Question

The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over the 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 13600 kg/m3, respectively. Show this using the systematic solution method.

Answer 1

Answer:

1) The absolute pressure equals = 96.98 kPa

2) Absolute pressure in terms of column of mercury equals 727 mmHg.

Explanation:

using the equation of pressure statics we have

$P(h)=P_{surface}-\rho gh...............(i)$

Now since it is given that at 6400 meters pressure equals 45 kPa absolute hence from equation 'i' we obtain

$P_{surface}=P(h)+\rho gh$

Applying values we get

$P_{surface}=45\times 10^{3}+0.828\times 9.81\times 6400$

$P_{surface}=96.98\times 10^{3}Pa=96.98kPa$

Now the pressure in terms of head of mercury is given by

$h_{Hg}=\frac{P}{\rho _{Hg}\times g}$

Applying values we get

$h=\frac{96.98\times 10^{3}}{13600\times 9.81}=0.727m=727mmHg$