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Answer 1 · 2018-01-17T15:57:54

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

So initial volume of air column $V_{1} = 10 a c m^{3}$ $V_1=10acm^3$

And initial pressure of the air column $P_{1} = (p - 72)$ $P_1=(p-72)$ cm of Hg.

Again final volume of air column $V_{2} = 8 a c m^{3}$ $V_2=8acm^3$

And initial pressure of the air column $P_{2} = (p - 71)$ $P_2=(p-71)$ cm of Hg.

By Boyle's law we can write.

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

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

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

$\Rightarrow 2 p = 152$ $=>2p=152$

$\Rightarrow p = \frac{152}{2} = 76$ $=>p=152/2=76$ cm of Hg.