Entropy Properties: Correct & Incorrect Expressions

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Guffie
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hello,

i have been given two formulae for the entropy of an ideal gas undergoing a reversible process.

the correct expression is:

S = Nk(s0-ln[itex]({\frac{N_0}{N}})^{5/2} + ln({\frac{T}{T_0}})^{3/2} + ln({\frac{V}{V_0}}))[/itex]

where s_0 is a constant, N the number of particles, T the temperature and V the volume.

and the incorrect is (derived assuming N is constant)

S=[itex]Nk (ln({\frac{T}{T_0}})^{3/2} + ln({\frac{V}{V_0}}))[/itex]

my question is,

the incorrect expression allows for negative entropies which is not possible, is this the only reason property of entropy this equation doesn't satisfy?

the other thing is showing that this isn't a problem with the correct expression, the only way to show this is to say that s_0 must be larger then the negative terms (the negation will depend on the system, if N > N_0 then V > V_0, so the ln(V/V_0) will be +, then there's two possibilities for T/T_0.)

is that the only way to show that the correct expression is always positive?

are there any other properties the incorrect expression doesn't satisfy that the correct expression does satisfy?
 
on Phys.org
Yes, there are several properties of entropy that the incorrect expression does not satisfy. The correct expression must satisfy the second law of thermodynamics, which states that the entropy of a system can never decrease over time. The incorrect expression allows for negative entropies, which violates this law. Additionally, the correct expression takes into account the number of particles in a system, while the incorrect expression does not. This means that the incorrect expression does not accurately account for the behavior of an ideal gas undergoing a reversible process. Finally, the correct expression includes a constant (s_0) which accounts for the entropy of the system at its initial state, while the incorrect expression does not.