What energy is this? Related to monoatomic ideal gas

[fluidistic]

For a monoatomic ideal gas, if $c_v$ is constant then the internal energy is worth $U=\frac{3nRT}{2}$. This is a state equation.
From the fundamental equation $S(U,V,n)= \frac {nS_0}{n_0}+nR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {n}{n_0} \right ) ^{-5/2} \right ]$, one could get $U(S,V,n)$ which would also be a fundamental equation.
But $U(S,V,n)$ is not equal to $U(T)=\frac{3nRT}{2}$. In other words, U(S,V,n) is not the internal energy of the gas. Then what is it? The total energy? I thought that the total energy of the gas was the internal energy... I'm confused.

 

[Andrew Mason]

For a monoatomic ideal gas, if $c_v$ is constant then the internal energy is worth $U=\frac{3nRT}{2}$. This is a state equation.
From the fundamental equation $S(U,V,n)= \frac {nS_0}{n_0}+nR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {n}{n_0} \right ) ^{-5/2} \right ]$, one could get $U(S,V,n)$ which would also be a fundamental equation.
But $U(S,V,n)$ is not equal to $U(T)=\frac{3nRT}{2}$. In other words, U(S,V,n) is not the internal energy of the gas. Then what is it? The total energy? I thought that the total energy of the gas was the internal energy... I'm confused.

The equations are equivalent. In differential form:

dU = nCvdT = TdS - PdV + μΔN

Assuming N does not change, and dividing by T:

nCvdT/T = dS - PdV/T = dS - nRdV/V

Integrating from T₀ to T:

ΔS = nCvln(T/T₀) + nRln(V/V₀) = nRln(T/T₀)^3/2 + nRln(V/V₀)

Without the change in n, that is just: $S(U,V,n)= nR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \right ]$ which is what you started with.

AM

 

[fluidistic]

The equations are equivalent. In differential form:

dU = nCvdT = TdS - PdV + μΔN

Assuming N does not change, and dividing by T:

nCvdT/T = dS - PdV/T = dS - nRdV/V

Integrating from T₀ to T:

ΔS = nCvln(T/T₀) + nRln(V/V₀) = nRln(T/T₀)^3/2 + nRln(V/V₀)

Without the change in n, that is just: $S(U,V,n)= nR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \right ]$ which is what you started with.

AM

I see yes, but you had to use the other state equation "PV=nRT" to replace P by nRT/V when you integrate with respect to T.
In other words you had to use some extra information to reach the fundamental equation. Because a single state equation is not enough to reach the fundamental equation (you need at least 2, though you could use 3, the other one being the one with mu), they should not be equivalent.
From the fundamental equation you can get any of the 3 state equations without any extra information. But you can't reach the fundamental equation with any of the 3 state equation alone. In that sense the fundamental equation is not equivalent to a single state equation but to at least a set of 2 of them.
So this would mean that U(S,V,n) has more information than U(T). I'm not sure what energy then is U(S,V,n)...

 

[Andrew Mason]

I see yes, but you had to use the other state equation "PV=nRT" to replace P by nRT/V when you integrate with respect to T.

In other words you had to use some extra information to reach the fundamental equation. Because a single state equation is not enough to reach the fundamental equation (you need at least 2, though you could use 3, the other one being the one with mu), they should not be equivalent.
From the fundamental equation you can get any of the 3 state equations without any extra information. But you can't reach the fundamental equation with any of the 3 state equation alone. In that sense the fundamental equation is not equivalent to a single state equation but to at least a set of 2 of them.
So this would mean that U(S,V,n) has more information than U(T).

You just need to know that that U/U₀ = T/T₀, which is true only because it is an ideal gas.

dU = nCvdT is true only for an ideal gas. But dU = TdS-PdV is true for any gas.

I'm not sure what energy then is U(S,V,n)...

It is the internal energy of the gas. It is the total kinetic and potential energies of all the gas molecules.

AM

 

[fluidistic]

You just need to know that that U/U₀ = T/T₀, which is true only because it is an ideal gas.

dU = nCvdT is true only for an ideal gas. But dU = TdS-PdV is true for any gas.

It is the internal energy of the gas. It is the total kinetic and potential energies of all the gas molecules.

AM

Ok I think I'm finally understanding it...
If the gas was a Van der Waals gas then I would still get that U(S,V,n) is the internal energy, which would correspond to kinetic+potential energies.
All in all, despite U(T) contains less information than U(S,V,n), if one completes the information according to the system one has (i.e. give the information that the gas is ideal or Van der Waals for instance) then one get that U(T) is in fact equivalent to U(S,V,n) in the sense that they represent the same energy; the total energy of the system.
Thanks for the replies.

 

[Andrew Mason]

All in all, despite U(T) contains less information than U(S,V,n), if one completes the information according to the system one has (i.e. give the information that the gas is ideal or Van der Waals for instance) then one get that U(T) is in fact equivalent to U(S,V,n) in the sense that they represent the same energy; the total energy of the system.

But for non-ideal gases, such as Van der Waals gases, U is not a function of T alone. U depends on T and V.

AM

 

[fluidistic]

But for non-ideal gases, such as Van der Waals gases, U is not a function of T alone. U depends on T and V.

AM

Oh right. In that case U(T,V,n) would be equivalent to U(S,V,n) if I use "an extra" state equation that tells me it's a Van der Waals gas (like the one that relates P to V).
The expression "U(T,V,n)" makes me think of the Helmholtz free energy.