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

  • I
  • Thread starter Thread starter jaumzaum
  • Start date Start date
  • Tags Tags
    Gases Mixing
AI Thread Summary
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.
jaumzaum
Messages
433
Reaction score
33
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!
 
Science news on Phys.org
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?
 

Thread 'Condensate rate of water for an air conditioning cooling coil'

I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
View full post »
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'

Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'

Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
View full post »

Similar threads

I Movable Wall in an Adiabatic System: Same Final Temperature?
Replies
8
Views
1K
I Focus Problem for Entropy Change in Irreversible Adiabatic Process
2
Replies
60
Views
9K
Mixing two ideal gases with different V, T at constant pressure
Replies
16
Views
3K
Chemistry How does it work to mix two gases reversibly in this device?
Replies
1
Views
1K
Mixing two gases in an isolated system and calculating final p and T
Replies
4
Views
2K
I How to calculate the mass of gas in a tank?
3
Replies
109
Views
7K
I Irreversible adiabatic processes & entropy change (clarification needed)
Replies
22
Views
5K
B Energy seems to disappear when gas is compressed?
Replies
3
Views
2K
I Graphical difference between adiabatic and isothermal processes.
Replies
1
Views
3K
I Entropy after removing partition separating gas into two compartments
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top