Making sense of Differentiation in Thermodynamics

1. May 16, 2012

Calcifur

Hey there guys,

So I've been doing some Thermodynamics revision particularly involving the equation pV$^{\gamma}$=constant , which is the adiabatic equation of state.

Now in my notes it says:

"we can differentiate this to obtain a relation between changes in volume and pressure:

V$^{\gamma}$dp+$\gamma$pV$^{\gamma-1}$dV=0"

Now it might be because I'm still half asleep but I don't understand this action, particularly the need for the dV at the end.
Can someone tell me why it is not just:

udv+vdu
=pdV$^{\gamma}$+V$^{\gamma}$dp
=p${\gamma}$V$^{\gamma-1}$+Vdp

Many thanks in advance

2. May 16, 2012

meldraft

$$d(pV^x)=pd(V^x)+V^x dp=pxV^{x-1}dV+V^x dp$$

3. May 16, 2012

Calcifur

but why does pd(V$^{x}$)=pxV$^{x-1}$dV

The part I'm struggling to understand is the why there is a dV at the end.

Why is it not just pd(V$^{x}$)=pxV$^{x-1}$ ?

Thanks

4. May 16, 2012

DrDu

dV^x=(dV^x/dV) dV=x V^(x-1) dV

5. May 16, 2012

meldraft

Exactly, or in words, don't forget that you are calculating infinitesimal amounts. It's the same reason that d(p) is dp and not 1.

6. May 16, 2012

HallsofIvy

That's just bad Calculus. For any variable x, $d(x^n)= nx^{n-1}dx$. There is always a 'd something' in a differential. If you want to get rid of it, you have to specify which variable you are differentiating with respect to. If p and V are both functions of some other variable, say 't', then
$$\frac{d(pV^\gamma}{dt}= \frac{dp}{dt}V^\gamma+ \gamma pV^{\gamma- 1}\frac{dV}{dt}$$

7. May 16, 2012

Calcifur

So when you differentiate V$^{x}$,
you must multiply the differential of V$^{x}$ with the differential of V alone?

Can anyone tell me what rule this is? I understand what happens, I'm just struggling to understand why.

Many thanks.

8. May 16, 2012

DrDu

9. May 16, 2012

Calcifur

Can it be done like this?

The way I see it is you have to treat the differential of pV$^{\gamma}$ as two separate differentials and by that I mean:

$\frac{d(pV ^{\gamma})}{dp}$=V$^{\gamma}$ and $\frac{d(pV^{\gamma})}{dV}$=p(${\gamma}$V$^{\gamma-1}$)

Which can be reformed so that:

$d(pV^{\gamma})$=V$^{\gamma}$dp and $d(pV^{\gamma})$=p(${\gamma}$V$^{\gamma-1}$)dV

Which can then be equalised:

V$^{\gamma}$dp=p(${\gamma}$V$^{\gamma-1}$)dV

And thus:

V$^{\gamma}$dp - p(${\gamma}$V$^{\gamma-1}$)dV=0

If this is correct then why is it minus when in the notes it says plus?

10. May 16, 2012

DrDu

$d (pV^\gamma)=\partial pV^\gamma/\partial p|_V\, dp+\partial pV^\gamma/\partial V|_p \, dV$
Now you are considering an adiabatic process in the course of which the product pV^gamma does not change, hence $d(pV^\gamma)=0$.

11. May 16, 2012

meldraft

No, this comes from the product rule of differentiation. Consider that you have a function:

$$f(x)=u(x)v(x)$$

then, take its logarithm and differentiate:

$$ln[f(x)]=ln[u(x)v(x)]=ln[u(x)]+ln[v(x)]$$

$$\frac{d}{dx}(ln[f(x)])=\frac{d}{dx}(ln[u(x)]+ln[v(x)])=\frac{d}{dx}(ln[u(x)])+\frac{d}{dx}(ln[v(x)])$$

then do the calculations:

$$\frac{1}{f}\frac{df}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}$$

but, since f=u v, we can multiply on both sides by uv:

$$\frac{u v}{f}\frac{df}{dx}=\frac{u v}{u}\frac{du}{dx}+\frac{u v}{v}\frac{dv}{dx}$$

and we get:

$$\frac{df}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$$

If you replace in your problem $f(x)=pV^x$ and $u(x)=p(x),v(x)=V^x$ you have the proof.

Edit: The "no" was for Calcifur's post :)

12. May 16, 2012

nonequilibrium

There's also a confusion in semantics here: "differential" does not mean the same as "derivative"; the former is an infinitesimal quantity, the latter is not; both are related by an infinitesimal factor $\mathrm d x$

13. May 16, 2012

meldraft

Indeed. Note however that even if you omit the dx from the proof above, it's still valid for differentials (which is the case we are discussing here, thanks mr. vodka!)

14. May 16, 2012

Calcifur

Ok, I think I finally understand it now! Thanks to everyone who gave their input! I think I was overcomplicating it!