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The pressure of the earth鈥檚 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.

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

$P{V}^{纬}=\text{const.}$ 鈥︹�� (1)

And

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

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

Isothermal compression is so slow that the temperature of the gas doesn鈥檛 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,

$P{V}^{纬}=\text{const.}$

$V^{纬}dP+纬V^{纬鈭�1}PdV=0$ 鈥︹�� (3)

Similarly, differentiate equation (2) on both sides,

$T^{\left(\frac{f}{2}\right)}dV+\frac{f}{2}T\left(\frac{f}{2}鈭�1\right)VdT=0$ 鈥︹�� (4)

Divide equation (3) by $V^{纬鈭�1}$on both sides

$\begin{array}{l}\frac{V^{纬}dP+纬V^{纬鈭�1}PdV}{V^{纬鈭�1}}=0\\ \frac{V^{纬}dP}{V^{纬鈭�1}}+纬\frac{V^{纬鈭�1}}{V^{纬鈭�1}}PdV=0\end{array}$

$VdP+纬PdV=0$ 鈥︹�� (5)

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

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

$\boxed{TdV+\frac{f}{2}VdT=0}$ 鈥︹�� (6)

Rearrange equation (5)

$dP=鈭�\frac{纬PdV}{V}$ 鈥︹�� (7)

Rearrange equation (6)

$dT=鈭�\frac{f}{2}\frac{TdV}{V}$ 鈥︹�� (8)

Divide equation (7) by equation (8)

$\frac{dT}{dP}=\frac{2}{f}\frac{TdV}{V}脳\frac{V}{纬PdV}$

$\frac{dT}{dP}=\frac{2}{f}\frac{T}{纬P}$ 鈥︹�� (9)

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

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

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

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

Barometric equation is

$\frac{dP}{dz}=鈭�\frac{mg}{kT}P$ 鈥︹�� (1)

Here,$m$ is mass,$k$ is Boltzmann constant,$T\text{鈥塧苍诲鈥�}P$ is temperature, and pressure respectively.

The relation between pressure, temperature, and volume is

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

Here,$f$ is the degree of freedom,$T\text{鈥塧苍诲鈥�}P$ are temperature and pressure respectively.

Isothermal compression is so slow that the temperature of the gas doesn鈥檛 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)

$\frac{dP}{P}=鈭�\frac{mg}{kT}dz$ 鈥︹�� (3)

From equation (2), it is found that

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

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

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

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