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")#