Question

Assuming air is an incompressible fluid, enter an expression for an estimate of the density of air, in terms of the defined quantities and the acceleration due to gravity, g.

Answer 1

Answer:

Density of Air = ($P_{1}$  - $P_{2}$ )/(g x h)

Density of Air =  1.27 Kg/$m^{3}$

Explanation:

Note: This question is not complete, and lacks its first part in which it contains important data to solve for the density of air. But, I have found the similar question and its data. So. I will be solving the question for the sack of understanding and concept.

Missing part: A weather balloon has an absolute-pressure sensor attached. On the ground the sensor reads $P_{1}$  = 1.01x $10^{5}$ Pa. At a height of h = 950 m, the sensor reads $P_{2}$=8.92x$10^{4}$ Pa.

Solution:

Let

$P_{1}$ be the pressure of the balloon at ground.

$P_{2}$ be the pressure of the balloon at height h = 950 m

g = acceleration due to gravity,

In order to derive the expression, we need to find the pressure difference:

Pressure difference = ΔP

ΔP = $P_{1}$  - $P_{2}$

As we know that,

Pressure difference = density x acceleration due to gravity x height.

So,

ΔP = $P_{1}$  - $P_{2}$  = (Density of Air) x (g) x (h)

We need expression for the density of air, so,

Density of Air = ΔP / (g x h)

Hence, the expression is:

Density of Air = ($P_{1}$  - $P_{2}$ )/(g x h)

Now, we can calculate the density of air as well, by putting the values given above in the data.

$P_{1}$  =  1.01 x $10^{5}$

$P_{2}$ = 8.92 x $10^{4}$

g = 9.8 m/s

h = 950 m

So,

Density of Air = ((1.01 x $10^{5}$) - (8.92 x $10^{4}$) )/ (9.8 x 950)

Density of Air =  1.27 Kg/$m^{3}$