Question

I have a Calculus test about derivatives tomorrow, and I am still struggling with the application problems that the professor assigned.

Here's one of them, that I still can't figure out. I know what I have to find, but I don't know how where to begin, how to proceed:

"Charles' law for gases states that if the pressure remains constant, then the relationship between the volume V that a gas occupies and its temperature T (in Celsius) is given by V = V0(1 + (1/273)(T)). Find the average rate of change of T with respect to V."

Answer 1

I know that I have to find the derivative of T with respect to V, that's obvious. But where does the pressure come in, and what would be the value of V0 (V sub zero or however it's usually called in English)? Does V0 actually represent a function? Does that 0 mean that temperature is 0 or simply that it's the initial value of V?

Answer 2

just do a derivative on it

Answer 3

Vo is the initial value i beleive; its just a constsant

Answer 4

http://upload.wikimedia.org/wikipedia/commons/e/e4/Charles_and_Gay-Lussac%27s_Law_animated.gif

Answer 5

It might be easier to start with the whole mess rearranged to show T as a function of V, instead of what you have. Then, you just take the derivative.

Answer 6

\[V _{0}\]is a constant.

Answer 7

So I should first try to leave T alone at one side?

Answer 8

Yep. The only other variable is V, so you can think of T as T(V). Then you can take T'(V).

Answer 9

V = V0(1 + (1/273)(T))

V' = V0'(1 + (1/273)T ) + V0(1 + (1/273)T )'

V' = 0(1 + (1/273)T ) + V0(1/273) T'

V' = V0(1/273) T'

system froze ...

Answer 10

V wrt V = 1 soo V' = 1

V' = V0(1/273) T'

1 = V0(1/273) T' ; and divide off the Vo/273

273/Vo = T'

Answer 11

Nice work.

Answer 12

The other method works just the same; since it's a linear function, we just move some constants and constant factors around to get something of the form T=mV + B. When you take the derivative, you get the slope of the line.

Answer 13

V' = V0'(1 + (1/273)T ) + V0(1 + (1/273)T )'

V' = 0(1 + (1/273)T ) + V0(1/273) T'

In the second line, where did the 1 + of the second term go? That is, it was V0(1 + (1/273)T)' in the previous step; now it's V0(1/273)T'

Answer 14

I think it's really more like\[V=V _{0}(1+T/273)=V _{0}+V _{0}T/273\]\[dV/dV=1=V _{0}/273*dT/dV\]\[273/V _{0}=dT/dV\]

Answer 15

Keep in mind that \[V _{0}\]is a constant, so it's derivative is zero.

Answer 16

I see. I was getting there in my notebook. :) I'll try again. Thanks. :)