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 450 cm3 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? (Round your answer to the nearest whole number.)

Answer 1

Answer:

$\dfrac{dV}{dt}=40.17\ cm^3/min$

Explanation:

Given that

PV¹°⁴ = C

V= 450 cm³

P=80 KPa

dP/dt = - 10 KPa/min

PV¹°⁴ = C

$PV^{\gamma}=C$

By differentiating with respect to time t

$\dfrac{dP}{dt}V^{\gamma}+\gamma PV^{\gamma-1}\dfrac{dV}{dt}=0$

$\dfrac{dP}{dt} + \dfrac{\gamma P}{V}\dfrac{dV}{dt}=0$

Here γ = 1.4

Now by putting the values

$\dfrac{dP}{dt} + \dfrac{\gamma P}{V}\dfrac{dV}{dt}=0$

$-10 + \dfrac{1.4\times 80}{450}\dfrac{dV}{dt}=0$

$\dfrac{dV}{dt}=40.17\ cm^3/min$