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 320 cm3, and the pressure is 95 kPa (kPa = kiloPascals) and is decreasing at a rate of 11 kPa/minute. At what rate is the volume increasing at this instant?

Answer 1

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

P$V^{1.4}$=C

we need to find dV/dt, so we will differentiate the above equation

$V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}$

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

lets solve for dV/dt, we will have

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

$\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}$  (plugged in all the values at the instant)

$\frac{dV}{dt}$ = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

Answer 2

Answer:

26.47 cm³/minute

Explanation:

We first differentiate the expression to obtain the rate of change of volume with time dV/dt

$PV^{1.4} = C\\ d(PV^{1.4})/dt = dC/dt\\1.4PV^{0.4}\frac{dV}{dt} + V^{1.4}\frac{dP}{dt} = 0\\1.4PV^{0.4}\frac{dV}{dt} = -V^{1.4}\frac{dP}{dt}\\ \frac{dV}{dt} = -\frac{V}{1.4P} \frac{dP}{dt}$

Now when V = 320 cm, P = 95 kPa and dP/dt = -11 kPa/minute since it is decreasing.

We substitute these values into the expression for dV/dt. So,

$\frac{dV}{dt} = -\frac{V}{1.4P} \frac{dP}{dt} = -\frac{320 cm^{3} }{1.4 X 95 kPa} X- 11 kPa/minute\\= 26.47 cm^{3}/minute$