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

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SUMMARY

The discussion focuses on calculating equilibrium conditions after mixing two different gases in adiabatic containers. The key equations involved are the Clapeyron Law, which is synonymous with the ideal gas law, and the conservation of energy equation, represented as dU1 + dW1 + dU2 + dW2 = 0. The internal energy of a monoatomic gas is given by E = (3/2)nRT, which is crucial for determining the final temperature. It is established that entropy cannot be assumed constant during this irreversible transformation.

PREREQUISITES
  • Understanding of the Clapeyron Law (Ideal Gas Law)
  • Knowledge of conservation of energy principles
  • Familiarity with internal energy calculations for monoatomic gases
  • Basic concepts of adiabatic processes
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  • Study the derivation and applications of the Clapeyron Law in thermodynamics
  • Learn about the conservation of energy in irreversible processes
  • Explore the calculation of work done by gases during transformations
  • Investigate the implications of entropy changes in gas mixtures
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Students and professionals in thermodynamics, chemical engineering, and physical chemistry who are involved in gas behavior analysis and equilibrium calculations.

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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