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Answer 1 · 2018-01-17T15:57:54

Let the cross section of the barometric tube be $acm^2$ and atmospheric pressure is $p$ cm of Hg.

So initial volume of air column $V_1=10acm^3$

And initial pressure of the air column $P_1=(p-72)$ cm of Hg.

Again final volume of air column $V_2=8acm^3$

And initial pressure of the air column $P_2=(p-71)$ cm of Hg.

By Boyle's law we can write.

$(p-72)*10a=(p-71)*8a$

$=>10p-720=8p-568$

$=>10p-8p=720-568$

$=>2p=152$

$=>p=152/2=76$ cm of Hg.