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 andU 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")