Question

7. Atmospheric pressure at sea level is 76 cm of mercury. Calculate the
vertical height of air column exerting the above pressure. Assume the density of
air 1.29 kg m-' and that of mercury is 13,600 kg m-? Why is the height
calculated by you far less than actual height of atmosphere​

Answer 1

Answer:

Explanation:

here is your answer

Answer 2

Given :-

Height of the mercury = 76 cm

Density of the air =  $\sf 1.29 \ kg \ m^{-1}$

Density of the mercury = $\sf 13600 \ kg \ m^{-3}$

To Find :-

Why is the height  calculated by you far less than actual height of atmosphere​.

Solution :-

We know that,

• h = Height
• hg = Barometer

Given that,

Height of mercury = 76 cm

Density of air =  $\sf 1.29 \ kg \ m^{-1}$

Density of mercury = $\sf 13600 \ kg \ m^{-3}$

According to the question,

Height of the mercury column = $\sf h_H_{g}$ = 76 cm = 0.76 cm

Verticles height of air column = $\sf h_{air}$

Density of mercury = $\sf \rho _{air} =13600 \ kgm^{-3}$

Density of air = $\sf \rho _{air}=1.29 \ kgm^{-3}$

Then,

Pressure due to air column = Pressure due to Hg column

$\sf \rho _{air} \times h_{air} \times g=h_{H_g} \times g$

$\sf h_{air}=\dfrac{\rho _{H_g} \times h_{H_g}}{\rho _{air}}$

Substituting their values,

$\sf h_{air}=\dfrac{136000 \times 0.76}{1.29}$

$\sf =8012.4 \ m$