At the surface of the Earth, atmospheric pressure is normal, and in the mine – 780 mm Hg.

At the surface of the Earth, atmospheric pressure is normal, and in the mine – 780 mm Hg. Determine the depth in the mine.

Given:

P1 = 101325 Pascal – atmospheric pressure on the Earth’s surface (normal atmospheric pressure);

P2 = 780 mm Hg – atmospheric pressure in the mine;

ro = 1.3 kg / m3 (kilogram per cubic meter) – air density;

g = 10 Newtons per kilogram – acceleration of gravity (approximate value).

It is required to determine h (meter) – the depth of the mine.

Let’s convert the units of measurement of pressure to the SI system:

P2 = 780 mm Hg column = 780 * 133.322 = 103991 Pascal.

Find the pressure difference:

dP = P1 – P2 = 103991 – 101325 = 2666 Pascal.

Then the depth of the mine will be equal to:

h = dP / (ro * g) = 2666 / (1.3 * 10) = 2666/13 = 205 meters.

Answer: the depth of the mine is 205 meters.