Thermodynamic equilibrium for systems only open to particle exchange

Click For Summary
SUMMARY

The discussion centers on the conditions for thermodynamic equilibrium in systems open to particle exchange, specifically focusing on hydrogen (H2) and neon (Ne) gases. It establishes that equilibrium requires the equality of chemical potentials and temperatures in the two subvolumes. The chemical potential for an ideal gas component is defined as $$\mu_j=\mu_j^0(T,P^0)+RT\ln{(p_j/P^0)}$$, where $$P_0$$ is a reference pressure. The participants clarify that while the initial temperatures are equal, the necessity for them to remain equal throughout the process depends on whether the connecting wall is adiabatic or diathermal.

PREREQUISITES
  • Understanding of thermodynamic equilibrium concepts
  • Familiarity with chemical potential equations for ideal gases
  • Knowledge of ideal gas behavior and properties
  • Basic principles of energy conservation in thermodynamic systems
NEXT STEPS
  • Study the implications of adiabatic vs. diathermal walls on thermodynamic processes
  • Explore the derivation and applications of the chemical potential equation for ideal gases
  • Investigate the role of mass transfer in maintaining equilibrium in thermodynamic systems
  • Review case studies involving ideal gas mixtures and their equilibrium conditions
USEFUL FOR

Researchers, physicists, and students in thermodynamics, particularly those focusing on gas behavior and equilibrium conditions in multi-component systems.

EE18
Messages
112
Reaction score
13
Homework Statement
I am working Callen Problem 3.4-13 which reads:
An impermeable, diathermal, and rigid partition divides a container into two subvolumes, each of volume ##V##. The subvolumes contam, respectively, one mole of H2 and three moles of Ne. The system is maintained at constant temperature ##T##. The partition is suddenly made permeable to H2, but not to Ne, and equilibrium is allowed to reestablish. Find the mole numbers and the pressures.
Relevant Equations
See below.
I am only interested in the initial equilibrium conditions, and I am struggling to convince myself whether that should correspond to the equality of chemical potentials for H2 or an equality of temperatures as well. My work is as below:

We take both gases as simple ideal (this is only relevant for later, and as mentioned no worries about this part). We could write ##dS## in terms of all of the possible ##dX_i## for extensive parameters ##X_i## which can be independently varied in a virtual process from the final constrained equilibrium, but it should be clear that the only possible parameters which can be varied are ##dU_A## and ##dN_{2,A}## (we let 1 denote Ne and 2 H\textsubscript{2} and ##A## and ##B## denote the two subvolumes) since ##dU_A = -dU_B## and ##dN_{2,A} = -dN_{2,B}## and the other parameters cannot vary. One further has, from conservation of energy, that in this constrained equilibrium the only possible virtual processes involve mass transfers so that we arrive at ##dU_A = \mu_{2,A}dN_{2,A}## (this is the chemical work noted in Chapter 1.8). Thus our equilibrium maximization condition is
$$0 = dS = \left(\frac{1}{T_A} - \frac{1}{T_B}\right)\mu_{2,A}dN_{2,A} + \left(\frac{\mu_{2,A}}{T_A} - \frac{\mu_{2,B}}{T_B}\right)dN_{2,A}$$
which, since ##dN_{2,A}## is arbitrary, implies that
$$ 0 =\left(\frac{1}{T_A} - \frac{1}{T_B}\right)\mu_{2,A} + \left(\frac{\mu_{2,A}}{T_A} - \frac{\mu_{2,B}}{T_B}\right) $$
But I can't go further than this. I can see that the equality of temperatures and chemical potentials is sufficient for this condition, but is it necessary?
 
Last edited:
Physics news on Phys.org
Isn't the temperature T equal for the two chambers and constant over the process? Isn't the condition for equilibrium of H2 equal chemical potential in the two chambers? What is the equation for the chemical potential of an ideal gas component in a mixture?
 
  • Like
Likes Lord Jestocost, vanhees71 and Greg Bernhardt
Chestermiller said:
Isn't the temperature T equal for the two chambers and constant over the process? Isn't the condition for equilibrium of H2 equal chemical potential in the two chambers? What is the equation for the chemical potential of an ideal gas component in a mixture?
Oy, I completely skipped over the part about each subvolume being at ##T##, you are right. Can you comment incidentally on what would occur if not (as i thought was the case)? How does the condition I've arrived imply the equality of temperatures in that case?
 
EE18 said:
Oy, I completely skipped over the part about each subvolume being at ##T##, you are right. Can you comment incidentally on what would occur if not (as i thought was the case)? How does the condition I've arrived imply the equality of temperatures in that case?
Please answer my questions first.
 
Chestermiller said:
Please answer my questions first.
Isn't the temperature T equal for the two chambers and constant over the process?
Yes, equal initially.. It's actually not clear to me why it must necessarily remain so throughout the process (I know that ##U## is only a funciton of ##T## in an ideal gas, but there's nothing here which makes it clear a a priori that the temperatures in the subvolumes can't change as there's mass transfer, so perhaps my initial question stands)? Actually I see that the system is maintained at ##T##, so this is moot. But if it were an isolated system it would be interesting to think about.

Isn't the condition for equilibrium of H2 equal chemical potential in the two chambers?
I am trying to prove the relevant equilibrium. I agree that this is certainly the case if the connecting wall is not adiabatic. As I've shown here, it's not as clear if not. I think this is a similar sort of problem as Callen's adiabatic piston.

What is the equation for the chemical potential of an ideal gas component in a mixture?
I quote from an earlier problem I worked: Using the definition of the partial pressure ##P_j## from Problem 3.4.11, we can change the first term to be of the form below:
$$\mu_j := RT\ln \frac{P_jv_0}{RT} + f(T).$$
 
EE18 said:
Isn't the temperature T equal for the two chambers and constant over the process?
Yes, equal initially.. It's actually not clear to me why it must necessarily remain so throughout the process (I know that ##U## is only a funciton of ##T## in an ideal gas, but there's nothing here which makes it clear a a priori that the temperatures in the subvolumes can't change as there's mass transfer, so perhaps my initial question stands)? Actually I see that the system is maintained at ##T##, so this is moot. But if it were an isolated system it would be interesting to think about.

Isn't the condition for equilibrium of H2 equal chemical potential in the two chambers?
I am trying to prove the relevant equilibrium. I agree that this is certainly the case if the connecting wall is not adiabatic. As I've shown here, it's not as clear if not. I think this is a similar sort of problem as Callen's adiabatic piston.
Off hand, it is not clear to me the conditions for equilibrium if the partition is adiabatic rather than diathermal. But clearly, as you correctly deduced, for diathermal, the condition is equal temperatures.
EE18 said:
What is the equation for the chemical potential of an ideal gas component in a mixture?
I quote from an earlier problem I worked: Using the definition of the partial pressure ##P_j## from Problem 3.4.11, we can change the first term to be of the form below:
$$\mu_j := RT\ln \frac{P_jv_0}{RT} + f(T).$$
The equation is $$\mu_j=\mu_j^0(T,P^0)+RT\ln{(p_j/P^0)}$$where ##P_0## is a reference pressure (typically 1 bar), ##\mu_j^0(T,P^0)## is the molar free energy of pure species j at T and P^0, and ##p_j## is the partial pressure of species j in the mixture (or pure). The condition of equilibrium is equal chemical potentials in the two chambers for H2, or, equivalently, equal partial pressures. So the pressure of H2 in the chamber with pure H2 must be equal to the partial pressure of the H2 in the chamber with the mixture of Ne and H2.
 

Similar threads

Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
6K
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K