Dereive an expression for an isobaric/isothermal change in entropy

[trelek2]

I already dereived the following expressions for the Van der Waals gas:
molar energy:
U=3RT - \frac{aP}{RT}
and the expression for the PV, where V is the molar volume:
PV= RT +(b - \frac {a}{RT})P
Using these and the central equation,
TdS = dU + PdV

i am to dereive an expression for dS in two cases:
1) as the substance undergoes an isobaric change from temperature T_{i} to T_{f} at pressure P.
2) as the substance undergoes an isothermal change from pressure P_{i} to P_{f} at temperature T.

 

[gabbagabbahey]

Using these and the central equation,
TdS = dU + PdV

i am to dereive an expression for dS in two cases:
1) as the substance undergoes an isobaric change from temperature T_{i} to T_{f} at pressure P.
2) as the substance undergoes an isothermal change from pressure P_{i} to P_{f} at temperature T.

Okay; and your attempt at a solution is...?

 

[trelek2]

On the example of the isobaric:
dS = \frac {dU}{T}+\frac{P}{T}dV
using the expression for the molar internal energy:
dU = 3RdT+ \frac {aP}{RT^{2}}dT - \frac {a}{RT}dP
and using the expression for the PV product:
V= \frac {RT}{P} +b - \frac{a}{RT}
then dV=\frac {R}{P}dT+ \frac{a}{RT^{2}}dT- \frac{RT}{P^{2}}dP
Substituting into dS (taking dP=0):
dS= 4R \frac{dT}{T}+\frac {2aPdT}{RT^{3}}
and integrating:
dS=4Rln( \frac{T_{f}}{T_{i}})+ \frac {aP}{R(T_{i}-T_{f})}
I have no idea whether this is correct though...

 

[gabbagabbahey]

On the example of the isobaric:
dS = \frac {dU}{T}+\frac{P}{T}dV
using the expression for the molar internal energy:
dU = 3RdT+ \frac {aP}{RT^{2}}dT - \frac {a}{RT}dP
and using the expression for the PV product:
V= \frac {RT}{P} +b - \frac{a}{RT}
then dV=\frac {R}{P}dT+ \frac{a}{RT^{2}}dT- \frac{RT}{P^{2}}dP
Substituting into dS (taking dP=0):
dS= 4R \frac{dT}{T}+\frac {2aPdT}{RT^{3}}

Looks good so far!

and integrating:
dS=4Rln( \frac{T_{f}}{T_{i}})+ \frac {aP}{R(T_{i}-T_{f})}
I have no idea whether this is correct though...

When you integrate dS from the initial state to the final state, you get \Delta S\equiv S_f-S_i, not dS.

 

[Gerenuk]

I found slightly a different equation for the van-der-Waals gas. Also in many places you seem to introduce the ideal gas law for a moment. That's where little differences come from, but maybe the difference isn't big though.

The van-der-Waals equation as I found it is
<br /> \left(p+\frac{a}{V^2}\right)(V-b)=RT<br />
In any case for a van-der-Waals gas the energy is
<br /> U=f(T)-\frac{a}{V}<br />
You have assumed that f(T)=3RT, which is strictly speaking an addition to the van-der-Waals equation, but probably justified.

Now
<br /> T\mathrm{d}S=\mathrm{d}U+p\mathdm{d}V<br /> =f&#039;\mathrm{d}T+\frac{a}{V^2}\mathrm{d}V+p\mathrm{d}V<br /> =f&#039;\mathrm{d}T+\frac{RT}{V-b}\mathrm{d}V<br />
so
<br /> \mathrm{d}S=f&#039;\frac{\mathrm{d}T}{T}+\frac{R\mathrm{d}V}{V-b}<br />
<br /> \Delta S=\int \frac{\mathrm{d}f}{T}+R\ln\left|\frac{V_1-b}{V_0-b}\right|<br />
With your assumption f(T)=3RT this gives
<br /> \Delta S=3R\ln\frac{T_1}{T_0}+R\ln\left(\frac{V_1-b}{V_0-b}\right)<br />
Anyone agrees if that is OK? This derivation has not made the ideal gas law substitutions pV=RT that seem to be in your calculation.

 

[Andrew Davies]

http://theory.phy.umist.ac.uk/~judith/stat_therm/node51.html