Is entropy a state function? How? Prove it?

1 Answer
Jan 16, 2017

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#.


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#