Pressure

 ATMOSPHERE AND ATMOSPHERIC PRESSURE

The earth is surrounded by a gaseous envelope extending upto a few thousand kilometres.This gaseous envelope is made of 78% nitrogen,21% oxygen and a small amount of carbob dioxide,water vapour,etc and this gaseous envelope is called earth’s atmosphere.The density of the atmosphere goes on decreasing as one goes up the surface of earth
The pressure exerted by the atmosphere is called atmospheric pressure.
Obviously,it will be maximum at the surface of the earth and will decrease as we move up the surface of earth.The atmospheric pressure is about 
(or Pa)
Thus, at sea level
    1 atm=pressure exerted  by 0.76m of mercury column
        = \( 0.76 \times 13.6 \times 10^3 \times 9.8 = 1.013 \times 10^5 Nm^{ - 2} \)
Thus, 1 atm= \( 1.013 \times 10^5 Nm^{ - 2} = 1.013 \times 10^6 dyncm^{ - 2} \)
The atmospheric pressure is also expressed in terms of the following units:
    1 torr=pressure exerted by 1mm mercury column=133.28Pa
    1 bar=10⁵Pa

HEIGHT OF THE ATMOSPHERE
Let us now calculate the height of the air column that would exert a pressure of \( 1.01 \times 10^5 Nm^{ - 2} \) 
on the surface of earth i.e., just as much that a mercury column of height 0.76m exerts .Since the density of air goes on decreasing as we go up,to determine the value of height of the air column ,the relation P=\(h\rho g\) cannot be applied directly .However taking the density of air as to be uniform(=1.3kg _m^-3),the height h of the air column is given by
        \( \begin{gathered} h \times 1.3 \times 9.8 = 0.76 \times \left( {13.6 \times 10^3 } \right) \times 9.8 \hfill \\ (or)h = \frac{{0.76 \times 13.6 \times 10^3 }} {{1.3}} = 7951m \approx 8km \hfill \\ \end{gathered} \)  

Following two  assumptions have been made in calculating the height of the atmosphere:
(i)The acceleration due to gravity does not change appreciably upto a height of the order of 8km
(ii)The temperature of the atmosphere is same throughout.