Question

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 300 cm3, the pressure is 180 kPa, and the pressure is increasing at a rate of 30 kPa/min. At what rate is the volume decreasing at this instant?

Answer 1

Answer:

The gas is decreasing at a rate of 50 cubic centimeters per minute.

Explanation:

The Boyle's Law is represented by the following expression:

$P\cdot V = k$ (1)

Where:

$P$ - Pressure, in kilopascals.

$V$ - Volume, in cubic centimeters.

$k$ - Proportionality constant, in kilopascal-cubic centimeters.

By definitions of rate of change and implicit differentiation, we derive the following differential equation:

$\dot P \cdot V + P\cdot \dot V = 0$ (2)

Where:

$\dot P$ - Rate of change of the pressure, in kilopascals per minute.

$\dot V$ - Rate of change of the volume, in cubic centimeters per minute.

Then, we clear the rate of change of the volume within (2):

$P\cdot \dot V = -\dot P\cdot V$

$\dot V = -\left(\frac{\dot P}{P} \right) \cdot V$

If we know that $P = 180\,kPa$, $\dot P = 30\,\frac{kPa}{min}$ and $V = 300\,cm^{3}$, then the rate of change of the volume is:

$\dot V = -\left(\frac{\dot P}{P} \right) \cdot V$

$\dot V = -50\,\frac{cm^{3}}{min}$

The gas is decreasing at a rate of 50 cubic centimeters per minute.