Question

Adiabatic law for the expansion of air is

PV^1.4=C,P(pressure) in lb/in^2 ,V(volume)in cubic inches, and C(constant). At a specific instant, the

pressure is 40 lb/in^2 and is increasing at the rate of 8 lb/in^2each second. If

C=5/16

, what is the rate of change of the volume at this instant?

Answer 1

$At\space pv^{1.4}=c$

$At\space p=40\frac{lb}{in^2},\frac{dp}{dt}=8\frac{dp}{in^2 sec}$

$c=\frac{5}{16}\\v^{1.4}=\frac{c}{p}$

$v^{1.4}=\frac{\frac{5}{16}}{40}=\frac{1}{128}$

$V^{\frac{14}{10}}=\frac{1}{27}\implies v^{\frac{7}{5}}=2^{-7}\\\implies v=(2^{-7})\times\frac{5}{7}=2^{-5}[v=[\frac{1}{2^5}]=\frac{1}{32}$

Now

$\frac{d}{dt}(v^{1.4})=\frac{d}{dt}(\frac{c}{p})=c\frac{d}{dt}(p^{-1})$

$1.4v^{1.4-1}(\frac{dv}{dt})=c[-1p^{-1-1}]\frac{dp}{dt}$

$1.4v^{0.4}(\frac{dv}{dt})=\frac{-c}{p^2}(\frac{dp}{dt})$

put $v=\frac{1}{32}=\frac{1}{2^5}\space and \space c=\frac{5}{16},\space p=40, \frac{dp}{dt}=8$

$1.4(\frac{1}{25})^{0.4}\frac{dv}{dt}=\frac{-5}{16(40)^2}\times 8$

$\frac{1.4}2^{5\times 0.4}\frac{dv}{dt}=\frac{-5\times8}{16\times40\times40}$

$\frac{1.4}{4}\frac{dv}{dt}=\frac{-1}{16\times 40}$

$\frac{dv}{dt}=\frac{-1}{16\times14}=\frac{-1}{224}$

So rate of change of volume at this instant $=\frac{-1}{224}\space \frac{in^3}{sec}$