IGCSE Physics 0625: Using a U-Tube Manometer with Δp = ρgΔh

A U-tube manometer contains water of density $1000\text{ kg/m}^3$ . One arm is connected to a laboratory gas supply pipe; the other arm is open to the atmosphere, where the pressure is $1.0\times10^{5}\text{ Pa}$ . Take $g = 10\text{ N/kg}$ .

With the gas supply turned on, the water level in the arm connected to the gas supply is pushed down, and the level in the open arm rises, until there is a vertical difference of $12\text{ cm}$ between the two water levels.

(a) Explain, using the difference in the two water levels, why the gas pressure must be greater than atmospheric pressure. [2]

(b) Calculate the pressure due to this $12\text{ cm}$ difference in water levels, using $\Delta p = \rho g \Delta h$ . [3]

(d) The technician repeats the experiment with the manometer filled with mercury (density $13\,600\text{ kg/m}^3$ ) instead of water, connected to the same gas supply. State and explain how the height difference between the two mercury levels would compare with the $12\text{ cm}$ found using water. [2]

Part (a): Why the gas pressure exceeds atmospheric pressure

In the arm connected to the gas supply, the gas pushes down on the water surface; in the open arm, only atmospheric pressure pushes down. The water level in the gas-side arm has been pushed down relative to the open arm, meaning the gas is pushing harder on its side than the atmosphere is pushing on the open side. If the gas pressure were equal to or less than atmospheric pressure, the gas-side level could not be lower than the open-side level. So the fact that the gas-side level is pushed down shows the gas pressure is greater than atmospheric pressure.

Part (b): Pressure due to the height difference

Convert the height difference to metres:

$12\text{ cm} = 0.12\text{ m}$

Use $\Delta p = \rho g \Delta h$ with $\rho = 1000\text{ kg/m}^3$ and $g = 10\text{ N/kg}$ :

$\Delta p = 1000 \times 10 \times 0.12 = 1200\text{ Pa}$

Part (c): Pressure of the gas supply

The gas pressure equals atmospheric pressure plus the extra pressure needed to push the water level down by the height difference:

$p_{\text{gas}} = p_{\text{atmosphere}} + \Delta p = 1.0\times10^{5}\text{ Pa} + 1200\text{ Pa} = 101\,200\text{ Pa} = 1.012\times10^{5}\text{ Pa}$

Part (d): Repeating with mercury instead of water

The gas supply is unchanged, so it still produces the same pressure difference of $1200\text{ Pa}$ between the two arms. Rearranging $\Delta p = \rho g \Delta h$ for the height difference:

$\Delta h = \frac{\Delta p}{\rho g}$

For mercury ( $\rho = 13\,600\text{ kg/m}^3$ ):

$\Delta h_{\text{mercury}} = \frac{1200}{13\,600 \times 10} = \frac{1200}{136\,000} \approx 0.0088\text{ m} = 8.8\text{ mm}$

This is far smaller than the $12\text{ cm}$ ( $120\text{ mm}$ ) found using water. Mercury is $13.6$ times denser, so it only needs $1/13.6$ of the height difference to produce the same pressure difference. This is why a lighter liquid such as water is often preferred for measuring small gas pressures: it gives a larger, easier-to-read height difference.

Final answers

(a) The gas-side level is pushed below the open-side level, which can only happen if the gas pressure is greater than atmospheric pressure
(b) Pressure due to the height difference $=$ $1200\text{ Pa}$
(c) Gas supply pressure $=$ $1.012\times10^{5}\text{ Pa}$
(d) With mercury, the height difference would be much smaller (about $8.8\text{ mm}$ , compared with $12\text{ cm}$ for water), since a denser liquid needs less height to produce the same pressure difference