1. The isothermal atmosphere. Consider a tall column of gaseous nitrogen standin

Question

1. The isothermal atmosphere. Consider a tall column of gaseous nitrogen standing in the earths gravitational field, its height is several kilometers, say. The gas is in thermal equilibrium at a uniform temperature T, but the local number density n(z), the number of molecules per cubic meter at height z, will be greater near the bottom.

a) note figure 7.6 and then show why the equation:

P(z)= P(z+(delta)z)+mgn(z)(delta)z

must hold. Start with newton's second law applied to the gas in the height interval z to z+(delta)z.

b) Convert the equation to a differentual equation and solve it for P(z), given that the pressure is P(0) at the base. Recall that the ideal gas law may be used to re-express n(z) in terms of more convenient quantities. Take g to be a constant.

c) Describe the behaviour with height of the local number density. At what height has the number density dropped to one half of it's value at the surface? Provide both algebraic and numerical answers.

the chemical potential hna ptentia." Am. Phys. 35, 483-7 (1967) Problems wosphere, Consider a tall column of gaseous nitrogen standingin field, its height is several kilometers, say. The gas is in thermal l at I. The isotherma the Earth's gravitational equilibrium at a uniform temperature, but the local number density n(2), the of molecules per cubic meter at height z, will be greater near the bottom. a) Note figure 7.6 and then explain why the equation must bold. Start with Newton's second law applied to the gas in the height interval (b) Convert the equation to a differential equation and solve it for PE), given that the pressure is P(0) at the base. Recall that the ideal gas law may be used to re-express nz) in terms of more convenient quantities. Take g to be a constant. (c) Describe the behavior with height of the local number density. At what height has the number density dropped to one-half of its value at the surface? Provide both algebraic and numerical answers. Vipire 1 6 Sketch for analyzing the iso othermal atmosphere

Explanation / Answer

given for the isothermal atmosphere

a. P(z) = P(z + dz) + mg*n(z) dz

consider

a amsll layer of atmosphere, height dz

pressure difference on both sides is P(z) - P(z + dz)

now, this pressure times area of cross section gives us force due to pressure

F = (P(z) - P(z + dz))A

but this force must be equal to the weight of this layer

hence

weight W = Mg

if mass of each molecule was m

then

number density n

n*A*dz*m = M

hence

from force balance

n*A*dz*m *g = (P(z) - P(z + dz))A

hence

P(z) = P(z + dz) + mg*n(z) dz

b. P(z) = P(z + dz) + mg*n(z) dz

hence

-dP/dz = mg*n(z)

now from ideal gas law

P = n(z)RT/Ao

where Ao is avogadro's constant

hence

RTdP/PAo = -dz

integrating from z = 0 to z = z

(RT/Ao)ln(P/Po) = -z

P = Po*e^(-z/kT)

where k is boltzmann's constna

c. as P decreases with z, hence n(z) also decreases with increase in z

now, for n(z) = 0.5no

P(z) = 0.5Po

ln(2) = z/kT

z = kT*ln(2)

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