Can't Avogadro Diffuse?

I request someone to clear the following doubt of mine:

Consider a closed room of volume 100 litres at STP. 1 mole of a gas (say oxygen) is introduced inside the room. I can conclude the following by my knowledge:

1) Avogadro's Law states that 1 mole of any gas at STP will occupy 22.4 litres. Hence, V[gas] = 22.4 litres.

2) Due to diffusion, the gas will move from a region of high concentration to a region of low concentration. Hence, the gas will occupy the volume of the entire room .Therefore, V[gas] = 100 litres.

Since, 22.4 litres $\neq$ 100 litres, something mus be wrong with my line of reasoning. Could anyone kindly point out the fallacy?

#Chemistry

Easy Math Editor

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Comments

Yes, you see Avogadro stated that 1 mole of any gas occupies 22.4 L at STP. When you diffuse it out, the conditions no longer will be that of STP , as temperature will increase as volume has increased. Incidentally, V is proportional to T, so accordingly Volume becomes 100 L and there is no STP. So, Avogadro's Law doesnot work here directly. So, that's the fallcy

Sarthak Rout - 4 years, 5 months ago

Thank you very much for providing me with this answer. You have relieved me of a painful sensation which I just couldn't get over. Perhaps it stands proof to the fact that laws of chemistry are amicable (in this case, the Ideal Gas Law with Molar Volume Of Gases).

Nilabha Saha - 4 years, 5 months ago

You are welcome.

Sarthak Rout - 4 years, 5 months ago

Suppose that the closed room is a basketball (balloon with rigid-ish walls). What happens as we pump air into it?.

Calvin Lin Staff - 4 years, 5 months ago

I thank you a lot for guiding me to a brilliant exploration here on Brilliant.

Nilabha Saha - 4 years, 5 months ago

The first thing is you have applied ( pv=mRT) to find the volume that will be occupied by a certain amount of oxygen which is (1 mole ) at STP conditions ، you have obtained that it will be 22.4 liters ، then you have put that amount of oxygen in a bigger room (100 liters volume) so you should apply another law which is (P1V1=P2V2), The value of P2 is what you will have as you kept the T is constant ........ let's approximate that you have 1 bar at STP so the P2 will be decreased around 0.2 bar in your 100 litter room which means that you have create vacuum state ..

Thank you ...

Mahmood Alsaad - 4 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$