- #1

- 79

- 1

## 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 20

^{o}C. Then the tube is flipped over and heated to 40

^{o}C. The column of mercury shifts by h = 8cm. What was the initial height of column of the gas if the atmospheric pressure is 10

^{5}Pa?

## Homework Equations

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 p

_{ox}is oxygen pressure, p

_{merc}is mercury pressure, and p

_{a}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 p

_{0}, 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 p

_{0}. Any ideas?