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March 16, 2026, 10:46:53 pm

Author Topic: Molar Volume question  (Read 1587 times)  Share 

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tigerclouds

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Molar Volume question
« on: April 02, 2020, 04:44:06 pm »
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Hi there,
This question always comes up for me and I've never reached a satisfying conclusion. Why is it that different gases with the same mole, pressure and temperature have the same volume? Why is it that a propane molecule, for example, has the same volume as methane under the same conditions, when a propane molecule is longer and therefore occupies more space? I understand how PV=nRT comes into this and that all the variables are proportional but I'm hoping someone can offer a more logical explanation or analogy?

Owlbird83

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Re: Molar Volume question
« Reply #1 on: April 02, 2020, 06:35:53 pm »
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Hi there,
This question always comes up for me and I've never reached a satisfying conclusion. Why is it that different gases with the same mole, pressure and temperature have the same volume? Why is it that a propane molecule, for example, has the same volume as methane under the same conditions, when a propane molecule is longer and therefore occupies more space? I understand how PV=nRT comes into this and that all the variables are proportional but I'm hoping someone can offer a more logical explanation or analogy?

The size difference of these gas molecules is way too tiny to have any effect on the volume they occupy. The space between molecules is massive compared to the tiny molecules, so being slightly larger won't change the volume.  :)
Someone correct me if I'm wrong.
2018: Biology
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tigerclouds

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Re: Molar Volume question
« Reply #2 on: April 02, 2020, 08:07:53 pm »
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The size difference of these gas molecules is way too tiny to have any effect on the volume they occupy. The space between molecules is massive compared to the tiny molecules, so being slightly larger won't change the volume.  :)
Someone correct me if I'm wrong.
Even if you're comparing methane with like a 40-carbon molecule? I find that hard to conceptualize.

whys

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Re: Molar Volume question
« Reply #3 on: April 02, 2020, 10:12:26 pm »
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All gases take up the same volume as the container they are in. They have such high kinetic energy that they are constantly colliding, so can easily expand to take up the shape of the container, and in the same way can also be compressed. Avogadro's law states: "For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant." Similarly, all ideal gases have a volume of 22.4L/mol at STP according to Avogadro's law.

Take this formula into consideration:
PV = NRT

We can rearrange this into V = NRT/P

If N, T and P are the same for 2 different gases, then the volume will be the same for the 2 gases as R is always the same for any gas (ideal gas constant).

Hope this helps! ;D
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angrybiscuit

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Re: Molar Volume question
« Reply #4 on: April 02, 2020, 10:22:40 pm »
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Even if you're comparing methane with like a 40-carbon molecule? I find that hard to conceptualize.
I think you need to visualise just how tiny molecules are compared to the space they occupy. Changes in volume as a result of changes in their molecular size is so minuscule it's insignificant. Furthermore, as molecules get larger, the less likely they are to be found as a gas. So I don't believe there could be something like a 40-carbon molecule and still be a gas.
Hope this clears it up a little :)
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tigerclouds

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Re: Molar Volume question
« Reply #5 on: April 03, 2020, 03:27:21 pm »
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Thanks guys, that clarifies it a bit more. I think I was confusing pressure with volume, because I assumed the larger the molecules naturally are the more pressure there would be inside the space. I’m assuming that’s wrong, because like you said the change is too small to have an impact. I’ve attached a photo that I’ve been relying on to understand the topic, would you say it’s inaccurate because it portrays the larger molecules as taking a significant amount of space when they’re actually not? Because I think that photo has been my source of confusion. Thanks again for your help :)