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3 years ago

13

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 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

2 answers:

DiKsa [7]3 years ago

5 0

Answer:

$\frac{dV}{dt}=21.21cm^3/min$

Explanation:

We are given that

$PV^{1.4}=C$

Where C=Constant

$\frac{dP}{dt}=-7KPa/minute$

V=420 cubic cm and P=99KPa

We have to find the rate at which the volume increasing at this instant.

Differentiate w.r.t t

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

Substitute the values

$(420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0$

$1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)$

$\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}$

$\frac{dV}{dt}=21.21cm^3/min$

kupik [55]3 years ago

3 0

Answer:

$\dot V=2786.52~cm^3/min$

Explanation:

Given:

initial pressure during adiabatic expansion of air, $P_1=99~kPa$

initial volume during the process, $V_1=420~cm^3$

The adiabatic process is governed by the relation $PV^{1.4}=C$ ; where C is a constant.

Rate of decrease in pressure, $\dot P=7~kPa/min$

Then the rate of change in volume, $\dot V$ can be determined as:

$P_1.V_1^{1.4}=\dot P.\dot V^{1.4}$

$99\times 420^{1.4}=7\times V^{1.4}$

$\dot V=2786.52~cm^3/min$

$\because P\propto\frac{1}{V}$

$\therefore$ The rate of change in volume will be increasing.

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Explanation:

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