Get Answers to all your Questions

header-bg qa

a) Pressure decreases as one ascends the atmosphere. If the density of air is \rho, what is the change in pressure dp over a differential height dh?

b) Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is \rho _{0}.

c) If p_{0}=1.03 \times 10^{5}N/m^{2}, \rho _{0}=1.29\frac{kg}{m^{3}} and g is 9.8 \frac{m}{s^{2}} at what height will the pressure drop to (1/10) the value at the surface of the earth?

d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.


Answers (1)

a) Let us assume that there is a packet which has a thickness of dh. In a liquid, at any point, the pressure is working in all the directions in an equal manner. The force as a result of the pressure is opposed by the buoyant force of the liquid.

(P + dP) A - P.A = - Vpg

dP A = - Adhpg

dP = - pgdh -----------(1) {since pressure decreases as height increases, the negative sign is present here}


b) let us assume Po to be the density of air on Earth.

At any point p, the pressure is directly proportional to the density


p = \frac{P}{P_{0}}p_0

now, dP = pgdh

dP = \frac{P}{P_{0}}p_{0}gdh

\frac{dP}{P} = -p_{0}\frac{g}{P_{0}}dh

on integrating both sides we get,

\log \frac{P}{P_{0}} = \frac{-p_{0}}{P_0}gh -----------(3)

\frac{P}{P_{0}} =e ^{\frac{-p_{0}}{P_0}gh}


we know that, \log \frac{P}{P_{0}} = \frac{-p_{0}}{P_0}gh

but, P=\frac{P_{0}}{10} (given)

\log \frac{\frac{P_{0}}{10}}{P_{0}}=\frac{-p_0gh}{P_{0}}

\log \frac{1}{10}=\frac{-p_0gh}{P_{0}}

h=\frac{P_{0}}{p_0g}\log 10

h = \frac{1.013 \times 10^5 \times 2.303}{1.29 \times 9.8} = 18.4 km


the temperature does not remain constant at greater height. It only remains constant near the surface of the Earth.


Posted by


View full answer