# Dereive an expression for an isobaric/isothermal change in entropy

1. Nov 2, 2009

### 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 $$T$$.

2. Nov 2, 2009

### gabbagabbahey

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

3. Nov 2, 2009

### 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.....

4. Nov 2, 2009

### gabbagabbahey

Looks good so far!

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

Last edited: Nov 2, 2009
5. Nov 2, 2009

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

Now
$$T\mathrm{d}S=\mathrm{d}U+p\mathdm{d}V =f'\mathrm{d}T+\frac{a}{V^2}\mathrm{d}V+p\mathrm{d}V =f'\mathrm{d}T+\frac{RT}{V-b}\mathrm{d}V$$
so
$$\mathrm{d}S=f'\frac{\mathrm{d}T}{T}+\frac{R\mathrm{d}V}{V-b}$$
$$\Delta S=\int \frac{\mathrm{d}f}{T}+R\ln\left|\frac{V_1-b}{V_0-b}\right|$$
With your assumption $f(T)=3RT$ this gives
$$\Delta S=3R\ln\frac{T_1}{T_0}+R\ln\left(\frac{V_1-b}{V_0-b}\right)$$
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.

Last edited: Nov 2, 2009
6. Mar 1, 2010

### Andrew Davies

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

Last edited by a moderator: May 4, 2017