# Helmholtz potential for mixture of simple ideal gases

• Othin
In summary, the Helmholtz potential for a mixture of simple ideal gases is the sum of the individual potentials. The fundamental equation of a monatomic gas in the Helmholtz potential is:F=NRT[\frac{F_0}{N_0RT_0} - ln(\frac{T}{T_0}^{3/2} \frac{V}{V_0} (\frac{N}{N_0})^{-1})]
Othin

## Homework Statement

Two subsystems within a 20 l cylinder are separated by an internal piston. Each of them is initially composed of 1 mole of component 1 and one mole of component 2, both of which will be treated as a monatomic ideal gas. The cylinder has diathermal walls and is in contact with a thermal reservoir (meaning its temperature should be constant and equal to $T_r = 373 K$ . The piston is permeable to component 1 but impermeable to component 2. The piston is in the center. We're to prove that the Helmholtz potential for this system is:
$F=N\frac{T}{T_o} - \frac{3}{2}NRTln\frac{T}{T_0} - N_1RTln(\frac{VN_o}{V_0N_1}) - N_2^1RTln(\frac{V^1N_0}{V_0N_2^1}) - N_2^2RTln(\frac{V^2N_0}{V_0N_2^2})$
Where $T_o, f_o, V_o , N_0$ are attributes of a standard state. N is the total mole number and superscripts represent the subsystem, while subscripts refer to components ($N_2^1$ is the mole number of component2 at subsystem 1 and so on).
Afterwards, we must calculate the work required to push the piston to a position such that $V^1=5$ and $V^2=15$ in two ways : first by direct integration (the relation dW= PdV is valid, for the process is assumed to be carried out quasi-statically)

## Homework Equations

We know that the Helmholtz potential of a mixture of simple ideal gases is the is the sum of the individual potentials. The fundamental equation of a monatomic gas in the Helmholtz potential is:
$F=NRT[\frac{F_0}{N_0RT_0} - ln(\frac{T}{T_0}^{3/2} \frac{V}{V_0} (\frac{N}{N_0})^{-1})]$

## The Attempt at a Solution

The Helmhholtz potential is additive, so I wrote both F as a sum of the potentials of each subsystem. Namely:
$F^1=N_1^1RT[\frac{F_{011}}{RN_{01}T_0} - \frac{3}{2}ln\frac{T}{T_0} -ln(\frac{V^1}{V_{01}}\frac{N_{011}}{N_1^1})] + N_2^1RT[\frac{F_{012}}{RN_{021}T_0} - \frac{3}{2}ln\frac{T}{T_0} - ln(\frac{V^1}{V_{01}}\frac{N_{012}}{N_2^1})]$
And
$F^2=N_1^2RT[\frac{F_{021}}{RN_{02}T_0} - \frac{3}{2}ln\frac{T}{T_0} -ln(\frac{V^2}{V_{01}}\frac{N_{021}}{N_1^2})] + N_2^2RT[\frac{F_{022}}{RN_{022}T_0} - \frac{3}{2}ln\frac{T}{T_0} - ln(\frac{V^2}{V_{01}}\frac{N_{02}}{N_2^2})]$
Summing then, I find :
$F-Nfo\frac{T}{T_0} - \frac{3}{2}RTln\frac{T}{T_0} - N_1^1RTln(\frac{V^1N_0}{V_0N_1^1}) - N_1^1RTln(\frac{V^2N_0}{V_0N_1^2}) - N_1^2RTln(\frac{V^1N_0}{V_0N_2^1}) - N_1^2RTln(\frac{V^2N_0}{V_0N_2^2})$
Where I've grupped all constants together. That's still different from the result I'm expected to prove, and the only different part seems to regard component one, which oddly enough is the component the piston is permeable to. Can't that be a mistake from the book? Even if it is, my expression wouldn't agree with the one I need, so I'd still need help.
As for the second part, the variation of F is vanishing identically for me. If there's a mistake and the membrane is actually permeable to 1 instead of 2, I get an expression which makes sense, but comes in terms of $N_2^1 and N_2^2$ . I wasn't able to eliminate both of them to get a number for the work

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Shouldn't ##N_1^1/V^1=N_1^2/V^2=N_1/V##?

Chestermiller said:
Shouldn't ##N_1^1/V^1=N_1^2/V^2=N_1/V##?
That would allow me to group the those two terms! But why? Is it because of Dalton's partial pressure law? If i write the ideal gas law for $N_1^1$ and $N_1^2$I guess the RT's will cancel out...

Othin said:
That would allow me to group the those two terms! But why? Is it because of Dalton's partial pressure law? If i write the ideal gas law for $N_1^1$ and $N_1^2$I guess the RT's will cancel out...
Yes. If there is a (semi-permeable) membrane that allows passage of one component, but not the other, the chemical potentials (partial pressures) of that component on both sides of the membrane must be the same.

Othin
What about the integration? I figure I must use $dw= P^1dV^1 + P^2dV^2$, but if I write $W= \int_{10}^5 N^1RT dV^1 + \int_{10}^{15} N^2RT dV^2$ , where both N's are treated as constants, it gives the wrong answer. I can't think of another relation that would enable the integration.

Othin said:
What about the integration? I figure I must use $dw= P^1dV^1 + P^2dV^2$, but if I write $W= \int_{10}^5 N^1RT dV^1 + \int_{10}^{15} N^2RT dV^2$ , where both N's are treated as constants, it gives the wrong answer. I can't think of another relation that would enable the integration.
No. N1 will change between the two compartments as you move the piston (so that the partial pressures of this species in the two compartments are always equilibrated).

Chestermiller said:
No. N1 will change between the two compartments as you move the piston (so that the partial pressures of this species in the two compartments are always equilibrated).
. Does that mean the work from $N_1^1$ and $N_1^2$ will cancel each other? So the net work is $\int_{10}^5 \frac{N_2^1RT}{V^1} dV^1 + \int_{10}^{15}\frac{N_2^2RT}{V^2} dV^2$? but that gives RTln(¾), so I'm still missing something.

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Othin said:
. Does that mean the work from $N_1^1$ and $N_1^2$ will cancel each other? So the net work is $\int_{10}^5 \frac{N_2^1RT}{V^1} dV^1 + \int_{10}^{15}\frac{N_2^2RT}{V^2} dV^2$? but that gives RTln(¾), so I'm still missing something.
When I substitute into this equation, I get the given answer.

Othin
Chestermiller said:
When I substitute into this equation, I get the given answer.
I'm dumb, RTln(3/4) is just the answer with changed signs, which makes sense because I calculated the work delivered by the work source when I should calculate the work delivered to it

Othin said:
I'm dumb, RTln(3/4) is just the answer with changed signs, which makes sense because I calculated the work delivered by the work source when I should calculate the work delivered to it
I don't think you're dumb. The only help I really gave you was how to address the semi-permeable membrane.

Othin

## What is the Helmholtz potential for a mixture of simple ideal gases?

The Helmholtz potential for a mixture of simple ideal gases is a thermodynamic quantity that characterizes the equilibrium state of the mixture. It is defined as the difference between the total internal energy of the system and the product of temperature and entropy.

## How is the Helmholtz potential related to the Gibbs free energy?

The Helmholtz potential and the Gibbs free energy are related by a Legendre transformation. This means that they are different mathematical representations of the same underlying thermodynamic system, and can be used interchangeably to describe the equilibrium state of a mixture of simple ideal gases.

## What is the significance of the Helmholtz potential for mixture of simple ideal gases in thermodynamics?

The Helmholtz potential plays a crucial role in understanding the thermodynamic behavior of mixtures of simple ideal gases. It is a fundamental quantity that is used to calculate other important thermodynamic variables such as the internal energy, entropy, and pressure of the system.

## How is the Helmholtz potential calculated for a mixture of simple ideal gases?

The Helmholtz potential for a mixture of simple ideal gases can be calculated using the ideal gas law and statistical mechanics. By considering the interactions between the gas molecules and applying the appropriate equations, the Helmholtz potential can be determined for a given mixture of gases at a specific temperature and pressure.

## How does the Helmholtz potential change with changes in temperature and pressure?

As with other thermodynamic quantities, the Helmholtz potential also changes with changes in temperature and pressure. At constant temperature, the Helmholtz potential decreases with increasing pressure. At constant pressure, the Helmholtz potential increases with increasing temperature. These relationships can be explained using the ideal gas law and statistical mechanics.

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