Question

A bulb of constant volume is attached to a very thin manometer tube as shown in figure. Gas start leaking through a small hole in the bulb causing change in pressure as:

$\frac{dP}{dt}=-k{P}^2\text{----- 1}$ ref. image

Where k is constant and P is pressure at any instant.
Initial height difference 'h' was 76 cm and after 10 min 'h' was 38 cm. Calculate the value of k in unit of $at{m}^{-1}h{r}^{-1}$.

Answer 1

Solution:-

$\frac{dP}{dt}=-k{P}^2\left(\text{Given}\right)$

$\Rightarrow \frac{dP}{P^2}=-kdt$

Integrating both sides, we have

Here,

For pressure - $P_0\to P$

For time - $t_0\to t$

${\int }_{P_0}^P\frac{dP}{P^2}={\int }_{t_0}^t-kdt$

${[-\frac{1}{P}]}_{P_0}^P={[-kt]}_{t_0}^t$

${\left[\frac{1}{P}\right]}_{P_0}^P=k{\left[t\right]}_{t_0}^t$

$[\frac{1}{P}-\frac{1}{P_0}]=k[t-t_0]$

Here,

$P_0=76+76=152\text{cm Hg}=2atm$

$P=76+38=114\text{cm Hg}=\frac{3}{2}atm$

$t_0=0$

$t=10\text{min}=\frac{1}{6}hr$

$\therefore (\frac{2}{3}-\frac{1}{2})=k(\frac{1}{6}-0)$

$\Rightarrow \frac{1}{6}=\frac{k}{6}$

$\Rightarrow k=1{atm}^{-1}{hr}^{-1}$

Hence the value of $k$ is $1{atm}^{-1}{hr}^{-1}$.