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+\gammapV^{\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+\gammapV^{\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!