Question

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^1.4 = C, where C is a constant. Suppose that at a certain instant the volume is 400 cm^3 and the pressure is 80 kPa and is decreasing at a rate of 10 kPa/min. At what rate is the volume increasing at this instant?

Answer 1

The volume increases at 35.71cm^3/min

Explanation:

Given

`PV^(1.4) = C`

`V = 400cm^3`

`P =80kPa`

`(dP)/(dt) =-10kPa/min`

Required

Rate at which volume increases

`PV^(1.4) = C`
`V = 400cm^3`
`P =80kPa`

Differentiate:
`PV^(1.4) = C`

`P*(dV^(1.4))/(dt) +V^(1.4)*(dP)/(dt) = (d)/(dt)C`

By differentiating C, we have:

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

Rewrite as:

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

Solve for
`(dV)/(dt)`

`P*(1.4)*V^(0.4)* (dV)/(dt) =- V^(1.4)*(dP)/(dt)`

`(dV)/(dt) =- (V^(1.4)*(dP)/(dt) )/(P*(1.4)*V^(0.4))`

Substitute values

`(dV)/(dt) =- (400^(1.4)*-10 )/(80*(1.4)*400^(0.4))`

`(dV)/(dt) =(400*10 )/(80*1.4)`

`(dV)/(dt) =(4000 )/(112)`

`(dV)/(dt) =35.71cm^3/min`