Question

A $10.0\text{cm}$ column of air is trapped by a column of 4 cm long $Hg$ in a capillary tube of uniform bore when the tube is held horizontally at $1\text{atm}$. The length of the air column when the tube is held vertically with the open end up is:
Given: $P_{atm}=76$ cm of Hg

• A. $3.53\text{cm}$
• B. $10.52\text{cm}$
• C. $9.50\text{cm}$
• D. $4.61\text{cm}$

Answer 1

The correct option is C $9.50\text{cm}$

When open end is up, the weight of mercury represents an increase in the pressure on the gas.
$4\text{cm}Hg=40\text{mm}Hg$

Thus, final pressure $=760+40=800\text{mm}Hg$

Also, $\text{Volume}=\text{Area}\times \text{length}$
$P_1{V}_1=P_2{V}_2$ (by Boyle's law)
$P_1\times A\times l_1=P_2\times A\times l_2$
($A=$ area of cross section = constant)
So, $P_1{l}_1=P_2{l}_2$
$\Rightarrow$ $760\times 10=800\times l_2$
$\Rightarrow$ $l_2=9.50\text{cm}$
Length of air column $=9.50\text{cm}$