Can Avogadro's Law Be Applied to a Gas with Zero Volume and Zero Molecules?

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Discussion Overview

The discussion revolves around the applicability of Avogadro's law and the ideal gas law (PV = nRT) in scenarios where the volume and number of moles of gas approach zero. Participants explore the implications of these laws under conditions of negligible particles and pressure.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes a scenario involving a movable piston and questions how Avogadro's law applies when both volume and moles of gas are zero, suggesting that pressure should remain constant at zero.
  • Another participant argues that thermodynamic laws like PV = nRT are statistically valid only with a large number of particles, implying that they do not hold in cases with few or no particles.
  • A follow-up question is posed regarding whether the graph of volume versus number of moles should include the origin, given that both n and V equal zero results in zero pressure.
  • Participants suggest graphing V/n versus n to analyze behavior at zero volume and zero moles, noting that division by zero leads to undefined results.
  • One participant speculates that if n cannot be zero, the graph should start from a point slightly greater than n=0.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of Avogadro's law and the ideal gas law under conditions of zero volume and zero moles. There is no consensus on how these laws should be interpreted in such scenarios.

Contextual Notes

Participants highlight limitations in applying gas laws when the number of particles is very low or nonexistent, indicating that the laws may not be applicable in these extreme cases.

sgstudent
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My understanding of Avogrado's law is this: say I have a movable piston at a certain volume. When I pump air in, the pressure increases as the frequency of collision increases so the force acting on the piston increases. This would cause the movable piston to move up causing the volume to increase and causing the pressure to drop back to its initial amount.

But I was thinking if the volume of the piston was 0 with 0 moles of gas in it, the starting pressure would have to be 0. So when I pump more air into it, using Avogrado's law the pressure should remain constant. However, I don't think this is possible as now my pressure can't stay at 0 as nRT/V is always more than 0.

So how can the law hold here?

Thanks for the help :)
 
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Thermodynamic laws such as ##PV = nRT## are only valid in a statistical sense, when there are a large number of particles in the system (because fluctuations in the pressure, volume or temperature are small when there are a large number of particles). In the situations you describe, there are either no particles or only a few, so the laws can't be expected to hold.
 
Mute said:
Thermodynamic laws such as ##PV = nRT## are only valid in a statistical sense, when there are a large number of particles in the system (because fluctuations in the pressure, volume or temperature are small when there are a large number of particles). In the situations you describe, there are either no particles or only a few, so the laws can't be expected to hold.

Ohh so the law would only be accepted when say I have already 100Pa and I add in more air? So when drawing the graph for volume of gas versus number of moles should I touch the origin? Because when n=0 and V=0, P=0 but when I add in more gas where n>0, P would also now be greater than 0.
 
Why don't you graph V/n versus n and see what happens to the graph at 0 volume and 0 molecules?

PV = nRT applies to a gas of molecules, not to a gas that does not exist.
In other words, if you double nothing what do you get as an answer?
 
256bits said:
Why don't you graph V/n versus n and see what happens to the graph at 0 volume and 0 molecules?

PV = nRT applies to a gas of molecules, not to a gas that does not exist.
In other words, if you double nothing what do you get as an answer?

Hi 256bits when I graphed it I got an undefined number as we cannot divide a number by 0. If I were to double 0 I would still get 0. But how does this link to this here?

I would guess that I can't have n=0 then? Meaning when I draw my line it should start from slightly greater than n=0?
 

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