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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?

User Whiteulver
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1 Answer

19 votes
19 votes

Answer:

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

Step-by-step 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((\dot P)/(P) \right) \cdot V

If we know that
P = 180\,kPa,
\dot P = 30\,(kPa)/(min) and
V = 300\,cm^(3), then the rate of change of the volume is:


\dot V = -\left((\dot P)/(P) \right) \cdot V


\dot V = -50\,(cm^(3))/(min)

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

User Silentnuke
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