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:

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

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

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

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

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

Answer 2

Part(a): The average rate of change of $\frac{dP}{dV}$ is $-0.0164 lb/in^{2}$

Part(b): The required answer is $\frac{dV}{dP} \infty \frac{1}{P^{2}}$

Part(a):

Given,

$PV=800$

Differentiating the above equation with respect to $V$.

$\frac{dP}{dV}=-\frac{P}{V}$

Now, at $V=200 in^{3}$ and $P=4 lb/in^2$

At $V=250 in^3$ and $P=\frac{80}{25} lb/in^2$

Hence, we can write,

$\frac{dP}{dV}$ at $V=200$ we get,

$\frac{-1}{200}=\frac{-1}{50}$

Again $\frac{dP}{dV}$ at $V=250$ we get,

$\frac{-80}{25\times 250}$

So, the average rate of change of $\frac{dP}{dV}$ is,

$\frac{-\frac{1}{50}- \frac{8}{25\times25} }2\\=-\frac{1}{100}-\frac{1}{625}\\=-0.0164 lb/in^2$

Part(b):

Given,$PV=800$ then,

$V=\frac{800}{P}$

Now,$\frac{dV}{dP} =-\frac{800}{P^2}$

Then the above implies,

$\frac{dV}{dP} \infty \frac{1}{P^{2}}$

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