Question #7b328

1 Answer
Jun 22, 2017

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 #"1.44 J/K"# as a result, but you'll have to supply temperatures if you want a nonisothermal calculation.


Write the total differential of the entropy as a function of temperature #T# and volume #V#:

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

Once we find out what these derivative terms mean, we can integrate #dS# from the initial to final state to find #DeltaS#.

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 #T# and #V#:

#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 #(2)# and #(3)# into #(1)#, this results in:

#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, #PV = nRT#, and assuming helium is ideal:

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

Plugging #(4)# into #(1)#, this means:

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

And finally, integrating #(5)# over the initial and final states gives the change in entropy for this process, which #(i)# heats/cools the gas at constant volume and then #(ii)# expands/contracts the gas at constant temperature.

#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")#