I am trying to figure this out but the book is not very specific with its terms so I am not sure what is actually meant by the author.
"Boyle's Law states that when a sample of gas is compressed at constant temperature, the pressure P and volume V satisfy the equation PV=C, where C is constant. Suppose that at a certain instant the volume is 600 $cm^3$ the pressure is 150 kPa and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?"
I am trying to find the rate of change of V for PV=C when the volume is 600 where P is Pressure and the rate of change of P is 20. C is constant. I am not sure what is meant by constant. Something that is constant will have a derivative of 0 but that does not give me a correct answer. Am I missing anything?
Best Answer
If $PV=C$, then taking derivatives with respect to time we get $$\begin{align*} \frac{d}{dt}PV &= \frac{dC}{dt}\\ \frac{dP}{dt}V + P\frac{dV}{dt} &= \frac{dC}{dt}. \end{align*}$$
If $C$ is constant, then $\frac{dC}{dt}=0$. We are interested in the case when $V=600$, $\frac{dP}{dt} = 20$. We are also told that $P=150$ at this moent. Substituting that we get $$20(600) + 150\frac{dV}{dt} = 0,$$ so $$\frac{dV}{dt} = -\frac{12000}{150} = -80.$$ What are the units? The units of volume are cubic centimeters, and the units of time are minutes. So the answer is that $\frac{dV}{dt}$ is $-80\ \mathrm{cm}^3/\mathrm{min}$.
Note that it makes sense that the volume is decreasing: since pressure times volume is constant, and pressure is increasing, the volume must be decreasing (that is, the derivative should be negative, which it is).