Why does the gas with the smallest molar mass have the highest pressure?

AI Thread Summary
In a scenario where equal masses of Xenon, Argon, and Neon are placed in flasks of equal volume and temperature, Neon, having the smallest molar mass, will exert the highest pressure. This is because the number of moles of gas is inversely proportional to its molar mass; thus, a lower molar mass results in a greater number of moles for the same mass. The ideal gas law supports this conclusion, indicating that pressure is directly related to the number of moles. The discussion also touched on notation conventions for molar mass and gas mass, but this does not affect the underlying physics. Ultimately, Neon will have the highest pressure among the three gases.
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Homework Statement


If equal masses of Xenon, Argon and Neon are placed in separate flasks of equal volume and same temperature, which one of the following statements is correct:
a) The pressure of Neon flask is greatest.
b) The pressure of Argon flask is greatest.
c) The pressure of Xenon flask is greatest.
d) The pressure in all 3 flasks is the same.

Homework Equations


None

The Attempt at a Solution


The answer stated was Neon because it contains the smallest molar mass out of the three. The smallest molar mass will contain the greatest number of moles. Since pressure is directly proportional to the number of moles in a gas, the pressure of Neon flask is the greatest.

I know that the pressure is directly proportional to the number of moles. But why does the smallest molar mass contain the greatest number of moles?
 
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Because it says equal masses of each gas are in the flask. If you have two piles with the same mass, and one pile consists of bowling balls and one consists of ping-pong balls, which pile has more balls?
 
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Ping-pong balls! Thank you for the wonderful analogy. It makes perfect sense now.
 
codcodo said:

Homework Equations


None
Just to point out that there is a relevant equation. The ideal gas law ##PV = RnT##. You can rewrite this as
$$
P = \frac{RnT}{V}.
$$
Since ##n = M/m##, where ##m## is the molar mass and ##M## is the total mass, you would find
$$
P = \frac{RMT}{V} \frac{1}{m}.
$$
Since ##R##, ##M##, ##T##, and ##V## were assumed to be the same, the gas with the smallest molar mass ##m## will have the highest pressure.
 
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Orodruin said:
Since ##n = M/m##, where ##m## is the molar mass and ##M## is the total mass

Funny, I would switch M and m (using the capital letter to mark molar mass and the small one for the gas mass). I feel like it is an accepted convention (but I can be wrong).
 
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Borek said:
Funny, I would switch M and m (using the capital letter to mark molar mass and the small one for the gas mass). I feel like it is an accepted convention (but I can be wrong).
That may be, it is not my direct field so I may have introduced conventions contrary to what is usually used. Of course, the notation has no impact on the physics.
 
@Orodruin Sure thing, what you wrote is perfectly correct. Actually I thought about posting exactly the same, just assumed OP already got it.
 
Borek said:
Funny, I would switch M and m (using the capital letter to mark molar mass and the small one for the gas mass). I feel like it is an accepted convention (but I can be wrong).

I use M for molar mass and m for gas mass.
 
Orodruin said:
Just to point out that there is a relevant equation. The ideal gas law ##PV = RnT##. You can rewrite this as
$$
P = \frac{RnT}{V}.
$$
Since ##n = M/m##, where ##m## is the molar mass and ##M## is the total mass, you would find
$$
P = \frac{RMT}{V} \frac{1}{m}.
$$
Since ##R##, ##M##, ##T##, and ##V## were assumed to be the same, the gas with the smallest molar mass ##m## will have the highest pressure.
Thank you.
 
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