Statistical weight (thermodynamics)

AI Thread Summary
The discussion focuses on deriving the equation of state relating pressure (P), volume (V), number of particles (N), and temperature (T) using the statistical weight Ω. The participant initially proposes using the entropy formula s = klnΩ to derive the relationship, leading to P = k2bNVT. However, they are advised to double-check their derivative due to a factor of V within the logarithm, which requires careful differentiation. Additionally, the terms E and N in the equation (ds/dV)E,N indicate that energy (E) and number of particles (N) should be held constant during differentiation. The conversation emphasizes the importance of accurate mathematical handling in thermodynamic equations.
quietrain
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Homework Statement


find the equation of state which gives the relationship between P , V , N and T

C and b are constants

Ω = CebNV2(EV)N

Homework Equations



P/T = (ds/dV)E,N

The Attempt at a Solution



so i just use s = klnΩ ?

then i get ds/dV = k2bNV

so P = k2bNVT ? that's it?

btw, what does the E and N in (ds/dV)E,N tells me? is it mean keep E and N constant? or the differential is specified by the terms E and N?

thanks!
 
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quietrain said:

Homework Statement


find the equation of state which gives the relationship between P , V , N and T

C and b are constants

Ω = CebNV2(EV)N

Homework Equations



P/T = (ds/dV)E,N

The Attempt at a Solution



so i just use s = klnΩ ?

then i get ds/dV = k2bNV

so P = k2bNVT ? that's it?

That seems like the correct method, but double check your derivative. You have a factor of V inside the logarithm, so you should have a second term in your derivative corresponding to differentiating the log(EV) and not V^2


btw, what does the E and N in (ds/dV)E,N tells me? is it mean keep E and N constant?

Yes, that is what it means.
 
ah i see thank you!
 
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