Question

Cloud formation condition

Consider a simplified model of cloud formation. Hot air in contact with the earth’s surface contains water vapour. This air rises connectively till the water vapour content reaches its saturation pressure. When this happens, the water vapour starts condensing and droplets are formed. We shall estimate the height at which this happens. We assume that the atmosphere consists of the diatomic gases oxygen and nitrogen in the mass proportion $21 : 79$ respectively. We further assume that the atmosphere is an ideal gas, $g$ the acceleration due to gravity is constant and air processes are adiabatic. Under these assumptions one can show that the pressure is given by

$p = p_{0} {(\frac{T_{0} - τ Z}{T_{0}})}^{α}$

Here $p_{0}$ and $T_{0}$ is the pressure and temperature respectively at sea level $(z = 0), τ$ is the lapse rate (magnitude of the change in temperature $T$ with height $z$ above the earth’s surface, i.e. $τ > 0$ ).

(a) Obtain an expression for the lapse rate $Γ$ in terms of $γ, R, g$ and $m_{a}$ . Here $γ$ is the ratio of specific heat at constant pressure to specific heat at constant volume; $R$ , the gas constant; and $m_{a}$ , the relevant molar mass.

(b) Estimate the change in temperature when we ascend a height of one kilometre ?

(c) Show that pressure will depend on height as given by Eq. (1). Find an explicit expression for exponent $α$ in terms of $γ$ .

(d) According to this model what is the height to which the atmosphere extends? Take $T_{0} = 300 K$ and $p_{0} = 1 atm .$

Accepted Answer

$(A)$ According to question, Lapse rate $τ = \frac{d T}{d z}$

For an adiabatic process, $P T^{\frac{γ}{1 - γ}} = Constant$

Differentiate the above expression w.r.t $z$

$p_{0}$

Using Ideal gas equation, $P V = n R T$

$P m_{a} = ρ R T \Rightarrow \frac{P}{T} = \frac{ρ R}{m_{a}} . . . . (3)$

From $(2)$ and $(3)$ , we get

$\Rightarrow \frac{ρ R}{m_{a}} \frac{γ}{1 - γ} \frac{d T}{d z} = ρ g \Rightarrow \frac{d T}{d z} = \frac{m_{a} g (1 - γ)}{R γ} = \frac{m_{a} g (γ - 1)}{R γ} \Rightarrow τ = \frac{m_{a} g (γ - 1)}{R γ} . . . . (4)$

$(B)$ From $(4)$ , $\frac{∆ T}{∆ z} = \frac{m_{a} g (γ - 1)}{Rγ}$

Molar mass of the gas mixture is $m_{a} = \frac{21 (32) + 79 (28)}{100} = 28.84 \times 10^{- 3} kg {mol}^{- 1}$

As the air is mixture of two diatomic gases, hence, the $γ = \frac{7}{5}$ .

$∆ T = \frac{28.84 \times 10^{- 3} (10) (\frac{7}{5} - 1)}{8.314 (\frac{7}{5})} (1000) = 9.91 K$

Hence, temperature change is $9.91 K$ .

$(C)$ From $(1)$ , $τ = |\frac{∆ T}{∆ z}| = |\frac{T - T_{0}}{z - 0}| = \frac{T_{0} - T}{z} \Rightarrow T = T_{0} - τ z$

For adiabatic process, $P T^{\frac{γ}{1 - γ}} = Constant$

$\Rightarrow P T^{\frac{γ}{1 - γ}} = P_{0} {T_{0}}^{\frac{γ}{1 - γ}} \Rightarrow P = P_{0} {(\frac{T_{0}}{T})}^{\frac{γ}{1 - γ}} = P_{0} {(\frac{T_{0}}{T_{0} - T z})}^{\frac{γ}{1 - γ}} = P_{0} {(\frac{T_{0} - T z}{T_{0}})}^{\frac{1 - γ}{γ}}$

Hence, $α = \frac{1 - γ}{γ}$ .

$(D)$ We know that temperature of atmosphere is decreasing with height. Minimum temperature possible is $0 K$ .

From $T = T_{0} - τ z \Rightarrow 0 = T_{0} - τ z \Rightarrow z = \frac{T_{0}}{τ}$

From previous calculations, we know that $τ$ is $9.91 K {km}^{- 1}$ .

$\Rightarrow z = \frac{300}{9.91} = 30.27 \approx 30.3 km$

Important Questions on Thermodynamics

Learn and Prepare for any exam you want !