Question

A bulb of constant volume is attached to a manometer tube open at other end as shown in the figure. The manometer is filled with a liquid of density $(1/3{)}^{rd}$ that of mercury. Initially $h$ was $228cm$. Through a small hole in the bulb, gas leaked. Assuming the pressure decreases as $\frac{dP}{dt}=-kP$ and the value of $h$ is $114cm$ after $14min$. What is the value of $k$ (in $hou{r}^{-1}$)? (ln$\left(4/3\right)=0.28$ and density of $Hg=13.6g/mL$)

• A. 0.6
• B. 1.2
• C. 2.4
• D. none of the above

Answer 1

Answer: (B)

1.2

$P=P_0{e}^{-kt};P_0=\frac{228}{3}+76=152cmHg$
At $t=14min;P=\frac{114}{3}+76=114cmHg$
$\Rightarrow 152e^{-kt}=114$
$\Rightarrow k=\frac{1}{t}ln\frac{152}{114}=\frac{1}{14}\times 0.28$
$=0.02\times 60h{r}^{-1}=1.2h{r}^{-1}$