Solve the problem.
-Boyle's law states that if the temperature of a gas remains constant, then PV $=c$, where $P$ is the pressure, $\mathrm{V}$ is the volume, and $\mathrm{c}$ is a constant. Given a quantity of gas at constant temperature, if $\mathrm{V}$ is decreasing at a rate of $9 \mathrm{in} .3 / \mathrm{s}$, at what rate is $\mathrm{P}$ increasing when $\mathrm{P}=70 \mathrm{lb} / \mathrm{in}^{2}$ and $\mathrm{V}=80$ in. 3 ?
A) $\frac{63}{8} \mathrm{lb} / \mathrm{in}^{2}{ }^{2} \mathrm{~s}$
B) $\frac{72}{7} \mathrm{lb} / \mathrm{in}^{2}{ }^{-\mathrm{s}}$
C) $\frac{49}{64} \mathrm{lb} / \mathrm{in}^{2}{ }^{-\mathrm{s}}$
D) $\frac{5600}{9} \mathrm{lb} /$ in. $^{2}-\mathrm{s}$

Question

Quizplus · Accepted Answer

By Boyle's law, $PV=c$. Differentiating both sides with respect to $t$, we get$$\frac{d}{dt}(PV)=\frac{d}{dt}(c)$$$$P\frac{dV}{dt}+V\frac{dP}{dt}=0$$$$\frac{dP}{dt}=-\frac{P}{V}\frac{dV}{dt}$$When $P=70 \mathrm{lb} / \mathrm{in}^{2}$ and $V=80$ in. 3 ,$$\frac{dP}{dt}=-\frac{70}{80}(-9)=\frac{63}{8} \mathrm{lb} / \mathrm{in}^{2}{ }^{-\mathrm{s}}$$

Solve the Problem =c=c=c , Where PPP Is the Pressure V\mathrm{V}V Is the Volume, And

Solve the Problem $=c$ , Where $P$ Is the Pressure $\mathrm{V}$ Is the Volume, And