# Is entropy a state function? How? Prove it?

##### 1 Answer

Essentially, this shows a derivation of entropy and that a state function can be written as a total derivative,

From the first law of thermodynamics:

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

where

Solving for

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

since

#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 **state function**.

Let

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

But since

#0 ne (nR)/V#

However, if we multiply through by *integrating factor*, we would get a new function of

#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, **state function** **entropy**, which in this case is a function of

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

and it can be shown that for the definition of the *total derivative* of

#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#