# Question #7b328

##### 1 Answer

Well, you haven't specified the conditions (what temperature? What pressure?), but you also haven't said anything about how the temperature changes (or doesn't change??)... I guess I'll have to assume constant temperature... shame on me.

#DeltaS_"isothermal"(T,V) = C_Vcancel(ln|T_2/T_1|)^(0) + nRln|V_2/V_1|#

I get

Write the **total differential** of the entropy as a function of temperature

#dS = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV# #" "bb((1))#

Once we find out what these derivative terms mean, we can integrate

The first derivative term is:

#((delS)/(delT))_V = 1/T ((delU)/(delT))_V = C_V/T# ,#" "bb((2))# where

#C_V# is the constant-volume heat capacity and#U# is the internal energy.

The second term is gotten from the Maxwell Relation for the Helmholtz energy as a function of the natural variables

#dA = -SdT - PdV#

The cross-derivatives are equal because the Helmholtz free energy is a *state function*:

#((delS)/(delV))_T = ((delP)/(delT))_V# #" "bb((3))#

Plugging

#dS = C_V/TdT + ((delP)/(delT))_VdV#

Now, we can't just leave a derivative term here. Evaluating it first would make this easier. Using the **ideal gas law**,

#((delP)/(delT))_V = del/(delT)[(nRT)/V]_V = (nR)/V# #" "bb((4))#

Plugging

#dS = C_V/TdT + nR cdot 1/VdV# #" "bb((5))#

And finally, integrating

#color(green)(DeltaS(T,V)) = int_((1))^((2)) dS = int_(T_1)^(T_2) C_V/TdT + nR int_(V_1)^(V_2) 1/VdV#

#= color(green)(C_Vln|T_2/T_1| + nRln|V_2/V_1|)#

Evidently, we cannot do this without knowing temperature values... So we assume you are somehow implying constant temperature conditions all the way through...

#=> color(blue)(DeltaS_"isothermal") = C_Vcancel(ln|T_2/T_1|)^(0) + nRln|(2cancel(V_1))/cancel(V_1)|#

#= (1 cancel"g He" xx cancel"1 mol"/(4.0026 cancel"g"))("8.314472 J/"cancel"mol"cdot"K")ln2#

#=# #color(blue)("1.44 J/K")#