Question

In Problem 1.16 you calculated the pressure of the earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 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 $\left|dT/dz\right|$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

$\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{P}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{f}\mathbf{+}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{P}}$

b. Assume that $\mathit{d}\mathit{T}\mathbf{/}\mathit{d}\mathit{z}$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 1.16(b) to find a formula for $\mathit{d}\mathit{T}\mathbf{/}\mathit{d}\mathit{z}$in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately $\mathbf{–}\mathbf{10}\mathbf{°}\mathit{C}\mathbf{/}\mathit{k}\mathit{m}$. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.

Short answer

1. The required differential equation is $\boxed{\frac{dT}{dP}=\left(\frac{2}{f+2}\right)\frac{T}{P}}$.
2. The required expression is $\boxed{\frac{dT}{dz}=-\frac{2mg}{k\left(f+2\right)}}$.

Step-by-step solution

Part a. Step 1. Given.

The pressure of the earth’s atmosphere is a function of altitude with the temperature being constant. The temperature of the bottommost 10-15 km of the atmosphere decreases with increasing altitude due to heating from the ground.

If the temperature gradient$\left|\frac{dT}{dz}\right|$ exceeds a certain critical value then convection will occur.

Part a. Step 2. Formula used.

For adiabatic expansion, the relation between pressure, volume, and temperature is

$P{V}^{\gamma }=\text{const.}$ …… (1)

And

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

Here,$P$ is the pressure of the gas,$V$ is the volume of the gas,$T$ is the temperature in Kelvin,$\gamma$ is the adiabatic exponent, and$f$ is the degree of freedom $\left(\gamma =\frac{f+2}{2}\right)$.

Part a. Step 3. Calculation.

Isothermal compression is so slow that the temperature of the gas doesn’t rise at all and in adiabatic compression, the process is so fast that no heat escapes from the gas during the process. Most real compression processes will be somewhere between these extremes usually closer to the adiabatic approximation.

Differentiate equation (1) on both sides,

$P{V}^{\gamma }=\text{const.}$

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

Similarly, differentiate equation (2) on both sides,

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

Divide equation (3) by $V^{\gamma -1}$on both sides

$\begin{array}{l}\frac{V^{\gamma }dP+\gamma V^{\gamma -1}PdV}{V^{\gamma -1}}=0\\ \frac{V^{\gamma }dP}{V^{\gamma -1}}+\gamma \frac{V^{\gamma -1}}{V^{\gamma -1}}PdV=0\end{array}$

$VdP+\gamma PdV=0$ …… (5)

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

$\begin{array}{c}\frac{T^{\left(\frac{f}{2}\right)}dV+\frac{f}{2}T\left(\frac{f}{2}-1\right)VdT}{T^{\left(\frac{f}{2}-1\right)}}=0\\ \frac{T^{\left(\frac{f}{2}\right)}dV}{T^{\left(\frac{f}{2}-1\right)}}+\frac{f}{2}\frac{T^{\left(\frac{f}{2}-1\right)}}{T^{\left(\frac{f}{2}-1\right)}}VdT=0\end{array}$

$\boxed{TdV+\frac{f}{2}VdT=0}$ …… (6)

Rearrange equation (5)

$dP=-\frac{\gamma PdV}{V}$ …… (7)

Rearrange equation (6)

$dT=-\frac{f}{2}\frac{TdV}{V}$ …… (8)

Divide equation (7) by equation (8)

$\frac{dT}{dP}=\frac{2}{f}\frac{TdV}{V}\times \frac{V}{\gamma PdV}$

$\frac{dT}{dP}=\frac{2}{f}\frac{T}{\gamma P}$ …… (9)

Substitute$\frac{f+2}{f}$ for$\gamma$ in equation (9)

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

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

Part a. Step 4. Conclusion.

Hence the required differential equation is $\boxed{\frac{dT}{dP}=\left(\frac{2}{f+2}\right)\frac{T}{P}}$.

Part b. Step 1. Given.

Barometric equation is

$\frac{dP}{dz}=-\frac{mg}{kT}P$ …… (1)

Here,$m$ is mass,$k$ is Boltzmann constant,$T\text{\hspace{0.17em}and\hspace{0.17em}}P$ is temperature, and pressure respectively.

Part b. Step 2. Formula.

The relation between pressure, temperature, and volume is

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

Here,$f$ is the degree of freedom,$T\text{\hspace{0.17em}and\hspace{0.17em}}P$ are temperature and pressure respectively.

Part b. Step 3. Calculation.

Isothermal compression is so slow that the temperature of the gas doesn’t rise at all and in adiabatic compression, the process is so fast that no heat escapes from the gas during the process. Most real compression processes will be somewhere between these extremes usually closer to the adiabatic approximation.

Simplify equation (1)

$\frac{dP}{P}=-\frac{mg}{kT}dz$ …… (3)

From equation (2), it is found that

$dT=\frac{2T}{f+2}\frac{dP}{P}$ …… (4)

Substitute $-\frac{mg}{kT}dz$for $\frac{dP}{P}$in equation (4)

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

Part b. Step 4. Conclusion.

Hence, the required expression is $\boxed{\frac{dT}{dz}=-\frac{2mg}{k\left(f+2\right)}}$.