Mixing two different gases

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.
  • #1
jaumzaum
434
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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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  • #2
Can I say the entropy is constant?
 
  • #3
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.
 
  • #4
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?
 

1. What happens when you mix two different gases?

When two different gases are mixed, they will form a homogeneous mixture. This means that the gases will be evenly distributed throughout the mixture, and there will be no visible separation between the two gases.

2. How do you determine the resulting pressure when mixing two different gases?

The resulting pressure of a mixture of two gases can be determined using the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of moles present.

3. Can two different gases react when mixed together?

Yes, two different gases can react when mixed together if the conditions are right. For example, if the gases are at the correct temperature and pressure, and if they are chemically reactive, they may undergo a chemical reaction and form a new compound.

4. What factors can affect the mixing of two different gases?

The mixing of two different gases can be affected by factors such as temperature, pressure, and the properties of the gases themselves. For example, gases with similar molecular weights will mix more easily compared to gases with very different molecular weights.

5. How does the mixing of two different gases affect their individual properties?

When two different gases are mixed, their individual properties may change. For example, the boiling point and melting point of the mixture may be different from the individual gases, and the density and solubility of the mixture may also be different. These changes are due to the interactions between the molecules of the two gases.

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