Pressure: Question 10

Worked solution

Part (a): Atmospheric pressure at sea level

Convert the column height to metres:

$760\text{ mm} = 0.760\text{ m}$

Use $\Delta p = \rho g \Delta h$ with $\rho = 13\,600\text{ kg/m}^3$ and $g = 10\text{ N/kg}$:

$\Delta p = 13\,600 \times 10 \times 0.760 = 103\,360\text{ Pa} = 1.0336\times10^{5}\text{ Pa}$

Part (b): Atmospheric pressure at the top of the mountain

Convert the new column height to metres:

$700\text{ mm} = 0.700\text{ m}$

$\Delta p = 13\,600 \times 10 \times 0.700 = 95\,200\text{ Pa} = 9.52\times10^{4}\text{ Pa}$

Part (c): Decrease in atmospheric pressure

Subtract the two calculated pressures:

$103\,360\text{ Pa} - 95\,200\text{ Pa} = 8160\text{ Pa}$

Part (d): Why pressure is lower at the top of the mountain

Atmospheric pressure at any point is caused by the weight of the column of air above that point, acting over a given area. At the top of the mountain, the barometer is higher up, so there is a shorter column of air remaining above it than there is above a barometer at sea level. A shorter column of air has a smaller weight, and since pressure is weight per unit area, this smaller weight of air produces a lower atmospheric pressure at the mountain top than at sea level, which is exactly why the mercury column falls from $760\text{ mm}$ to $700\text{ mm}$.

Final answers

• (a) Atmospheric pressure at sea level $=$ $1.0336\times10^{5}\text{ Pa}$
• (b) Atmospheric pressure at the mountain top $=$ $9.52\times10^{4}\text{ Pa}$
• (c) Decrease in atmospheric pressure $=$ $8160\text{ Pa}$
• (d) A shorter column of air above the barometer at altitude has a smaller weight, giving a lower atmospheric pressure