Question

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

p=p(z), p=pressure, p(z)=elevation

Answer 1

Step-by-step explanation:

Expression for vertical fluid pressure is as follows.

`(dP)/(dh) = \rho * 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 = (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.

`(dP)/(dh) = \rho * g`

`(dP)/(dh) = (mP)/(kT) * g`

`(dP)/(P) =  (mg)/(kT)dh`

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

`ln(P_(h))/(P^(o)) = (-mgh)/(kT)`

or,
`P_(h) = P_(o)e^{(-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^{(-mgh)/(kT)}`.