Question

Is entropy a state function? How? Prove it?

Answer 1

Essentially, this shows a derivation of entropy and that a state function can be written as a total derivative, `dF(x,y) = ((delF)/(delx))_ydx + ((delF)/(dely))_xdy`.

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From the first law of thermodynamics:

`dU = deltaq_"rev" + deltaw_"rev"`,

where `q` is the heat flow, `w` is the work (which we define as `-int PdV`), and `delta` indicates that heat flow and work are inexact differentials (path functions).

Solving for `deltaq_"rev"` gives:

`deltaq_"rev" = dU - delw_"rev" = C_V(T)dT + PdV`,

since `((delU)/(delT))_V = C_V`, the constant-volume heat capacity. For an ideal gas, we'd get:

`deltaq_"rev"(T,V) = C_V(T)dT + (nRT)/VdV`

It can be shown that this is an inexact total derivative, indicative of a path function. Euler's reciprocity relation states that for the total derivative

`bb(dF(x,y) = M(x)dx + N(y)dy)`,

where `M(x) = ((delF)/(delx))_y` and `N(y) = ((delF)/(dely))_x`,

a differential is exact if `((delM)/(dely))_x = ((delN)/(delx))_y`. If this is the case, this would indicate that we have a state function.

Let `M(T) = ((delq_"rev")/(delT))_V = C_V(T)`, `N(V) = ((delq_"rev")/(delV))_T = (nRT)/V`, `x = T`, and `y = V`. If we use our current expression for `deltaq_"rev"`, we obtain:

`((delC_V(T))/(delV))_T stackrel(?" ")(=) ((del(nRT"/"V))/(delT))_V`

But since `C_V(T)` is only a function of `T` for an ideal gas, we have:

`0 ne (nR)/V`

However, if we multiply through by `1/T`, called an integrating factor, we would get a new function of `T` and `V` which is an exact differential:

`color(green)((deltaq_"rev"(T,V))/T = (C_V(T))/TdT + (nR)/VdV)`

Now, Euler's reciprocity relation works:

`((del[C_V(T)"/"T])/(delV))_T stackrel(?" ")(=) ((del(nR"/"V))/(delT))_V`

`0 = 0` `color(blue)(sqrt"")`

Therefore, this new function, `(q_"rev"(T,V))/T` can be defined as the state function `S`, entropy, which in this case is a function of `T` and `V`:

`color(blue)(dS(T,V) = (deltaq_"rev")/T)`

and it can be shown that for the definition of the total derivative of `S`:

`dS = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV`

`= ((delS)/(delT))_VdT + ((delP)/(delT))_VdV`
(where we've used a cyclic relation in the Helmholtz free energy Maxwell relation)

which for an ideal gas is:

`= (C_V)/TdV + (nR)/VdV`