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 900 cm3, the pressure is 100 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?

Answer 1

The value of rate of decrease of volume = - 3600
`(k pa)/(min)`

Explanation:

According to Boyle's law P V = C ------- (1)

Pressure ( P ) = 100 k pa = 10
`(N)/(cm^(2) )`

Volume ( V ) = 900
`cm^(3)`

Put these values in equation ( 1 ) we get,

⇒ C = 10 × 900 = 9000 N-cm = 90 N-m

Differentiate Equation ( 1 ) with respect to time we get,

⇒ V
`(dP)/(dt)` + P
`(dV)/(dt)` = 0

⇒ V
`(dP)/(dt)` = - P
`(dV)/(dt)`

⇒
`(dV)/(dt)` = -
`(V)/(P)`
`(dP)/(dt)` ---------- (2)

This equation gives the rate of decrease of volume.

Given that Rate of increase of pressure =
`(dP)/(dt)` = 40
`(k pa)/(min)`

C = 90 N-m

P =
`10^(5)` pa = 10
`(N)/(cm^(2) )`

V = 900
`cm^(3)`

Put all the above values in equation 2 we get,

⇒
`(dV)/(dt)` = -
`(900)/(10)` × 40

⇒
`(dV)/(dt)` = - 3600
`(k pa)/(min)`

This is the value of rate of decrease of volume.