Question

A storm front moves in and Rachel and Pam notice the column of mercury in the barometer rises only to 736 mm. Assume the density of mercury is 13, 000 kg/m 3

(a) What is the change in air pressure?
(b) What if their barometer was filled with water instead of mercury, how high does the column rise? Density of water = 1000 kg/m

Answer 1

a

`\Delta P = 7558.6 \ Pa`

b

`h_1 = 10 \ m`

Step-by-step explanation:

From the question we are told that

The position of the column of mercury in the barometer is
`h = 736 \ mm = 0.76 \ m`\

The density of mercury is
`\rho = 13,000 \ kg / m^3`

Generally the pressure of the atmosphere at that column is mathematically represented as

`P = \rho * g * h`

=>
`P =13 000 * 9.8 * 0.736`

=>
`P = 93766.4 \ Pa`

Generally the atmospheric pressure at sea level (Generally the pressure before the change in level of the mercury column) is
`P_a = 101325 \ Pa`

Generally the change in air pressure is mathematically represented as

`\Delta P = P_a - P`

=>
`\Delta P = 101325 - 93766.4`

=>
`\Delta P = 7558.6 \ Pa`

Generally the height which the column will rise to is mathematically evaluated as

`h_1 = (P)/( \rho_w * g )`

Here
`\rho_w` is the density of water with value
`\rho_w = 1000 \ kg/m^3`

So

`h_1 = ( 93766.4)/( 1000 * 9.8 )`

=>
`h_1 = 10 \ m`