Internal energy and thermodynamic entropyby SW VandeCarr Tags: energy, entropy, internal, thermodynamic 

#1
Feb1711, 07:32 AM

P: 2,490

As a practical matter, the internal energy of a system is treated as a state function of a the system and is concerned only with the kinetic energy of particles, not the potential energy. So for thermodynamic entropy, S=E/T, (S=entropy, T= absolute temperature) I'm considering E to be kinetic energy.
Now thermodynamic temperature has been defined as the average kinetic energy per unit particle. If so, we can say T=E/N where E is the internal (kinetic) energy of a system. Then, S=E/T=E/E/N=N. That is, S becomes a dimensionless number equal to (or direct function of) the number of particles in the system. The Boltzmann equation: S=k log W considers only the number of particles in a system since [tex]W=N!/\Pi N_{i}![/tex]. If S=f(N), then the entropy of a system of N particles would seem to be constant. That would mean it doesn't matter if you add or remove energy or dilute or concentrate the particles. S is always constant unless you add or remove particles. Is this true? 



#2
Feb1711, 11:55 AM

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[tex]KE_{avg} = 3kT/2[/tex] where k is the Boltzmann constant having units of Joules/degree K AM 



#3
Feb1711, 03:35 PM

P: 969

So you're saying:
dS=dE/T which is fine when entropy doesn't depend on volume (zero pressure always). Integrating both sides you get: [tex]S=\int dE/T [/tex] But T is a function of just E and not V (try to guess why), so it ought to be: [tex]S=\int dE/T(E) [/tex] But now you say that T(E)=aE, where 'a' is a proportionality constant. Then the integral becomes: [tex]S=\int dE/aE=\log E /a [/tex] So your entropy depends on the log of energy. And notice that this is the result for how the entropy of a solid varies with energy in the high energy limit! But it's not quite the exact result. The reason is that T is proportional to E only at high energies, whereas in the integral you assumed T is proportional to E even at low energies. But still the integral gives an accurate result when you have a lot of energy, since the integral is dominated by this high energy region (since this region is large), and a few inaccuracies at the beginning of the integral don't matter. 



#4
Feb1811, 05:20 AM

P: 2,490

Internal energy and thermodynamic entropyMy question is: Given that dU/dT is constant in an ideal gas, can we say that [tex]U=k(mv^2/2)[/tex] in an ideal gas? In other words, in an ideal gas, does entropy only depend on the number of particles: N? 



#5
Feb1811, 07:37 AM

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AM 



#6
Feb1811, 09:31 AM

P: 969

for the entropy of an ideal gas. It contains lots of variables besides N. Notice how it varies with energy as a log. T=(2/(3NK))E for a monotone ideal gas, so you would plug in "a"=2/(3NK) in the equation: [tex]S=\frac{log E}{a} [/tex] For an ideal gas you need to include volume. For a solid volume is not really important. As long as energy is proportional to temperature, you'll get an entropy dependence on energy as a log. 



#7
Feb1811, 01:55 PM

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#8
Feb1811, 03:48 PM

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[tex]\Delta S = \int dQ/T = \int dU/T + \int PdV/T[/tex] For a monatomic ideal gas dU = nCvdT = 3NkdT/2, and with no change in volume (PdV = 0) over a reversible path between states i and f: [tex]\Delta S_{if} = \int_{T_i}^{T_f} dU/T = 3Nk/2\int_{T_i}^{T_f} dT/T = \frac{3Nk}{2}\ln\left(\frac{T_f}{T_i}\right)[/tex] Since T = 2U/3Nk, we have: [tex]\Delta S = \frac{3Nk}{2}\ln\left(\frac{U_f}{U_i}\right)[/tex] Caution: this is only true if there is no change in volume over a reversible path between the two states i and f. AM 


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