Question

When air expands adiabatically (without gaining or losing heat), its pressure P P and volume V V are related by the equation P V 1.4 = C PV1.4=C, where C C is a constant. Suppose that at a certain instant the volume is 450 cm 3 450 cm3 and the pressure is 80 kPa 80 kPa and is decreasing at a rate of 10 kPa/min 10 kPa/min. At what rate is the volume increasing at this instant? (Round your answer to the nearest whole number.)

Answer 1

Answer:

Rate at which the volume is increasing, $\frac{dV}{dt} = 40.18 cm^3/min$

Explanation:

Volume, V = 450 cm³

Pressure, P = 80 kPa

Rate of decrease in pressure, $\frac{dP}{dt} = - 10 kPa/min$

Rate of increase in volume, $\frac{dV}{dt} = ?$

The equation relating the pressure, P and the volume,V

$PV^{1.4} = C$..............(1)

Differentiating both sides with respect to t (Using Products rule)

$1.4 PV^{0.4} \frac{dV}{dt} + V^{1.4} \frac{dP}{dt} = 0$..............(2)

Substitute the necessary parameters into equation (2)

$1.4 * 80*450^{0.4} \frac{dV}{dt} + 450^{1.4} *(-10) = 0\\\\1289.74 \frac{dV}{dt} - 51820.013 = 0\\\\1289.74 \frac{dV}{dt} = 51820.013\\\\\frac{dV}{dt} = \frac{51820.013}{1289.74}\\\\\frac{dV}{dt} = 40.18 cm^3/min$