Relation between energy annd pressure

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SUMMARY

The discussion centers on deriving the relationship between pressure and energy in gases, specifically through the kinetic theory of gases. Key equations include the average kinetic energy of a molecule, expressed as \( \frac{1}{2} mv^2 = \frac{3}{2} kT \), and the pressure equation \( P = \rho v_x^2 \). The derivation leads to the ideal gas law \( pV = nR_GT \), demonstrating that pressure times volume equates to energy. This relationship is crucial for understanding gas behavior at the molecular level.

PREREQUISITES
  • Kinetic theory of gases
  • Ideal gas law
  • Basic thermodynamics
  • Statistical mechanics
NEXT STEPS
  • Study the derivation of the ideal gas law using kinetic theory
  • Explore the concept of molecular velocity distributions
  • Learn about the implications of Avogadro's number in gas laws
  • Investigate the relationship between temperature and kinetic energy in gases
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Students of physics, chemists, and engineers interested in thermodynamics and gas behavior, as well as anyone seeking to deepen their understanding of the relationship between energy and pressure in gases.

Haseeb Ali
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Me and and my friend were having discussion about the motion of molecules of gas.We talked about their velocity ,kinetic energy and much more. He asked me to derive a relation between pressure and energy. I was unable to explain him that... Can anyone explain the relation?
 
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Haseeb Ali said:
He asked me to derive a relation between pressure and energy.
How would you begin? What information do you have?
 
Bystander said:
How would you begin? What information do you have?
P=density*Vx*2

T= 2/3k 1/2mv*2
 
You could read up on the gas law, e.g. here
Your equations feature in there, but one appears molecular to me and the other macroscopic:

##{1\over 2} mv^2 = {3\over 2} kT## average kinetic energy for a molecule.

##P = \rho v_x^2 = {mass\over V} v_x^2 ## may be ok if in the right context.

I can wrangle a bit with formulas: $$ {1\over 2} mv^2 = {3\over 2} kT\ \Rightarrow\ {1\over 2} mv_x^2 = {1\over 2} kT$$ (the energy is equally distributed over the three degrees of freedom),

##\displaystyle \ \Rightarrow\ v_x^2 = {kT\over m} ##

mass = number of molecules * mass of a molecule = number of moles * ##N_A## * mass of a molecule (Avogadro number); write mass = ##n\; N_A\; m##

Leaves

##pV = n\;N_A\;m \ {kT\over m} = n\; N_A \;kT##;

then use

##R_G = N_A \; k## (gas constant) to get the ideal gas law

##pV = nR_GT##.

pressure times volume has the dimension of energy.

It's not the whole story, but quite a big part of it.
 

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