Question

A narrow glass tube of length $100\text{cm}$ is closed at both ends. It lies horizontally with $20\text{cm}$ of mercury column in the middle dividing the tube into two compartments I and II of equal length. The air in each compartment is at standard temperature and pressure. The tube is then turned to a vertical position. By what distance will the mercury column be displaced?

Answer 1

$P_2-P_1=20\left(cmofHg\right)$----(1)

For chamber 1

$P_0{V}_0=P_1{V}_1$

$P_1=\frac{P_0{V}_0}{V_1}$---(2)

For chamber 1I

$P_0{V}_0=P_2{V}_2$

$P_2=\frac{P_0{V}_0}{V_2}$ ---(3)

We have $P_2-P_1=20\left(cmofHg\right)$

$\frac{P_0{V}_0}{V_2}-\frac{P_0{V}_0}{V_1}=20\left(cmofHg\right)$

$\frac{\left(76cmofHg\right)A\left(40cm\right)}{A(40-x)cm}-\frac{76\left(cmofHg\right)A40cm}{4(40+x)cm}=20cm$

$76\times 2(\frac{1}{40-x}-\frac{1}{40+x})=1$

$76\times 2\times 2x=1600-x^2$

$\Rightarrow x^2+304x-1600=0$

$x=5.17cm$