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")