- #1

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[tex] pV = \nu*R*T [/tex]

to

[tex] p*dV + V*dp = \nu * R * dT[/tex]

I think there's an application of the chain rule in here but I don't see exactly how it's working. Could someone show me the steps in between?

Thanks!

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- Thread starter wumple
- Start date

- #1

- 60

- 0

[tex] pV = \nu*R*T [/tex]

to

[tex] p*dV + V*dp = \nu * R * dT[/tex]

I think there's an application of the chain rule in here but I don't see exactly how it's working. Could someone show me the steps in between?

Thanks!

- #2

- 9

- 0

So in partial derivatives you can take the derivative with respect to one variable and then the derivative with respect to the next variable and so on. The sum of this being your answer. That is how you get that equation.

P(n,V,T) Derivative with respect to P= dP*V, the derivative with respect to V= dV*P and derivative with respect to T= n*R*dT.

So you end up with dP*V+dv*P=n*R*dT.

- #3

- 650

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In a sense, this can also be thought of as being an application of the product rule.

- #4

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thanks! I see it now

- #5

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So in partial derivatives you can take the derivative with respect to one variable and then the derivative with respect to the next variable and so on. The sum of this being your answer. That is how you get that equation.

P(n,V,T) Derivative with respect to P= dP*V, the derivative with respect to V= dV*P and derivative with respect to T= n*R*dT.

So you end up with dP*V+dv*P=n*R*dT.

Actually now I'm looking back at this and I'm not sure I follow exactly what you mean. What are you taking the derivative of? Each side independently and then summing them? As in...

[tex] \frac {\partial} {\partial p} (pv) [/tex]

and then

[tex] \frac {\partial} {\partial v} (pv) [/tex]

etc...and then summing the results, keeping everything on the original side?

Then why do you get dp * v? I always learned that that would be

[tex] \frac {\partial} {\partial p} (pv) = v [/tex]

Where did the dp come from?

Thanks!

- #6

HallsofIvy

Science Advisor

Homework Helper

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[tex]\frac{d(pV)}{dx}= \frac{d(NRT)}{dx}[/tex]

then, using the product rule, not the chain rule,

[tex]p\frac{dV}{dx}+ \frac{dp}{dx}V= NR\frac{dT}{dx}[/tex]

which, in "differential notation", is

[tex]pdV+ Vdp= NRdT[/tex]

- #7

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andp, V, and T are functions (N and R are constants, of course) ofsomeindependent variable, x (whatever it is- I started to write t for time but this is true for any parameter) then differentiating both sides with respect to x,

[tex]\frac{d(pV)}{dx}= \frac{d(NRT)}{dx}[/tex]

then, using the product rule, not the chain rule,

[tex]p\frac{dV}{dx}+ \frac{dp}{dx}V= NR\frac{dT}{dx}[/tex]

which, in "differential notation", is

[tex]pdV+ Vdp= NRdT[/tex]

Oh! ok that makes sense. But then what exactly do you mean by 'differential notation'? I see that if you 'multiply' by dx then that's what you get, but I know that that isn't really very rigorous and not completely correct since a differential isn't a fraction. Does the "[tex]d[/tex]" mean in general a derivative with respect to some arbitrary independent variable? I always thought it was an infinitesimal.

- #8

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- #9

HallsofIvy

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- #10

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More specifically, the differential df of a function f is the 1-form which assigns to any vector field x the number x(f). Differentials are rarely rigorously defined in an introductory calculus sequence as they require a different way of thinking about geometry which would confuse new students, and frankly it is unnecessary to have them formally introduced at that stage. So don't worry about it too much, just treat them using the same rules as derivatives and you'll be fine

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