Question

related rates: problem 3 (1 point) boyles law states that when a sample of gas is compressed at a constant temperature, the pressure ( p ) and the volume ( v ) satisfy the equation ( pv = c ), where ( c ) is a 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?

Explanation:

Step1: Differentiate the equation $PV = C$ with respect to time $t$

Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = P$ and $v = V$. We get $P\frac{dV}{dt}+V\frac{dP}{dt}=0$.

Step2: Rearrange the differentiated equation to solve for $\frac{dV}{dt}$

$\frac{dV}{dt}=-\frac{V}{P}\cdot\frac{dP}{dt}$.

Step3: Substitute the given values into the formula

We are given that $P = 150$ kPa, $V = 600$ $cm^3$, and $\frac{dP}{dt}=20$ kPa/min. Substituting these values into the formula $\frac{dV}{dt}=-\frac{600}{150}\times20$.

Step4: Calculate the value of $\frac{dV}{dt}$

$\frac{dV}{dt}=- 80$ $cm^3$/min.

Answer:

The volume is decreasing at a rate of $80$ $cm^3$/min.