# Question #57bf5

Jan 17, 2018

Let the cross section of the barometric tube be $a c {m}^{2}$ and atmospheric pressure is $p$ cm of Hg.

So initial volume of air column ${V}_{1} = 10 a c {m}^{3}$

And initial pressure of the air column ${P}_{1} = \left(p - 72\right)$ cm of Hg.

Again final volume of air column ${V}_{2} = 8 a c {m}^{3}$

And initial pressure of the air column ${P}_{2} = \left(p - 71\right)$ cm of Hg.

By Boyle's law we can write.

$\left(p - 72\right) \cdot 10 a = \left(p - 71\right) \cdot 8 a$

$\implies 10 p - 720 = 8 p - 568$

$\implies 10 p - 8 p = 720 - 568$

$\implies 2 p = 152$

$\implies p = \frac{152}{2} = 76$ cm of Hg.