Pressure (p) and volume (v) given in eqn $p v^{1.4} = k$ $pv^1.4=k$ where k is constant. At certain instant pressure is $\frac{25 g m}{c m^{3}}$ $(25gm)/(cm^3$ and volume is $32 c m^{3}$ $32cm^3$ . If volume is increasing at rate of $\frac{5 c m^{3}}{s}$ $(5cm^3)/s$ , how do you find the rate at which pressure is changing?

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Pressure (p) and volume (v) given in eqn $p v^{1.4} = k$ $pv^1.4=k$ where k is constant. At certain instant pressure is $\frac{25 g m}{c m^{3}}$ $(25gm)/(cm^3$ and volume is $32 c m^{3}$ $32cm^3$ . If volume is increasing at rate of $\frac{5 c m^{3}}{s}$ $(5cm^3)/s$ , how do you find the rate at which pressure is changing?

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Answer 1 · 2018-06-10T23:13:04

$- 5.468$ $-5.468$ grams $c m s^{2} {sec}^{- 1}$ $cms^2 sec^-1$

Explanation:

Assuming that the pressure should be in grams $c m s^{2}$ $cms^2$ and not $c m s^{3}$ $cms^3$ as in the question.

It is given that $p v^{1.4} = K$ $pv^1.4=K$ and we are told that at time $t$ $t$ , Pressure $P = 25$ $P= 25$ and at the same time $t$ $t$ , Volume= $32$ $32$ , from this information we can work out the constant $K$ $K$ .

Thus $25 [32^{1.4}] = K$ $25[32^1.4]=K$ , i.e $K = 3200$ $K = 3200$

So, at time $t$ $t$ , $[p v^{1.4} = 3200]$ $[pv^1.4=3200]$ ......... $[1]$ $[1]$ , differentiating .... $[1]$ $[1]$ implicitly with respect to pressure and time using the product rule, in this case [ $\frac{d}{d t} [p v] = v \frac{d p}{d t} + p \frac{d v}{d t}$ $d/dt[pv]=v[dp]/[dt]+p[dv]/[dt]$ ], where $p$ $p$ and $v$ $v$ and are both functions of $t$ $t$ .

$v^{1.4} \frac{d p}{d t} + p [1.4 v^{0.4}] \frac{d v}{d t} = 0$ $v^1.4[dp]/[dt] + p[1.4v^[0.4]][dv]/[dt]=0$ ......... $[2]$ $[2]$ . We know $\frac{d v}{d t}$ $[dv]/[dt]$ at time $t$ $t$ is $5$ $5$ and we also know that at time $t$ $t$ , $p$ $p$ = $25$ $25$ and that the volume $v = 32$ $v =32$ so we can substitute all these values into..... $[2]$ $[2]$ and solve for $\frac{d p}{d t}$ $[dp]/[dt]$

$\frac{d p}{d t} = - [[25] [1.4] [32^{0.4}] \frac{5}{32^{1.4}}$ $[dp]/[dt] = -[[25][1.4][32^0.4][5]/[32^1.4]$ This is the rate of change of pressure at time $t$ $t$ . and is equal to - $5.468 c m s^{2} {sec}^{- 1}$ $5.468 cms^2 sec^-1$ . Hope this helps.

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Pressure (p) and volume (v) given in eqn pv1.4=kpv^1.4=k where k is constant. At certain instant pressure is 25gmcm3(25gm)/(cm^3 and volume is 32cm332cm^3. If volume is increasing at rate of 5cm3s(5cm^3)/s, how do you find the rate at which pressure is changing?

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