Making sense of Differentiation in Thermodynamics

[Calcifur]

Hey there guys,

So I've been doing some Thermodynamics revision particularly involving the equation pV[itex]^{\gamma}[/itex]=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[itex]^{\gamma}[/itex]dp+[itex]\gamma[/itex]pV[itex]^{\gamma-1}[/itex]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[itex]^{\gamma}[/itex]+V[itex]^{\gamma}[/itex]dp
=p[itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex]+Vdp

Many thanks in advance

 

[meldraft]

[tex]d(pV^x)=pd(V^x)+V^x dp=pxV^{x-1}dV+V^x dp[/tex]

 

[Calcifur]

but why does pd(V[itex]^{x}[/itex])=pxV[itex]^{x-1}[/itex]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[itex]^{x}[/itex])=pxV[itex]^{x-1}[/itex] ?

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[itex]^{\gamma}[/itex]=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[itex]^{\gamma}[/itex]dp+[itex]\gamma[/itex]pV[itex]^{\gamma-1}[/itex]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[itex]^{\gamma}[/itex]+V[itex]^{\gamma}[/itex]dp
=p[itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex]+Vdp

Many thanks in advance

That's just bad Calculus. For any variable x, [itex]d(x^n)= nx^{n-1}dx[/itex]. 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
[tex]\frac{d(pV^\gamma}{dt}= \frac{dp}{dt}V^\gamma+ \gamma pV^{\gamma- 1}\frac{dV}{dt}[/tex]

 

[Calcifur]

So when you differentiate V[itex]^{x}[/itex],
you must multiply the differential of V[itex]^{x}[/itex] 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[itex]^{\gamma}[/itex] as two separate differentials and by that I mean:

[itex]\frac{d(pV ^{\gamma})}{dp}[/itex]=V[itex]^{\gamma}[/itex] and [itex]\frac{d(pV^{\gamma})}{dV}[/itex]=p([itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex])

Which can be reformed so that:

[itex]d(pV^{\gamma})[/itex]=V[itex]^{\gamma}[/itex]dp and [itex]d(pV^{\gamma})[/itex]=p([itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex])dV

Which can then be equalised:

V[itex]^{\gamma}[/itex]dp=p([itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex])dV

And thus:

V[itex]^{\gamma}[/itex]dp - p([itex]{\gamma}[/itex]V[itex]^{\gamma-1}[/itex])dV=0

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

 

[DrDu]

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

 

[meldraft]

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

[tex]f(x)=u(x)v(x)[/tex]

then, take its logarithm and differentiate:

[tex]ln[f(x)]=ln[u(x)v(x)]=ln[u(x)]+ln[v(x)][/tex]

[tex]\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)])[/tex]

then do the calculations:

[tex]\frac{1}{f}\frac{df}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}[/tex]

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

[tex]\frac{u v}{f}\frac{df}{dx}=\frac{u v}{u}\frac{du}{dx}+\frac{u v}{v}\frac{dv}{dx}[/tex]

and we get:

[tex]\frac{df}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}[/tex]

If you replace in your problem [itex]f(x)=pV^x[/itex] and [itex]u(x)=p(x),v(x)=V^x[/itex] 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 [itex]\mathrm d x[/itex]

 

[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 [itex]\mathrm d x[/itex]

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!