Question

An air column closed in a tube sealed at one end by a $Hg$ column having height $h$. When the tube is placed with open ends down, the height of the air column is $l_1$. If the tube is turned so that its open end is at the top, the height of the air column is $l_2$. What is the atmospheric pressure $\left(P_0\right)$?

• A. $P_0=\frac{h(l_1+l_2)}{l_2-l_1}\text{cm of}Hg$
• B. $P_0=\frac{h(l_1-l_2)}{l_1+l_2}\text{cm of}Hg$
• C. $P_0=76\text{cm of}Hg$
• D. $P_0=\frac{h(l_1+l_2)}{l_1-l_2}\text{cm of}Hg$

Answer 1

The correct option is D $P_0=\frac{h(l_1+l_2)}{l_1-l_2}\text{cm of}Hg$

$P_{gas}+h\text{cm of}Hg=P_0\text{cm of}Hg$
i.e., $P_{gas}=P_0\text{cm of}Hg-h\text{cm of}Hg$
$\Rightarrow P_{gas}=(P_0-h)\text{cm of}Hg...\left(i\right)$
$\text{and}{P}_{gas}=(P_0+h)\text{cm}Hg...\left(ii\right)$
By applying Boyle's law between $\left(i\right)$ and $\left(ii\right)$
$(P_0-h)\times l_1=(P_0+h)\times l_2$
$\Rightarrow$ $P_0{l}_1-h{l}_1=P_0{l}_2+h{l}_2$
$\Rightarrow$ $P_0(l_1-l_2)=h(l_1+l_2)$
So, $P_0=\frac{h(l_1+l_2)}{(l_1-l_2)}\text{cm of}Hg$