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

Answer:

The volume increases at 35.71cm^3/min

Step-by-step explanation:

Given

$PV^{1.4} = C$

$V = 400cm^3$

$P =80kPa$

$\frac{dP}{dt} =-10kPa/min$

Required

Rate at which volume increases

$PV^{1.4} = C$       $V = 400cm^3$       $P =80kPa$

Differentiate: $PV^{1.4} = C$  

$P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = \frac{d}{dt}C$

By differentiating C, we have:

$P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = 0$

Rewrite as:

$P*(1.4)*V^{0.4}* \frac{dV}{dt} + V^{1.4}*\frac{dP}{dt} = 0$

Solve for $\frac{dV}{dt}$

$P*(1.4)*V^{0.4}* \frac{dV}{dt} =- V^{1.4}*\frac{dP}{dt}$

$\frac{dV}{dt} =- \frac{V^{1.4}*\frac{dP}{dt} }{P*(1.4)*V^{0.4}}$

Substitute values

$\frac{dV}{dt} =- \frac{400^{1.4}*-10 }{80*(1.4)*400^{0.4}}$

$\frac{dV}{dt} =\frac{400*10 }{80*1.4}$

$\frac{dV}{dt} =\frac{4000 }{112}$

$\frac{dV}{dt} =35.71cm^3/min$