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The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13,600 kg/m3, respectively.

The local atmospheric pressure in the city in kPa is ?

The local atmospheric pressure in the city in mmHg is ?

Respuesta :

Answer:

Local atmospheric pressure in kPa = 97.932 kPa

Local atmospheric pressure in mm Hg = 734.78 mm Hg

Explanation:

Given:

  • [tex]P_a[/tex] = absolute pressure = 46 kPa = 46000 kPa
  • [tex]\rho_a[/tex] = density of air = [tex]0.828\ kg/m^3[/tex]
  • [tex]\rho_h[/tex] = density of mercury = [tex]13600\ kg/m^3[/tex]
  • [tex]h[/tex] = altitude of the plane = 6400 m

Assume:

  • [tex]P_h[/tex] = pressure decrease due to air at height h
  • [tex]P_{atm}[/tex] = local atmospheric pressure

As we go up the altitude, the pressure decreases. So, the reading of the pressure will be equal to the decrease in pressure due to height subtracted from the local atmospheric pressure.

[tex]\therefore Absolute\ pressure = Atmospheric\ pressure - Pressure\ decrease\\\Rightarrow P_a=P_{atm}-P_h\\\Rightarrow P_{atm}=P_a+P_h\\\Rightarrow P_{atm}=46000+\rho_a g h\\\Rightarrow P_{atm}=46000+0.828\times 9.8\times 6400\\\Rightarrow P_{atm}=97932.16\ Pa\\\Rightarrow P_{atm}=97.932\ kPa[/tex]

Now, conversion of the local atmospheric pressure is quite simple. let us assume that  height of H mm of Hg is required to exert a pressure equivalent to the local atmospheric pressure.

[tex]\therefore P_{atm}= \rho_h g H\\\Rightarrow H = \dfrac{P_{atm}}{\rho_h g}\\\Rightarrow H = \dfrac{97932.16\ Pa}{13600\ kg/m^3 \times 9.8\ m/s^2}\\\Rightarrow H = 0.73478\ m\\\Rightarrow H =734.78\ mm[/tex]

Hence, 734.78 mm Hg is the local atmospheric pressure in mm Hg.

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