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

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SUMMARY

Avogadro's Law, which relates the volume of a gas to the number of moles, cannot be applied when both the volume and the number of moles are zero. In such cases, the pressure is also zero, making the application of the ideal gas law (PV = nRT) invalid. The discussion highlights that thermodynamic laws are statistically valid only with a significant number of particles present, as fluctuations become negligible. Therefore, when graphing the relationship between volume and number of moles, the graph cannot touch the origin, indicating that n must be greater than zero for the law to hold.

PREREQUISITES
  • Understanding of Avogadro's Law
  • Familiarity with the Ideal Gas Law (PV = nRT)
  • Basic knowledge of thermodynamics
  • Graphing skills for mathematical relationships
NEXT STEPS
  • Research the statistical mechanics underlying thermodynamic laws
  • Explore the implications of zero volume and zero moles in gas laws
  • Learn about the limitations of the Ideal Gas Law in real-world applications
  • Investigate graphical representations of gas laws and their behaviors at limits
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Students of physics, chemists, and anyone studying thermodynamics or gas laws will benefit from this discussion, particularly those interested in the limitations and applications of Avogadro's Law.

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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