Mixing gas and find the temperature

In summary, the problem involves two different types of monoatomic ideal gas, A and B, in separate insulated compartments with an insulated partition between them. The initial temperature and pressure of A is known (Ta, P) and the initial temperature and pressure of B is known (Tb, P). The final temperature after removing the partition is to be found, assuming Tb>Ta and equal initial volumes for each compartment. The correct equation to use is the energy relation, with the capacity heat (C) being either isobaric (Cp) or isovolume (Cv). The equation \Delta U=c_Vn\Delta T=(c_P-R)n\Delta T always applies for an ideal gas, regardless of the process.
  • #1
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Homework Statement


Assume there are two different kind of monoatomic ideal gas A and B. n moles of A are placed in a insulated compartment while m moles of B are placed in another insulated compartment. There is an insulated partition to join this two compartment. If the initial temperature and pressure of A is known (Ta, P) and the initial temperature and pressure of B is know (Tb, P), find the final temperature after the partition was removed. Assuming Tb>Ta, two initial volume for each compartment are the same.

2. The attempt at a solution
I know I should apply the energy relation to find out final temperature. That is, the energy absorbed by the low temperature gas is identical to the energy released by the high temperature gas. If we know the capacity heat, it is easily to write down some thing like

n C (Tf-Ta) + m C (Tf-Tb) = 0

I think this relation is correct, right?

Now, I just wondering how to solve for C in this case. There is two different capacity for ideal gas: isobar (Cp) or isovolumn (Cv). However, since while the gas is mixing the pressure must change and volume will also be change, it seems that both Cp or Cv is not suitable. How to find the capacity heat?
 
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  • #2
Constant volume. The equation

[tex]
\Delta U=c_Vn\Delta T=(c_P-R)n\Delta T
[/tex]

always applies for an ideal gas, regardless of the process. (I know this can be confusing. But [itex]U[/itex] is a state variable; it is process independent. [itex]c_V[/itex] and [itex]c_P[/itex] are simply variables that are used in this state equation, not constraints on the type of process.) Does this make sense?
 
  • #3
Mapes said:
Constant volume. The equation

[tex]
\Delta U=c_Vn\Delta T=(c_P-R)n\Delta T
[/tex]

always applies for an ideal gas, regardless of the process. (I know this can be confusing. But [itex]U[/itex] is a state variable; it is process independent. [itex]c_V[/itex] and [itex]c_P[/itex] are simply variables that are used in this state equation, not constraints on the type of process.) Does this make sense?

Use Cv, I do can find the correct answer. Still quite confusing ... does it mean if only I know the temperature difference, no matter what process it is, the heat absorbed/released by the ideal gas always equal to [tex]Cv\Delta T[/tex] ? What also confuses me is: is this sense, why don't use [tex]C_p\Delta T[/tex] instead?
 
  • #4
[itex]C_V[/itex] is the right variable to use because (whether you knew it or not) you're solving the problem by splitting it into two reversible processes: (1) isothermal expansion to the final volume and (2) isochoric heating or cooling. The isochoric heating or cooling is where [itex]\Delta U=C_V\Delta T[/itex] comes in.

You could also solve the problem by assuming isobaric heating or cooling ([itex]\Delta H=C_P\Delta T[/itex]), and you'd get the same answer. It's a little more involved because an additional intermediate variable (the volume at the end of the isobaric process) comes into play. It's a good idea to try to solve the problem using a couple different approaches, though, to make sure everything's straight in your head.
 
  • #5
Mapes said:
[itex]C_V[/itex] is the right variable to use because (whether you knew it or not) you're solving the problem by splitting it into two reversible processes: (1) isothermal expansion to the final volume and (2) isochoric heating or cooling. The isochoric heating or cooling is where [itex]\Delta U=C_V\Delta T[/itex] comes in.

You could also solve the problem by assuming isobaric heating or cooling ([itex]\Delta H=C_P\Delta T[/itex]), and you'd get the same answer. It's a little more involved because an additional intermediate variable (the volume at the end of the isobaric process) comes into play. It's a good idea to try to solve the problem using a couple different approaches, though, to make sure everything's straight in your head.

Thanks a lot.
 
  • #6
Sorry to bother again. According to your hint, I want to extend the problem to find the total entropy changed. I know the free expansion is not a reversible process so we cannot use the definition of entropy to calculate the total change of entropy. But since entropy is state variable, for the gas in the left compartment, I imagine two processes such that these process has same initial state and final state with free expansion.

1) The gas expand from V to 2V with the temperature unchanged (isotherm)
2) the expanded gas being heated up by absorbing some amount of heat (isochoric)

For the first process, since temperature doesn't change, the internal energy doesn't change, so [tex]dQ = PdV[/tex], we find the change of entropy

[tex]
\Delta S_1 = \int_V^{2V}\frac{PdV}{T} = nR\int_V^{2V}\frac{dV}{V} = nR\ln 2
[/tex]

For second process, the volume doesn't change, hence, [tex]dU = dQ[/tex]

[tex]
\Delta S_2 = \int \frac{dQ}{T} = \int \frac{dU}{T} = nC_v\int_{T_a}^{T_f}\frac{dT}{T} = nC_v\ln \frac{T_f}{T_a}
[/tex]

the total change for the gas in the left compartment be

[tex]
\Delta S_{left} = \Delta S_1 + \Delta S_2 = nR\ln 2 + nC_v\ln\frac{T_f}{T_a}
[/tex]

For the gas in the right compartment, we can also design a similar process to calculate the entropy. So the total change of entropy is just the sum of change of entropy of left gas and that of right gas. Is this a correct approach to calculate the change of entropy in this case?

Thanks and sorry to bother again.
 
  • #7
Nice work!
 
  • #8
Mapes said:
Nice work!

:)
Mapes, thanks a lot for your help and encouragement. Wish you have a great Christmas!
 

Related to Mixing gas and find the temperature

1. How does mixing gas affect temperature?

When gases are mixed, the average kinetic energy of the molecules increases, causing an increase in temperature. This is due to the transfer of energy between the molecules during the mixing process.

2. Can mixing gas lead to a decrease in temperature?

Yes, mixing gases with different temperatures can lead to a decrease in temperature. The molecules with higher kinetic energy will transfer some of their energy to the molecules with lower kinetic energy, resulting in a decrease in overall temperature.

3. What is the ideal gas law and how does it relate to mixing gas?

The ideal gas law states that the pressure, volume, and temperature of a gas are all related. When gases are mixed, the pressure and volume may change, causing a change in temperature to maintain equilibrium. This is known as the Gay-Lussac's Law.

4. How can we determine the temperature of a gas after mixing?

We can determine the temperature of a gas after mixing by using the ideal gas law, which states that the product of pressure and volume is directly proportional to the absolute temperature. By measuring the pressure and volume of the mixed gases, we can calculate the temperature.

5. Is there a limit to how much gas can be mixed before the temperature becomes too high?

Yes, there is a limit to how much gas can be mixed before the temperature becomes too high. This is known as the critical temperature, which is the temperature at which a gas cannot be liquefied, regardless of the pressure. Beyond this point, further mixing will not result in a higher temperature.

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