# Effusion of particles from one box to another - pressure calculation

• unscientific
I think the answer you get is justN_{left} = \frac{N_0}{2} \left( 1 - e^{-\frac{A\langle v \rangle}{V} t} \right) orP_{left} = \frac{P_0}{2} \left( 1 - e^{-\frac{A\langle v \rangle}{V} t} \right) I was thinking the "2" came from the fact that there are two halves. But it's not quite that simple. It's a factor of 2 because the pressure times the volume is proportional to the number of molecules. So if the number of molecules in one half
unscientific

## Homework Statement

Suppose we have a container divided into equal halves. Right half is fixed at temperature ##T##, volume ##V_2##.

Initially it has pressure ##P_0##, a hole of area ##A## is opened between them.

Part (a): Show that the pressure in the right half is:

$$p = \frac{p_0}{2} \left[ 1 + exp \left(-\langle v \rangle\frac{A}{V}t \right) \right]$$

Part (b): Vessel has total volume ##10^{-4} m^3##, area of ##7.9 \times 10^{-9} m^2##, ##T = 290 K##. Find the time taken for the pressure on the left half to become half that in the right part.

## The Attempt at a Solution

Part (a)

$$dN = -A\phi_{(t)} dt$$
Using flux as ##\phi = \frac{1}{4} n\langle v \rangle##:
$$dN = - \frac{A}{4} n \langle v \rangle dt = -\frac{A}{4}\left( \frac{N}{V/2} \right) \langle v \rangle dt = -\frac{NA}{2V} \langle v \rangle dt$$

The graph looks like:

We solve for the dotted graph, then translate upwards by ##\frac{N_0}{2}##. Integrating, and translating upwards i get: ##p = \frac{p_0}{2} \left[ 1 + exp \left(-\langle v \rangle\frac{A}{2V}t \right) \right]##. Not sure why I'm missing a factor of ##2##..

Part (b)

I know the way to solve this question is to find the pressure on the left half as a function of time, using a process similar to part (a). The graph we want to solve for is (L):

But I'm not sure how to begin..

The hole is not a one-way passage. Particles can pass through the hole from either side.

TSny said:
The hole is not a one-way passage. Particles can pass through the hole from either side.

Yes but I think that makes the problem insanely difficult. I think we can assume molecules only go from right to left.

Part (a): Any idea why I'm missing by a factor of 2?

Part (b): I don't even know how to start..

Surely particles will need to come back through the hole from left to right. Otherwise, the entire gas would eventually end up in the left half.

I get the correct factor of 2 if I account for particles passing back through the hole from left to right. However, I did assume that the gas in the left side is always in thermal equilibrium with the gas in the right side. That would be the case if the wall separating the two halves is a good diathermal (heat conducting) wall.

I think part (b) is just a fairly direct application of the result from part (a). But I haven't worked out the numerical answer.

[EDIT: It seems to me you need to know the molecular mass to answer (b).]

Last edited:
TSny said:
Surely particles will need to come back through the hole from left to right. Otherwise, the entire gas would eventually end up in the left half.

I get the correct factor of 2 if I account for particles passing back through the hole from left to right. However, I did assume that the gas in the left side is always in thermal equilibrium with the gas in the right side. That would be the case if the wall separating the two halves is a good diathermal (heat conducting) wall.

I think part (b) is just a fairly direct application of the result from part (a). But I haven't worked out the numerical answer.

[EDIT: It seems to me you need to know the molecular mass to answer (b).]

Part (a)

$$dN_r = -A\phi_r dt + A \phi_l dt$$
$$= -\frac{AN_r \langle v \rangle}{2V} dt + \frac{AN_l \langle v_l \rangle }{2V} dt$$
$$= -\frac{AN_r \langle v \rangle}{2V} dt + \frac{A(N_0 - N_r) \langle v_l \rangle }{2V} dt$$

Assuming ##\langle v \rangle_l = \langle v \rangle_r## here (If not, ##T_r = \frac{4}{3} T_l##, so ##\langle v \rangle_l = \frac{2}{\sqrt 3} \langle v \rangle_r##):

$$dN = -\frac{A \langle v \rangle}{V} \left(\frac{N_0}{2} - N \right) dt$$

Integrating, I get ## N = \frac{N_0}{2} \left( 1 + e^{-\frac{A \langle v \rangle t}{V}} \right)##. Assuming temperature in the right side stays constant throughout,

$$p = \frac{p_0}{2} \left[ 1 + exp \left(-\langle v \rangle\frac{A}{V}t \right) \right]$$

Part(b)

I can say ##N_{left} = N_0 - N_{right} = \frac{N_0}{2} \left( 1 - e^{-\frac{A\langle v \rangle}{V} t} \right)##.

Thus,

$$\frac{P_{left}}{T_{left}} = \frac{P_0}{2T_0}\left( 1 - e^{-\frac{A\langle v \rangle}{V} t} \right)$$

Now, ##T_{left} = T_{right}## (Assumption made in part (a))

$$P_{left} = \frac{P_0}{2} \left( 1 - e^{-\frac{A\langle v \rangle}{V} t} \right)$$

Mass of nitrogen molecule is ## 4.65 \times 10^{-26} kg##
Average speed is ##468 ms^{-1}##
I get an answer of ##29.7 s##

Last edited:
That all looks good.

1 person

## 1. How is pressure calculated in the effusion of particles from one box to another?

Pressure is calculated by using the ideal gas law, which states that pressure is equal to the number of moles of gas (n) multiplied by the universal gas constant (R) and the absolute temperature (T), divided by the volume (V). This can be written as P = (nRT)/V. In the case of effusion, the pressure of the gas in the first box will decrease as particles escape into the second box, while the pressure in the second box will increase.

## 2. What factors can affect the rate of effusion of particles?

The rate of effusion can be affected by several factors, including the size of the particles, the temperature, and the pressure difference between the two boxes. Smaller particles will effuse faster than larger particles, and higher temperatures will also increase the rate of effusion. Additionally, a larger pressure difference between the two boxes will result in a faster rate of effusion.

## 3. How does the number of particles in each box affect the pressure calculation?

The number of particles in each box will affect the pressure calculation, as it is directly related to the number of moles (n) in the ideal gas law. The more particles present in a box, the higher the pressure will be. As particles effuse from one box to another, the number of particles in each box will change, and the pressure will be affected accordingly.

## 4. Can the effusion of particles from one box to another be reversed?

Yes, the effusion of particles from one box to another can be reversed. This can be achieved by increasing the pressure in the second box, which will cause particles to effuse back into the first box. The rate of effusion can also be slowed down by decreasing the pressure difference between the two boxes.

## 5. Is the effusion of particles from one box to another a spontaneous process?

Yes, the effusion of particles from one box to another is a spontaneous process. This means that it will occur naturally without any outside influence or energy input. The particles will move from an area of higher pressure to an area of lower pressure until the pressure is equalized in both boxes.

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