Q. 1.40 - An Introduction to Thermal Physics

Question

In Problem 1.16 you calculated the pressure of earth's atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottom most 10-15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.
(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation
$\frac{dT}{dP}=\frac{2}{f+2}\frac{T}{P}$
(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 1.16b to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately -10^oC/ km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.

Step-by-Step Solution

Answer

a) The differential equation is $\frac{dT}{dP}=\left(\frac{2}{f+2}\right)\frac{T}{P}$

b) The required expression is $\frac{dT}{dz}=-\frac{2mg}{k(f+2)}$

Part(a)Step1: Given information

The atmospheric pressure is function of altitude if temperature is constant. 

The temperature of the bottom most 10-15 km of the atmosphere decreases with increasing altitude due to heating from ground. If the temperature gradient $\left|\frac{dT}{dz}\right|$exceeds a certain critical value then convection will occur.

Part(a)Step2: Explanation

The relation between pressure, volume and temperature in an adiabatic system is given as 

$P{V}^{\gamma }=\text{ constant........................................(1) }$

and 

$V{T}^{\left(\frac{f}{2}\right)}=\text{ const................................................. (2) }$

Where 

P = pressure of the gas, 

V =  volume of the gas, 

T =  temperature in Kelvin, 

γ = adiabatic exponent 

f = degree of freedom  $\left(\gamma =\frac{f+2}{2}\right)$
Now differentiate equation 1 and 2 on both side we get 

$V^{\gamma }dP+\gamma V^{\gamma -1}PdV=0 ................\left(3\right)$

and 

$T^{\left(\frac{f}{2}\right)}dV+\frac{f}{2}{T}^{\left(\frac{f}{2}-1\right)}VdT=0. ...............................\left(4\right)$

Divide equation (3) by $V^{\gamma -1}$ on both sides and simplify in order to solve, we get

$$\begin{gathered} \frac{V^{\gamma }dP+\gamma V^{\gamma -1}PdV}{V^{\gamma -1}}=0 \\ \frac{V^{\gamma }dP}{V^{\gamma -1}}+\gamma \frac{V^{\gamma -1}}{V^{\gamma -1}}PdV=0 \\ VdP+\gamma PdV=0 .....................................\left(5\right) \end{gathered}$$

Divide equation (4) by $T^{\left(\frac{f}{2}-1\right)}$ on both sides and simplify, we get

$$\begin{gathered} \frac{T\left(\frac{f}{2}\right)dV}{{}_T\left(\frac{f}{2}-1\right)}+\frac{f}{2}\frac{T\left(\frac{f}{2}-1\right)}{\left.T^{\left(\frac{f}{2}-1\right.}\right)}VdT=0 \\ TdV+\frac{f}{2}VdT=0...................................\left(6\right) \end{gathered}$$

Now rearrange equation (5)  and equation(6) we get 

$dP=-\frac{\gamma PdV}{V}...................................\left(7\right)$

$dT=-\frac{f}{2}\frac{TdV}{V}.......................................\left(8\right)$

From equation 7 and 8 we get 

$$\begin{gathered} \frac{dT}{dP}=\frac{2}{f}\frac{TdV}{V}\times \frac{V}{\gamma PdV} \\ \frac{dT}{dP}=\frac{2}{f}\frac{T}{\gamma P}..................................\left(9\right) \end{gathered}$$

Substitute $\gamma =\frac{f+2}{f}$ in equation (9) we get

$$\begin{gathered} \frac{dT}{dP}=\frac{2}{f}\times \left(\frac{f}{f+2}\right)\frac{T}{P} \\ \frac{dT}{dP}=\left(\frac{2}{f+2}\right)\frac{T}{P} \end{gathered}$$

Part(b) Step1: Given information

The barometric equation as $\frac{dP}{dz}=-\frac{mg}{kT}P$

Part(b) STep2: Explanation

Relation between pressure, temperature and volume is identified in part (a) as below

$\frac{dT}{dP}=\left(\frac{2}{f+2}\right)\frac{T}{P}..................................\left(1\right)$

Where 

f = degree of freedom, 

T = Temperature  and 

P= pressure 

Generally Isothermal compression is very slow and the temperature of the gas doesn't rise at all .

But in adiabatic compression, the process is very fast and no heat escapes from the gas during the process. 

In case of most real compression processes will be between these extremes usually closer to the adiabatic approximation.

On simplification of barometric equation we get 

$\frac{dP}{P}=-\frac{mg}{kT}dz ..............................\left(2\right)$

From equation (1) and (2) we get 

$dT=\frac{2T}{f+2}\frac{dP}{P}.........................\left(3\right)$

Substitute $\frac{dP}{P}=-\frac{mg}{kT}dz$ in equation (3), we get

$$\begin{gathered} dT=\frac{2T}{f+2}\left(-\frac{mg}{kT}dz\right) \\ \frac{dT}{dz}=-\frac{2mg}{k(f+2)} \end{gathered}$$