Problem on Theory of Gases

Classical Mechanics Level 4

A chamber of volume 1000 cm 3 1000\text{ cm}^3 is evacuated by a pump at constant temperature whose evactuation rate is equal to 50 cm 3 s − 1 50\text { cm}^3\text{ s}^{-1} .

Find the time in which the pressure of the chamber reduces to e e times the original pressure.

Take e = 2.7182 … e = 2.7182 \ldots .


The answer is 20.

Vatsalya Tandon by Vatsalya Tandon , 21, India

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2 solutions

Sardor Yakupov
Jul 21, 2017

T = const, so

p 1 V 1 = p 2 V 2 { p }_{ 1 }{ V }_{ 1 }={ p }_{ 2 }{ V }_{ 2 } p V = ( p + d p ) ( V + s d t ) , pV=(p+dp)(V+sdt), , where s = d V d t s=\frac { dV }{ dt }

p V = p V + d p V + p s d t + s d p d t ; − V d p = p s d t + s d p d t , s d p d t = 0 ; − V d p = p s d t ; d p / p = − s d t / V ; p = C e − s t V pV=pV+dpV+psdt+sdpdt;\quad \\ -Vdp=psdt+sdpdt,\quad sdpdt=0;\\ -Vdp=psdt;\\ dp/p=-sdt/V;\\ p=C{ e }^{ -\frac { st }{ V } }

At t=0 p = p 0 p={ p }_{ 0 } , so C = p 0 p 0 / p = e s t V = 1 t = V s = 1000 50 = 20 s C={ p }_{ 0 }\\ { p }_{ 0 }/p=e\\ \frac { st }{ V } =1\\ t\quad =\frac { V }{ s } =\frac { 1000 }{ 50 } =20s

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Vatsalya Tandon
Nov 19, 2016

In a time d t dt gas expands from V V to V + d V V + dV .

We use-

P V PV = ( P − d P ) ( V + d V ) (P-dP)(V+dV)

d P P = − d V V \frac{dP}{P} = - \frac{dV}{V}

Also, d V d t \frac{dV}{dt} = evactuation rate

Integrating the expression, we get,

t = l n ( e ) . 1000 50 t= ln(e) . \frac{1000}{50}

t = 20 t = 20 seconds

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