# Oxygen and mercury in a tube problem

## Homework Statement

We have a tube with the top end open and the bottom end closed. There is some oxygen gas in the tube and on top of the oxygen there is 10 cm high column of mercury. The initial temperature is 20oC. Then the tube is flipped over and heated to 40oC. The column of mercury shifts by h = 8cm. What was the initial height of column of the gas if the atmospheric pressure is 105 Pa?

Ideal gas law

## The Attempt at a Solution

We can assume that the gas is ideal, from which we know that $\frac{pV}{T} = const$ thus $\frac{p_0 V_0}{T_0} = \frac{p_1 V_1}{T_1}$. When we flip the tube over we also know that $p_{ox} + p_{merc} = p_a$, where pox is oxygen pressure, pmerc is mercury pressure, and pa is atmospheric pressure. Also, we can state that $V_1 = V_0 + Sh$, $Sh_1 = Sh_0 + Sh$, thus we can rewrite the equation: $\frac{p_0 S h_0}{T_0} = \frac{(p_a - p_{merc}) S (h_0 + h)}{T_1} => \frac{p_0 h_0}{T_0} = \frac{(p_a - p_{merc}) (h_0 + h)}{T_1}$. The only additional unknown here is p0, which I don't know how to calculate. I was thinking of using the equation $p_0 V_0 = \frac{m}{M}RT_0$, but I don't have enough information to solve that equation for p0. Any ideas?

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Bystander
Homework Helper
Gold Member
Initial condition: capillary containing oxygen confined by 10 cm Hg column,
10 cm high column of mercury.
open to atmosphere.

So do you mean that $p_a = p_{merc} + p_0$? But what about the fact that the bottom end is sealed?

Bystander
Homework Helper
Gold Member
Bottom end.

• Rugile
Sooo, the pressure of the gas is caused by the atmospheric pressure and mercury, thus $p_0 = p_{merc} + p_a$ instead? I'm confused

Bystander