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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?

Sagot :

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:

[tex]P\cdot V = k[/tex] (1)

Where:

[tex]P[/tex] - Pressure, in kilopascals.

[tex]V[/tex] - Volume, in cubic centimeters.

[tex]k[/tex] - Proportionality constant, in kilopascal-cubic centimeters.

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

[tex]\dot P \cdot V + P\cdot \dot V = 0[/tex] (2)

Where:

[tex]\dot P[/tex] - Rate of change of the pressure, in kilopascals per minute.

[tex]\dot V[/tex] - Rate of change of the volume, in cubic centimeters per minute.

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

[tex]P\cdot \dot V = -\dot P\cdot V[/tex]

[tex]\dot V = -\left(\frac{\dot P}{P} \right) \cdot V[/tex]

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

[tex]\dot V = -\left(\frac{\dot P}{P} \right) \cdot V[/tex]

[tex]\dot V = -50\,\frac{cm^{3}}{min}[/tex]

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

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