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Suppose I have a thermally insulated cylinder containing carbon dioxide (assuming it is ideal) at 1 atm pressure(atmospheric pressure is 1 atm). Now let's consider two cases:

CASE-1: I cut open the top of the cylinder, now the gas inside experiences a uniform 1 atm pressure and exerts the same pressure too. As there is no pressure difference so carbon dioxide should not diffuse out!

CASE-2: I pinhole the cylinder, now it should follow Graham's law of effusion and hence a net movement of air molecules inside the cylinder should take place; but how can air rush in if the pressure is the same 1 atm exerted by gas inside.

Further, my teacher told that since the partial pressure of $\ce{CO2}$ inside is much more than that of $\ce{CO2}$ outside, so $\ce{CO2}$ would diffuse out, but my point is that how can $\ce{CO2}$ differentiate whether the pressure it is subjected to, is exerted by 'X' gas or 'Y' gas. Hence the concept of partial pressure doesn't apply here. $\ce{CO2}$ would only experience that some gas is exerting 1 atm pressure, it doesn't care about the type of gas as ideal gases only care about the amount.

So, would $\ce{CO2}$ diffuse out and why?

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    $\begingroup$ Remember the nature of gas molecules is dynamic, not static. The same pressure of 2 gases in contact means the number of molecules leaving each gas is compensated by molecules coming from the other gas, so the pressure difference remains constant. $\endgroup$
    – Poutnik
    Jun 18 at 4:55
  • $\begingroup$ physics.stackexchange.com/questions/565575/… has useful information. $\endgroup$
    – Jon Custer
    Jun 18 at 13:22
  • $\begingroup$ You got answers and comment. At the root of your concern is that you ignore that the molecules in gases are in translational motion. Following the same logic you have used, you would conclude that gases are rigid or something like that. $\endgroup$
    – Alchimista
    Jun 21 at 9:50
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Diffusion is a fundamental molecular phenomenon and it does not have a direction.

CASE-1 : I cut open the top of cylinder, now the gas inside experiencees a uniform 1 atm pressure and exerts the same pressure too. As there is no pressure difference so Carbon dioxide should not diffuse out!

Why not? Recall if you open a bottle of perfume in a room, eventually you will smell that somebody has opened a bottle. Diffusion is not driven by pressure. Diffusion in gases is pretty fast.

In the second case, diffusion will still occur but at a much slower rate because the area is small (a pinhole). In the first case, you had simply cut open the cylinder exposing a bigger area.

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  • $\begingroup$ Ok, I got the case-1 and case-2 but what about the partial pressure concept..is it correct to apply that in case of diffusion? $\endgroup$
    – user265825
    Jun 18 at 6:13
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    $\begingroup$ @user265825 The macroscopic net diffusion follows the opposite direction of the concentration gradient or partial pressure gradient. It means diffusion from the region of higher partial pressure is faster than in the opposite direction. See wikipedia.org - Fick's laws of diffusion. $\endgroup$
    – Poutnik
    Jun 18 at 7:30
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....how can $\ce{CO2}$ differentiate whether the pressure it is subjected to, is exerted by 'X' gas or 'Y' gas. Hence the concept of partial pressure doesn't apply here. $\ce{CO2}$ would only experience that some gas is exerting 1 atm pressure, it doesn't care about the type of gas as ideal gases only care about the amount.

Let's suppose we treat these gases ideally. In that case, you're right, the gases don't "know" about each other. Yet differential movement can still occur in spite of this.

Let's label the inside gas X, and the the outside gas Y. As ideal gases, each particle is essentially oblivious to the presence of the other gas particles. They can each be considered to operate independently.

Since the temperature and pressure inside and outside are the same, the concentrations of X and Y will be identical. Further suppose, for simplicity, that gases X and Y have identical masses, and thus have the same speed distributions.

Now mark a disk of area 1 mm^2 anywhere on the wall of your container. Mark both the inside and outside surface. On the inside, the X particles are going to strike that disk at a rate that depends on the temperature, concentration, and particle mass. Same with the Y particles striking that disk on the outside. Since the temperature, concentration, and mass are the same for the X and Y particles, they're striking that disk at the same rate. Let's assume the rate is 100 hits per second (you could actually calculate this, but I'm not going to bother).

So, at this disk, 100 Y particles are hitting the outside wall every second, and 100 X particles are hitting the inside wall every second. Now remove the disk, turning it into a hole.

What happens? Well, every X particle that hits it from the inside ends up exiting the container. And every Y particle hitting it from the outside ends up entering the container. Hence initially, you have only X particles exiting, and Y particles entering. Then as more Y particles enter the container, some Y particles exit along with the X's. This continues until the relative concentration of particles is the same on both sides of the hole.

What does this have to do with partial pressure? Assuming equal temperatures, partial pressure tells you the relative concentration of X and Y particles on either side of the hole. If the total pressure is 1 atm inside and out, then the initial partial pressure of X is 1 atm on the inside, and 0 atm on the outside. And the opposite for Y.

In summary, for the same temperature and particle mass, the relative movement of X and Y particles through the hole is determined by their relative concentrations which, for equal temperatures, is expressed by their relative partial pressures.

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