Question

According to Boyle's Law, if the temperature of a confined gas is held fixed, then the product of the pressure P and the volume V is a constant, suppose that, for a certain gas, PV=800 where P is measured in pounds per square inch and V is measured in cubic inches. Answer the following: (this is calculus)
A) Find the average rate of change of P as V increases from 200in^3 to 250in^3.
B) Express V as a function of P and show that the instantaneous rate of change of V with respect to P is inversely proportional to the square of P.

Answer 1

If PV = 800, then P can be written as a function of V,

P(V) = 800 / V

(a) The average rate of change of P as V increases from 200 to 250 in³ is then

(P (250) - P (200)) / (250 in³ - 200 in³) = (3.2 lb/in² - 4lb/in²) / (50 in³)

... = -0.016 (lb/in²)/in³

(Or -0.016 lb/in⁵, but I figure writing the rate as (units of pressure) per (unit volume) makes more sense.)

(b) We can also write V as a function of P :

V(P) = 800 / P

Take the derivative:

V'(P) = - 800 / P²

which immediately demonstrates that V'(P) ∝ 1 / P², as required. (The fish-looking symbol, ∝, means "is proportional to".)

If differentiating is supposed to be more involved, you can use the limit definition:

$V'(P)=\displaystyle\lim_{h\to0}\frac{V(P+h)-V(P)}h$

$V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800}{P+h}-\frac{800}P}h$

$V'(P)=\displaystyle\lim_{h\to0}\frac{\frac{800P-800(P+h)}{P(P+h)}}h$

$V'(P)=\displaystyle800\lim_{h\to0}\frac{-\frac h{P(P+h)}}h$

$V'(P)=\displaystyle-800\lim_{h\to0}\frac1{P(P+h)}=-\dfrac{800}{P^2}$