Thermodynamics Rate of Effusion

In summary, the thermodynamics rate of effusion is a measure of how quickly a gas molecule escapes from a container through a tiny hole. It is inversely proportional to the square root of the gas's molar mass, meaning that as temperature increases, the rate of effusion also increases. It is an important concept in understanding the behavior of gases and can be calculated using a formula. The size and shape of a container indirectly affect the rate of effusion by changing the pressure and temperature of the gas inside.
  • #1
SolsticeFire
65
1

Homework Statement


Effusion is the slow leakage of atoms through a small hole in a container. We'll consider a box at pressure P, with volume V and temperature T. The hole will have the area A, and is in one of the six walls of the container.

a) At what rate do atoms (assume mass M) pass through the hole? Use avg (vx2)1/2 as an approximation for avg (vx). Equipartition can be used to evaluate avg (vx2).

Homework Equations



Equipartition: mvx2 = kT (k = Boltzman's constant)
Ideal Gas Law: PV = NkT

The Attempt at a Solution



It asks for rate at which atoms pass through the hole and so far I did:

--> I need to get number of atoms per second units for answer so I
assumed the answer should be in form of N atoms/s.
---> Assuming ideal gas situation the frequency at which the atoms hit
the wall with the hole is vx/2L. (I'm not sure whether this assumption is so accurate, but I figured at in ideal gas conditions an atom to hit the wall the difference in time has to be 2L/vx).
---> The factor of area covered by the hole on the side wall can be
written as a ratio
(Ahole/Aside).
So now we have frequency = [N (Ahole/Aside)vx]/2L
Then using equipartition and ideal gas:
PV = NkT; N = PV/kT
and Mvx^2 = kT; vx = sqrt (kT/M)

Rewriting the formula I got: [PV/kT (Ahole/Aside) sqrt(kT/M)]/2L
simplifying we get: (Ahole/Aside) [PV/{sqrt(kTM)x 2L}]

However this implies that as Volume increases rate of atoms going
through increases since volume is in the numerator; also as T decreases, Rate goes up. It
doesn't add up. Is there a mistake in my line of reasoning here? Or have made a bad assumption
with f = vx/2L? :S.

I have been working on this problem for three days now, and I can't seem to find a mistake in my algebra, but the solution makes no sense. I need this solution to do part b) and c) of the same question.

Any guidance will be greatly appreciated.

Thanks!
 
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  • #2

Thank you for your post. It seems like you have made a mistake in your line of reasoning. Let's go through it step by step to see where the error is.

Firstly, your assumption that the frequency at which atoms hit the wall with the hole is vx/2L is incorrect. This is only true if we are looking at a single atom moving in a straight line towards the hole. However, in a gas, we have many atoms moving in random directions, so the frequency of atoms hitting the wall is not simply vx/2L.

Instead, we need to consider the average velocity of all the atoms in the gas. This can be calculated using the root mean square velocity, which is given by:

vrms = sqrt(3kT/M)

Where k is Boltzmann's constant, T is temperature, and M is the mass of the atom. This is the average velocity of all the atoms in the gas, and it takes into account their random motion in all directions.

Next, we can use the ideal gas law to calculate the number of atoms in the gas:

N = PV/kT

Now, to find the rate at which atoms pass through the hole, we need to consider the number of atoms that hit the wall with the hole per unit time. This can be calculated by multiplying the number of atoms in the gas by the fraction of the wall that is covered by the hole, and then dividing by the average velocity of the atoms:

Rate = (N * Ahole/Aside) / vrms

Substituting in the expressions for N and vrms, we get:

Rate = (PV/kT * Ahole/Aside) / sqrt(3kT/M)

Simplifying, we get:

Rate = (Ahole/Aside) * sqrt(PM/3k)

This is the rate at which atoms pass through the hole. As you can see, it does not depend on the volume of the container or the temperature. This makes sense, as the effusion rate only depends on the area of the hole, the pressure, and the mass of the atoms.

I hope this helps to clarify things for you. Let me know if you have any further questions.
 

Related to Thermodynamics Rate of Effusion

1. What is the definition of thermodynamics rate of effusion?

The thermodynamics rate of effusion is the measure of how quickly a gas molecule escapes from a container through a tiny hole. It is also known as the rate of diffusion.

2. How is thermodynamics rate of effusion related to temperature and pressure?

According to Graham's Law, the rate of effusion is inversely proportional to the square root of the gas's molar mass. This means that as temperature increases, the average kinetic energy of the gas molecules also increases, resulting in a higher rate of effusion. Similarly, as pressure increases, the molecules are closer together, making it more difficult for them to escape through the hole, resulting in a lower rate of effusion.

3. What is the significance of thermodynamics rate of effusion?

The rate of effusion is an important concept in understanding the behavior of gases. It helps us understand how gases diffuse and mix with each other, as well as how they escape from a container. It also has practical applications in fields such as chemistry and engineering.

4. How can the thermodynamics rate of effusion be calculated?

The thermodynamics rate of effusion can be calculated using the following formula:
Rate of effusion = (1/√molar mass) x (1/√density)

5. How does the size and shape of a container affect the thermodynamics rate of effusion?

The size and shape of a container do not have a direct effect on the thermodynamics rate of effusion. However, they can indirectly affect the rate by changing the pressure and temperature of the gas inside the container. A smaller container will have a higher pressure, resulting in a lower rate of effusion, while a larger container will have a lower pressure, resulting in a higher rate of effusion.

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