Question

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.

Answer 1

The elevation of the mountain top is approximately 13,972 feet.

Step-by-step explanation:

The difference in barometric pressure between the sea level and the top of the mountain represents the hydrostatic pressure exerted by the column of air. We can calculate the elevation of the mountain top by using the equation:

ΔP = ρgh

Where ΔP is the difference in pressure, ρ is the density of the air, g is the acceleration due to gravity, and h is the elevation. Rearranging the equation, we have:

h = ΔP / (ρg)

Substituting the given values, we have:

h = (30 - 29) in / (0.0075 lb/ft³ * 32.2 ft/sec²)

Simplifying the equation, we get:

h ≈ 13,972 ft

Therefore, the elevation of the mountain top is approximately 13,972 feet.

Answer 2

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

`\rho_m = 846lb/ft^3`

`g = 32.17405ft/s^2`

`h_1 = 1in = (1)/(12) ft`

For the air the defined properties would be

`\rho_a = 0.0075lb/ft^3`

`g = 32.17405ft/s^2`

`h_2 = ?`

We have for equilibrium that

`\text{Pressure change in Air}=\text{Pressure change in Mercury}`

`\rho_m g h_1 = \rho_a g h_2`

Replacing,

`(846)(32.17405)((1)/(12)) = (0.0075)(32.17405)(h_2)`

Rearranging to find
`h_2`

`h_2 = ((846)(32.17405)((1)/(12)) )/((0.0075)(32.17405))`

`h = 9400ft`

Therefore the elevation of the mountain top is 9400ft