Fyneman exercise I ch39-5

Q.

An atmosphere in which the pressure and density as a function of height satisfy the relation $P\rho^{-\gamma}=constant$ is called an adiabatic atmosphere.

a)

Show that the temperature of such an atmosphere decreases linearly with height, and find the constant of proportionality. This temperature gradient is called the adiabatic lapse rate. Evaluate this temperature gradient for the earth’s atmosphere.

b)

Use an argument based on energy considerations to show that an atmosphere having less or more temperature gradient than the adiabatic lapse rate will be stable or unstable against convection, respectively.

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A.

a)

From $P\rho^{-\gamma}=constant$,

$\rho^{-\gamma}dP-\gamma P\rho^{-\gamma-1}d\rho = 0$

$dP-\gamma P\rho^{-1}d\rho=0$

$\rho dP=\gamma Pd\rho\tag{1}$

For Ideal gas,

$P=\rho kT$

$dP = kTd\rho+\rho kdT\tag{2}$

Put $d\rho={\rho \over \gamma P}dP$ from eq.(1) to eq.(2)

$dP = kT({\rho \over \gamma P}dP)+\rho kdT$

$(1-{1 \over \gamma})dP=\rho kdT\tag{3}$

Using eq.(1) of previous problem

$dP + m g\rho dh = 0$

eq.(3) becomes

$(1-{1 \over \gamma})(-mg\rho)dh=\rho kdT$

${dT \over dh}=({1 \over \gamma} -1){mg \over k}\tag{4}$

RHS of eq.(4) is constant, which means temperature decrease linearly, because $\gamma$ is greater than 0.

For earth’s atmosphere,

symbol	meaning	value
$\gamma$	Earth’s Heat capacity ratio	1.4
$k$	Boltzmann constant	$1.38 \times10^{-23}$ J/K
$m$	Molecular Mass in Air	$28.97$ kg/kmol

${dT \over dh}=({1 \over 1.4}-1){28.97\times 10^{-3}\times9.8 \over 1.38 \times 10^{-23}\times 6.02\times 10^{23}}\approx-9.8^\circ C/km$

b)

Atmospheric convection is is a means by which thermal energy is distributed. So, more temperature gradient means more energy to distribute. Then convection occurs more actively.