Question

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

Answer 1

`\dot V=2786.52~cm^3/min`

Step-by-step explanation:

Given:

initial pressure during adiabatic expansion of air,
`P_1=99~kPa`

initial volume during the process,
`V_1=420~cm^3`

The adiabatic process is governed by the relation
`PV^(1.4)=C` ; where C is a constant.

Rate of decrease in pressure,
`\dot P=7~kPa/min`

Then the rate of change in volume,
`\dot V` can be determined as:

`P_1.V_1^(1.4)=\dot P.\dot V^(1.4)`

`99* 420^(1.4)=7* V^(1.4)`

`\dot V=2786.52~cm^3/min`

`\because P\propto(1)/(V)`

`\therefore` The rate of change in volume will be increasing.

Answer 2

`(dV)/(dt)=21.21cm^3/min`

Step-by-step explanation:

We are given that

`PV^(1.4)=C`

Where C=Constant

`(dP)/(dt)=-7KPa/minute`

V=420 cubic cm and P=99KPa

We have to find the rate at which the volume increasing at this instant.

Differentiate w.r.t t

`V^(1.4)(dP)/(dt)+1.4V^(0.4)P(dV)/(dt)=0`

Substitute the values

`(420)^(1.4)* (-7)+1.4(420)^(0.4)(99)(dV)/(dt)=0`

`1.4(420)^(0.4)(99)(dV)/(dt)=(420)^(1.4)* (7)`

`(dV)/(dt)=((420)^(1.4)* (7))/(1.4(420)^(0.4)(99))`

`(dV)/(dt)=21.21cm^3/min`