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

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To calculate the equilibrium conditions after mixing two different gases in adiabatic containers, one must use the conservation of energy alongside the ideal gas law. The internal energy of a monoatomic gas, which depends solely on temperature, can be expressed as E = (3/2)nRT. Since the process is irreversible, calculating the work done by the gases can be complex, and one cannot assume constant entropy. Clarification on whether the gases are the same and if thermal equilibration is allowed is also essential for accurate calculations. Understanding these principles is crucial for determining the final pressure and temperature after mixing.
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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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