The density of air on the Earth decreases almost linearly with height from 1.22 kg m-3 at sea level to 0.74 kg m-3 at an altitude of 5000 m.

Question 3

The density of air on the Earth decreases almost linearly with height from 1.22 kg m^-3 at sea level to 0.74 kg m^-3 at an altitude of 5000 m.

Atmospheric pressure at the Earth’s surface on a particular day is 100 000 Pa. The value of g between the Earth’s surface and an altitude of 5000 m can be considered to have a constant value of 9.7 m s^-2.

What will be the atmospheric pressure at an altitude of 5000 m?

A 36 000 Pa                B 48 000 Pa                C 52 000 Pa                D 59 000 Pa

Reference: Past Exam Paper – June 2016 Paper 12 Q15

Solution:

Answer: C. 

Atmospheric pressure is caused by the weight of air above the point.

So, the pressure at sea level is due to the weight of air above the sea level. Similarly, the pressure at an altitude of 5000 m is due to the weight of air above a line at a height of 5000 m. The mass of air between the sea level and 5000 m DOES NOT cause a pressure at an altitude of 5000 m.

Let the pressure due to the weight of air between sea level and 5000 m be X.

Pressure at 5000 m = Pressure at sea level – X

Pressure X = hρg

Since the density varies linearly between the sea level and at 5000 m, we can calculate the average density by:

Average density = (1.22 + 0.74) / 2 = 0.98 kg m^-3

Pressure X = hρg = 5000 × 0.98 × 9.7 = 47 530 Pa

Pressure at 5000 m = 100 000 – 47 530 = 52 470 Pa