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3 years ago

11

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

1 answer:

Orlov [11]3 years ago

6 0

Answer:

a

$\Delta P = 7558.6 \ Pa$

b

$h_1 = 10 \ m$

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

$h_1 = \frac{P}{ \rho_w * g }$

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

So

$h_1 = \frac{ 93766.4}{ 1000 * 9.8 }$

=> $h_1 = 10 \ m$

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