In Problem 1.40 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process.

(a) Use the first law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by $dT=\frac{2}{7}\frac{T}{P}dP-\frac{2}{7}\frac{L}{nR}d{n}_w$

where nw is the number of moles of water vapor present, L is the latent heat of vaporization per mole, and I've assumed $f=7/5$for air.

(b) Assuming that the air is always saturated during this process, the ratio nw/n is a known function of temperature and pressure. Carefully express dnw/dz in terms of $dP/dzanddT/dz$, and the vapor pressure $P_{v}T$. Use the Clausius-Clapeyron relation to eliminate $dP/dT$.

(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, $dT/dz$, to the pressure gradient, $dP/dz$. Eliminate Figure 5.18. Cumulus clouds form when rising air expands adiabatically and cools to the dew point (Problem 5.44); the onset of condensation slows the cooling, increasing the tendency of the air to rise further (Problem 5.45). These clouds began to form in late morning, in a sky that was clear only an hour before the photo was taken. By mid-afternoon they had developed into thunderstorms. the latter using the "barometric equation" from Problem 1.16. You should finally obtain $\frac{dT}{dz}=-\left(\frac{2}{7}\frac{Mg}{R}\right)\frac{1+\frac{P_v}{P}\frac{L}{RT}}{1+\frac{2}{7}\frac{P_v}{P}{\left(\frac{L}{RT}\right)}^2}$

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(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (I bar) and $25°C$, then at atmospheric pressure and $0°C$. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?

Step by step solution

01

Part(a) step 1:Given information

We have been given that $dU=-PdV-Ld{n}_w$

02

Part(a) step 2: Simplify

The energy as a function of temperature of:

$U=\frac{f}{2}nRT$

$f$is the number of degree of freedom

$P{V}^{\gamma }=K$

$dT=\frac{2}{7}\frac{T}{P}dP-\frac{2}{7}\frac{L}{nR}d{n}_w$

03

Part(b) step1: Given information

We have been given that $\frac{n_w}{n}=\frac{P_g}{P}$

04

Part(b) Step 2: simplify

We are getting this in the end:

$\frac{d{n}_w}{dz}=\frac{\partial n_w}{\partial P}\frac{dP}{dz}+\frac{\partial n_w}{\partial T}\frac{dT}{dz}$

The partial derivatives are:

$\frac{\partial n_w}{\partial P}=\frac{\partial }{\partial P}\left(\frac{n{P}_g}{P}\right)=-\frac{n{P}_g}{P^2}$

05

Part(c) Step 1: Given information

We have been given that $\frac{dT}{dz}=\frac{2}{7}\frac{T}{P}\frac{dP}{dz}-\frac{2}{7}\frac{L}{nR}\frac{d{n}_w}{dz}$

06

Part(c) Step 2: Simplify

By substituting the results

$\frac{dT}{dz}\left(1+\frac{2}{7}\frac{L}{RP}\frac{d{P}_g}{dT}\right)=\frac{dP}{dz}\left(\frac{2}{7}\frac{T}{P}+\frac{2}{7}\frac{L{P}_g}{R{P}^2}\right)$

From the barometric equation:

$\frac{dP}{dz}=-\frac{MgP}{RT}$

Solve to get:

$\frac{dT}{dz}=-\frac{2Mg}{7R}\frac{\left(1+\frac{P_g}{P}\frac{L}{RT}\right)}{\left(1+\frac{2}{7}\frac{P_g}{P}{\left(\frac{L}{RT}\right)}^2\right)}$

07

Part(d) Step 1: Given information

We have been given that $P_g=C{e}^{-L/RT}$

08

Part(d) Step 2: Simplify

The ratio of vapour pressure

$\frac{L}{RT}=\frac{43.99\times {10}^3\mathrm{J}/\mathrm{mol}}{(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})(298\mathrm{K})}=17.755$

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