A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density `(1//3)^("rd")` that of mercury. Initially h was 228 cm.

Through a small hole in the bulb gas leaked assuming pressure decreases as `(dP)/(dt)=-kp.`If value of h is 114 cm after 14 minutes, what is the value of k (in `"hour"^(-1)`)?
[ Use : In (4/3) =0.28 and density of Hg =13.6 g/mL.].

Options

A. 0.6
B. 1.2
C. 2.4
D. None of these

Answer

B

Explanation

`P=P_(0)e^(-kt) , P_(0)=(228)/( 3) + 76= 152cm Hg`
At `t=14`min, `P =(114)/( 3) +76 = 114 cm Hg `
`:. 152 e^(-kt) = 114`
`rarr k = ( 1)/( t)ln"" ( 152)/( 114) = ( 1)/( 14) xx 0.28`
`= 0.0 2 xx 60hr^(-1) = 1.2 hr^(-1)`