#### Answer

(a) $P = \frac{mg}{A}$
(b) The column of air would be $7.99~km$ high.
(c) The height of $7.99~km$ found in part (b) is a lower limit on the height of the atmosphere.

#### Work Step by Step

(a) The gauge pressure at the bottom of a column is $P = \rho~g~h$
Let $m$ be the mass of the column of air. Then the force exerted on the area at the bottom of the column is the weight of the air $mg$:
$F = mg$
$F = \rho~V~g$
$F = \rho~A~h~g$
Note that $P = \rho~g~h = \frac{\rho~A~h~g}{A} = \frac{F}{A} = \frac{mg}{A}$
(b) We can find $h$:
$P = \rho~g~h$
$\rho~g~h = 101~kPa$
$h = \frac{101~kPa}{\rho~g}$
$h = \frac{1.01~\times 10^5~N/m^2}{(1.29~kg/m^3)(9.80~m/s^2)}$
$h = 7990~m = 7.99~km$
The column of air would be $7.99~km$ high.
(c) Since the atmosphere becomes less dense as we go higher, the actual atmosphere must be higher than the result in part (b) in order to create the measured air pressure on the earth's surface. Therefore, the height of $7.99~km$ found in part (b) is a lower limit on the height of the atmosphere.