# Homework Help: Oxygen and mercury in a tube problem

Tags:
1. Dec 26, 2014

### Rugile

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Ideal gas law

3. 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?

2. Dec 26, 2014

### Bystander

Initial condition: capillary containing oxygen confined by 10 cm Hg column,
open to atmosphere.

3. Dec 26, 2014

### Rugile

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

4. Dec 26, 2014

### Bystander

Bottom end.

5. Dec 26, 2014

### 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

6. Dec 26, 2014

### Bystander

Very good.

7. Dec 26, 2014

### Rugile

Thank you for the help!