Making sense of Differentiation in Thermodynamics

[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

 

[meldraft]

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

 

[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

 

[DrDu]

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

 

[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.

 

[HallsofIvy]

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

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
$$\frac{d(pV^\gamma}{dt}= \frac{dp}{dt}V^\gamma+ \gamma pV^{\gamma- 1}\frac{dV}{dt}$$

 

[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.

 

[DrDu]

http://en.wikipedia.org/wiki/Differential_forms

 

[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?

 

[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 $d(pV^\gamma)=0$.

 

[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 :)

 

[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 $\mathrm d x$

 

[meldraft]

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 $\mathrm d x$

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!)

 

[Calcifur]

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