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?

Note by Nilabha Saha
4 years, 7 months ago

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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

- 4 years, 7 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).

- 4 years, 6 months ago

You are welcome.

- 4 years, 6 months ago

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

Staff - 4 years, 7 months ago

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

- 4 years, 6 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 ...

- 4 years, 6 months ago