Question

What is the correlation between pressure and elevation for an ideal gas?

Answer 1

Explanation:

Expression for vertical fluid pressure is as follows.

               $\frac{dP}{dh} = \rho \times g$ ........... (1)

where,     p = pressure

            $\rho$ = density

               g = acceleration due to gravity

               h = height

Since, g is negative then it means that an increase in height will lead to decrease in pressure.

Also, expression for density according to ideal gas law is as follows.

                 $\rho = \frac{mP}{kT}$ .......... (2)

where,         m = average mass or air molecule

                    P = pressure at a given point

                    k = Boltzmann constant

                    T = temperature in kelvin

Substituting values from equation (2) into equation (1) as follows.

              $\frac{dP}{dh} = \rho \times g$

              $\frac{dP}{dh} = \frac{mP}{kT} \times g$

               $\frac{dP}{P} = \frac{mg}{kT}dh$

Now, on applying integration on both the sides we get the following.

                    $ln\frac{P_{h}}{P^{o}} = \frac{-mgh}{kT}$

or,                          $P_{h} = P_{o}e^{\frac{-mgh}{kT}}$                    

where,           $P_{h}$ = pressure at height h

                      $P_{o}$ = pressure at reference point    

Thus, we can conclude that the correlation between pressure and elevation for an ideal gas is  $P_{h} = P_{o}e^{\frac{-mgh}{kT}}$.