temperature of the mixed gases, considering factors such as conservation of energy and the behavior of gases during irreversible transformations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using the Clapeyron Law and conservation of energy to find the equilibrium conditions after mixing the gases.
  • Another participant questions the assumption of constant entropy during the process.
  • A different participant asserts that the Clapeyron Law is synonymous with the ideal gas law and emphasizes that the internal energy of a monatomic gas depends solely on temperature.
  • There is a query regarding the calculation of work done by the gases during the irreversible transformation.
  • One participant seeks clarification on whether the gases in the two containers are the same and if thermal equilibration is allowed.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions regarding entropy and the applicability of the Clapeyron Law. The discussion remains unresolved, with multiple competing perspectives on the equations and principles to apply.

Contextual Notes

Participants have not reached consensus on the assumptions regarding entropy and the specific equations to use, highlighting potential limitations in their approaches.

jaumzaum
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Consider that I have 2 adiabatic containers, one with a monoatomic gas at pressure P1, volume V1 and temperature T1 and another with pressure P2, volume V2, temperature T2. If I open a valvule and mix the two gases, how do I calculate the equilibrium pressure and temperature?

I know the final quantity of moles, and the final volume, so I need a system of 2 eq to calculate the temperature and pressure. One is the Clapeyron Law, what is the other? I mean, what variable remains constant in this system?

If I try to solve it using conservation of energy:
dU1+dW1+dU2+dW2=0

But how would I calculate the work done by the gases, as this is an irreversible transformation?Thanks!
 
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Can I say the entropy is constant?
 
You can definitely not assume that entropy is constant. Furthermore, the Clapeyron Law apparently is apparently just another name for the ideal gas law. I never heard the name Clapeyron Law before...

The other equation you need is just the conservation of energy. The internal energy of a monatomic gas is:
$$E = \frac{3}{2}nRT$$
This only depends on temperature. So with this you can compute the temperature, then with the ideal gas law you can compute the pressure.
 
jaumzaum said:
Consider that I have 2 adiabatic containers, one with a monoatomic gas at pressure P1, volume V1 and temperature T1 and another with pressure P2, volume V2, temperature T2. If I open a valvule and mix the two gases, how do I calculate the equilibrium pressure and temperature?

I know the final quantity of moles, and the final volume, so I need a system of 2 eq to calculate the temperature and pressure. One is the Clapeyron Law, what is the other? I mean, what variable remains constant in this system?

If I try to solve it using conservation of energy:
dU1+dW1+dU2+dW2=0

But how would I calculate the work done by the gases, as this is an irreversible transformation?Thanks!
Is the gas the same in the two containers? Are you allowing the two containers to equilibrate with one another thermally?
 

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