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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.

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

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

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,

纬 = adiabatic exponent

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

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

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 $V^{纬-1}$on both sides and simplify in order to solve, we get

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

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

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

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

From equation 7 and 8 we get

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

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

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

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

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

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