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

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

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

=>   $h_1 = 10 \ m$