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

• I
• jaumzaum
In summary: If so, then you can use the Clausius-Clapeyron equation which is:$$P_{eq} = P_{T1} + \frac{Q_{T1}}{V_{T1}}$$In summary, the equilibrium pressure and temperature are found by solving the Clausius-Clapeyron equation for pressure and temperature, respectively.
jaumzaum
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!

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?

• Thermodynamics
Replies
58
Views
4K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
771
• Thermodynamics
Replies
22
Views
2K
• Thermodynamics
Replies
3
Views
1K
• Advanced Physics Homework Help
Replies
5
Views
2K
• Engineering and Comp Sci Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
844
• Introductory Physics Homework Help
Replies
5
Views
1K