IPhO 2008 · Theory problem 3

Change of Air Temperature with Altitude, Atmospheric Stability and Air Pollution

Download PDFIPhO 2008

Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants’ concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.

We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above 119 m.

Let us consider the air as an ideal diatomic gas, with molar mass \(\mu=29\ \mathrm{g/mol}\).

Quasi equilibrium adiabatic transformation obey the equation \(pV^\gamma=\mathrm{const}\), where \(\gamma=\dfrac{c_p}{c_V}\) is the ratio between isobaric and isochoric heat capacities of the gas.

The student may use the following data if necessary:

The universal gas constant is \(R=8.31\ \mathrm{J/(mol.K)}\).

The atmospheric pressure on ground is \(p_0=101.3\ \mathrm{kPa}\)

The acceleration due to gravity is constant, \(g=9.81\ \mathrm{m/s^2}\)

The molar isobaric heat capacity is \(c_p=\dfrac72R\) for air.

The molar isochoric heat capacity is \(c_V=\dfrac52R\) for air.

Mathematical hints

a. \[ \int\frac{dx}{A+Bx} =\frac1B\int\frac{d(A+Bx)}{A+Bx} =\frac1B\ln(A+Bx) \]

b. The solution of the differential equation \(\dfrac{dx}{dt}+Ax=B\) (with \(A\) and \(B\) constant) is \[ x(t)=x_1(t)+\frac BA \] where \(x_1(t)\) is the solution of the differential equation \(\dfrac{dx}{dt}+Ax=0\).

c. \[ \lim_{x\to\infty}\left(1+\frac1x\right)^x=e \]

1. Change of pressure with altitude.

1.1. Assume that the temperature of the atmosphere is uniform and equal to \(T_0\). Write down the expression giving the atmospheric pressure \(p\) as a function of the altitude \(z\).

1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation

\[ T(z)=T(0)-\Lambda z \]

where \(\Lambda\) is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is \(-\Lambda\)).

1.2.1. Write down the expression giving the atmospheric pressure \(p\) as a function of the altitude \(z\).

1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of \(\Lambda\) does the free convection occur?

2. Change of the temperature of an air parcel in vertical motion

Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.

As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value \(p(z)\), with \(z\) - the altitude of the parcel center. The temperature in the parcel is uniform and equals to \(T_{\mathrm{parcel}}(z)\), which is generally different from the temperature of the surrounding air \(T(z)\). In parts 2.1 and 2.2, we do not make any assumption about the form of \(T(z)\).

2.1. The change of the parcel temperature \(T_{\mathrm{parcel}}\) with altitude is defined by \[ \frac{dT_{\mathrm{parcel}}}{dz}=-G. \] Derive the expression of \(G(T,T_{\mathrm{parcel}})\).

2.2. Consider a special atmospheric condition in which at any altitude \(z\) the temperature \(T\) of the atmosphere equals to that of the parcel \(T_{\mathrm{parcel}}\), \(T(z)=T_{\mathrm{parcel}}(z)\). We use \(\Gamma\) to denote the value of \(G\) when \(T=T_{\mathrm{parcel}}\), that is \[ \Gamma=-\frac{dT_{\mathrm{parcel}}}{dz} \] (with \(T=T_{\mathrm{parcel}}\)). \(\Gamma\) is called dry adiabatic lapse rate.

2.2.1. Derive the expression of \(\Gamma\)

2.2.2. Calculate the numerical value of \(\Gamma\).

2.2.3. Derive the expression of the atmospheric temperature \(T(z)\) as a function of the altitude.

2.3. Assume that the atmospheric temperature depends on altitude according to the relation \(T(z)=T(0)-\Lambda z\), where \(\Lambda\) is a constant. Find the dependence of the parcel temperature \(T_{\mathrm{parcel}}(z)\) on altitude \(z\).

2.4. Write down the approximate expression of \(T_{\mathrm{parcel}}(z)\) when \(|\Lambda z|\ll T(0)\) and \(T(0)\approx T_{\mathrm{parcel}}(0)\).

3. The atmospheric stability.

In this part, we assume that \(T\) changes linearly with altitude.

3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude \(z_0\), i.e. it has the same temperature \(T(z_0)\) as that of the surrounding air. If the parcel is moved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur:

  • The air parcel finds its way back to the original altitude \(z_0\), the equilibrium of the parcel is stable. The atmosphere is said to be stable.
  • The parcel keeps moving in the original direction, the equilibrium of the parcel is unstable. The atmosphere is unstable.
  • The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral.

What is the condition on \(\Lambda\) for the atmosphere to be stable, unstable or neutral?

3.2. A parcel has its temperature on ground \(T_{\mathrm{parcel}}(0)\) higher than the temperature \(T(0)\) of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of \(\Lambda\) and \(\Gamma\).

4. The mixing height

4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7: 00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula \(T(z)=T(0)-\Lambda z\) with different lapse rates \(\Lambda\) in the three layers \(0<z<96\ \mathrm{m}\), \(96\ \mathrm{m}<z<119\ \mathrm{m}\) and \(119\ \mathrm{m}<z<215\ \mathrm{m}\).

Consider an air parcel with temperature \(T_{\mathrm{parcel}}(0)=22^\circ\mathrm{C}\) ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of 96 m and 119 m.

4.2. Determine the maximal elevation \(H\) the parcel can reach, and the temperature \(T_{\mathrm{parcel}}(H)\) of the parcel.

\(H\) is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer.

Table 1

Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi.

Altitude, m Temperature, °C
5 21.5
60 20.6
64 20.5
69 20.5
75 20.4
81 20.3
90 20.2
96 20.1
102 20.1
109 20.1
113 20.1
119 20.1
128 20.2
136 20.3
145 20.4
153 20.5
159 20.6
168 20.8
178 21.0
189 21.5
202 21.8
215 22.0
225 22.1
234 22.2
246 22.3
257 22.3

5. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi.

Hanoi metropolitan area can be approximated by a rectangle with base dimensions \(L\) and \(W\) as shown in the figure, with one side taken along the south-west bank of the Red River.

Figure 1

It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are \(8\times10^5\) motorbikes on the road, each running on average 5 km and emitting 12 g of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate \(M\) during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides \(L\) of the rectangle) with velocity \(u\), passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere.

Also, we use the following rough approximate model:

  • The CO spreads quickly throughout the entire volume of the mixing layer above the Hanoi metropolitan area, so that the concentration \(C(t)\) of CO at time \(t\) can be assumed to be constant throughout that rectangular box of dimensions \(L\), \(W\) and \(H\).
  • The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind.
  • Before 7:00 am, the CO concentration in the atmosphere is negligible.

5.1. Derive the differential equation determining the CO pollutant concentration \(C(t)\) as a function of time.

5.2. Write down the solution of that equation for \(C(t)\).

5.3. Calculate the numerical value of the concentration \(C(t)\) at 8:00 a.m.

Given \(L=15\ \mathrm{km}\), \(W=8\ \mathrm{km}\), \(u=1\ \mathrm{m/s}\).

Official English text of the problem.

Answers

Write \(T_0=T(0)\), \(T_p=T_{\mathrm{parcel}}\), and \(T_{p0}=T_p(0)\). Temperature ratios use kelvin. In Part 5, \(t\) is the elapsed time since 7:00 am.

  • 1.1. \[ \boxed{p(z)=p_0\exp\!\left(-\frac{\mu gz}{RT_0}\right)} \]

  • 1.2.1. For \(\Lambda\ne0\) and \(T_0-\Lambda z>0\), \[ \boxed{p(z)=p_0\left(1-\frac{\Lambda z}{T_0}\right)^{\mu g/(R\Lambda)}} \] The limit \(\Lambda=0\) gives the result of 1.1.

  • 1.2.2. Using the density-increase criterion specified in this part, \[ \boxed{\Lambda>\frac{\mu g}{R} =3.42\times10^{-2}\ \mathrm{K\,m^{-1}} =34.2\ \mathrm{K\,km^{-1}}} \]

  • 2.1. \[ \boxed{G(T,T_p)=\frac{\mu g}{c_p}\frac{T_p}{T} =\frac{\gamma-1}{\gamma}\frac{\mu g}{R}\frac{T_p}{T}} \]

  • 2.2.1. \[ \boxed{\Gamma=\frac{\mu g}{c_p}=\frac{2\mu g}{7R}} \]

  • 2.2.2. \[ \boxed{\Gamma=9.78\times10^{-3}\ \mathrm{K\,m^{-1}} =9.78\ \mathrm{K\,km^{-1}}} \]

  • 2.2.3. \[ \boxed{T(z)=T_0-\Gamma z} \]

  • 2.3. \[ \boxed{ T_p(z)= \begin{cases} T_{p0}\left(1-\dfrac{\Lambda z}{T_0}\right)^{\Gamma/\Lambda}, &\Lambda\ne0,\\[6pt] T_{p0}\exp\!\left(-\dfrac{\Gamma z}{T_0}\right), &\Lambda=0. \end{cases}} \]

  • 2.4. To first order in height, \[ \boxed{T_p(z)\simeq T_{p0}-\Gamma\frac{T_{p0}}{T_0}z \simeq T_{p0}-\Gamma z} \]

  • 3.1. \[ \boxed{ \begin{aligned} \Lambda<\Gamma &: \text{stable},\\ \Lambda>\Gamma &: \text{unstable},\\ \Lambda=\Gamma &: \text{neutral}. \end{aligned}} \]

  • 3.2. In the intended mixing-height approximation, neglecting inertial overshoot, \[ \boxed{z_{\max}\simeq\frac{T_{p0}-T_0}{\Gamma-\Lambda}}, \qquad \Lambda<\Gamma. \] The corresponding expression without linearizing the parcel temperature is given in the solution.

  • 4.1. \[ \boxed{T_p(96\ \mathrm m)\simeq21.1^\circ\mathrm C,\qquad T_p(119\ \mathrm m)\simeq20.8^\circ\mathrm C} \]

  • 4.2. \[ \boxed{H\simeq144\ \mathrm m,\qquad T_p(H)\simeq20.6^\circ\mathrm C} \]

  • 5.1. \[ \boxed{\frac{dC}{dt}+\frac{u}{W}C=\frac{M}{LWH}, \qquad C(0)=0} \]

  • 5.2. During the rush hour, \[ \boxed{C(t)=\frac{M}{uLH}\left(1-e^{-ut/W}\right)} \]

  • 5.3. \[ \boxed{C(8{:}00\ \mathrm{am})\simeq2.2\ \mathrm{mg\,m^{-3}}} \] Keeping intermediate digits gives \(2.24\ \mathrm{mg\,m^{-3}}\).

Solution

1.1. Isothermal atmosphere

Hydrostatic equilibrium of a horizontal atmospheric layer requires \[ \frac{dp}{dz}=-\rho g. \] For ideal air, the mass density is \[ \rho=\frac{\mu p}{RT}. \] Thus, when \(T=T_0\), \[ \frac{dp}{p}=-\frac{\mu g}{RT_0}\,dz. \] Integrating from the ground, where \(p=p_0\), to altitude \(z\), \[ \ln\frac{p(z)}{p_0}=-\frac{\mu gz}{RT_0}. \] Therefore, \[ \boxed{p(z)=p_0e^{-\mu gz/(RT_0)}}. \]

1.2.1. Atmosphere with a constant temperature lapse rate

With \(T(z)=T_0-\Lambda z\), hydrostatic equilibrium becomes \[ \frac{dp}{p}=-\frac{\mu g}{R}\frac{dz}{T_0-\Lambda z}. \] For \(\Lambda\ne0\), integration gives \[ \ln\frac{p(z)}{p_0} =\frac{\mu g}{R\Lambda} \ln\frac{T_0-\Lambda z}{T_0}. \] Hence \[ \boxed{ p(z)=p_0\left(1-\frac{\Lambda z}{T_0}\right)^{\mu g/(R\Lambda)} }. \] This formula applies where the absolute temperature remains positive.

As \(\Lambda\to0\), \[ \frac{\mu g}{R\Lambda} \ln\left(1-\frac{\Lambda z}{T_0}\right) \longrightarrow-\frac{\mu gz}{RT_0}, \] recovering the isothermal result.

1.2.2. Density increasing with height

Using \(\rho=\mu p/(RT)\), \[ \frac{d\ln\rho}{dz} =\frac{d\ln p}{dz}-\frac{d\ln T}{dz} =-\frac{\mu g}{RT}+\frac{\Lambda}{T}. \] Therefore, \[ \frac{d\rho}{dz} =\frac{\rho}{T}\left(\Lambda-\frac{\mu g}{R}\right). \] Since \(\rho,T>0\), density increases upward precisely when \[ \boxed{\Lambda>\frac{\mu g}{R}}. \] Numerically, \[ \frac{\mu g}{R} =\frac{0.029(9.81)}{8.31} =0.0342347\ \mathrm{K\,m^{-1}}, \] so the requested condition is \[ \boxed{\Lambda>34.2\ \mathrm{K\,km^{-1}}}. \]

This is the density-inversion criterion stated in this question. The adiabatic parcel-instability criterion in Part 3 is less restrictive.

2.1. Temperature change of an adiabatic parcel

The parcel contains a fixed amount of gas, remains in pressure equilibrium with its surroundings, and exchanges negligible heat with them. Its adiabatic and ideal-gas relations are \[ pV^\gamma=\mathrm{const}, \qquad pV=nRT_p. \] Eliminating \(V\), \[ T_p^\gamma p^{1-\gamma}=\mathrm{const}, \] or \[ T_p p^{-(\gamma-1)/\gamma}=\mathrm{const}. \] Taking logarithmic derivatives, \[ \frac{1}{T_p}\frac{dT_p}{dz} =\frac{\gamma-1}{\gamma}\frac1p\frac{dp}{dz}. \]

The pressure gradient is set by the surrounding atmosphere, whose temperature is \(T\): \[ \frac1p\frac{dp}{dz}=-\frac{\mu g}{RT}. \] Consequently, \[ \frac{dT_p}{dz} =-\frac{\gamma-1}{\gamma}\frac{\mu g}{R}\frac{T_p}{T}. \]

Because \(c_p-c_V=R\) and \(\gamma=c_p/c_V\), \[ \frac{\gamma-1}{\gamma}=\frac{R}{c_p}. \] Thus \[ \boxed{G(T,T_p)=\frac{\mu g}{c_p}\frac{T_p}{T}}. \]

The factor \(T_p/T\) matters: the parcel’s cooling rate is not exactly constant when its temperature differs from that of its surroundings.

2.2.1. Dry adiabatic lapse rate

When \(T_p=T\), the temperature ratio in the preceding expression equals one: \[ \boxed{\Gamma=\frac{\mu g}{c_p}}. \] For the specified diatomic gas, \[ c_p=\frac72R, \] so \[ \boxed{\Gamma=\frac{2\mu g}{7R}}. \]

2.2.2. Numerical value

Substitution gives \[ \Gamma =\frac{2(0.029)(9.81)}{7(8.31)} =0.00978133\ \mathrm{K\,m^{-1}}. \] Therefore, \[ \boxed{\Gamma=9.78\ \mathrm{K\,km^{-1}}}. \]

2.2.3. Atmospheric temperature profile

Under the special condition \(T(z)=T_p(z)\), \[ \frac{dT}{dz}=-\Gamma. \] Since \(\Gamma\) is constant, \[ \boxed{T(z)=T(0)-\Gamma z}. \]

2.3. Parcel temperature in an atmosphere with constant \(\Lambda\)

From Part 2.1, \[ \frac{dT_p}{dz} =-\Gamma\frac{T_p}{T_0-\Lambda z}. \] Separating variables, \[ \frac{dT_p}{T_p} =-\Gamma\frac{dz}{T_0-\Lambda z}. \] Integrating, \[ \ln\frac{T_p(z)}{T_{p0}} =\frac{\Gamma}{\Lambda} \ln\left(1-\frac{\Lambda z}{T_0}\right). \] Hence, for \(\Lambda\ne0\), \[ \boxed{ T_p(z)=T_{p0} \left(1-\frac{\Lambda z}{T_0}\right)^{\Gamma/\Lambda} }. \]

If \(\Lambda=0\), the surrounding atmosphere is isothermal. Direct integration instead yields \[ \boxed{T_p(z)=T_{p0}e^{-\Gamma z/T_0}}. \]

For a layer beginning at altitude \(z_i\), the same result can be written \[ T_p(z)=T_p(z_i) \left[\frac{T(z)}{T(z_i)}\right]^{\Gamma/\Lambda}, \] which allows successive layers to be joined continuously.

2.4. Linear approximation

Expanding the result of Part 2.3 about \(z=0\), \[ T_p(z) =T_{p0}\left[ 1-\frac{\Gamma z}{T_0} +\frac{\Gamma(\Gamma-\Lambda)z^2}{2T_0^2} +\cdots \right]. \] To first order, \[ T_p(z)\simeq T_{p0}-\Gamma\frac{T_{p0}}{T_0}z. \] With \(T_{p0}\approx T_0\), \[ \boxed{T_p(z)\simeq T_{p0}-\Gamma z}. \]

Strictly, this is a small-height expansion. If \(\Lambda\) is very small, the stated condition \(|\Lambda z|\ll T_0\) alone does not ensure linearity; one also needs \(\Gamma|z|/T_0\ll1\). Both temperature changes are small compared with the absolute temperature in Part 4.

3.1. Stability of a displaced parcel

Let the parcel initially be at \(z_0\), with \[ T_p(z_0)=T(z_0)=T_*. \] Displace it upward by a small signed distance \(\xi\). At its initial equilibrium, its adiabatic temperature gradient is \(-\Gamma\), so \[ T_p(z_0+\xi)=T_*-\Gamma\xi+O(\xi^2). \] The environmental temperature is \[ T(z_0+\xi)=T_*-\Lambda\xi. \] The parcel’s temperature excess is therefore \[ T_p-T=(\Lambda-\Gamma)\xi+O(\xi^2). \]

At equal pressure, density is inversely proportional to absolute temperature. The parcel’s upward acceleration due to buoyancy and gravity is \[ a_z =g\left(\frac{\rho}{\rho_p}-1\right) =g\left(\frac{T_p}{T}-1\right) \simeq\frac{g}{T_*}(\Lambda-\Gamma)\xi. \]

Thus:

  • If \(\Lambda<\Gamma\), acceleration opposes the displacement: the atmosphere is stable.
  • If \(\Lambda>\Gamma\), acceleration reinforces the displacement: the atmosphere is unstable.
  • If \(\Lambda=\Gamma\), the parcel remains at the same temperature and density as its surroundings: the atmosphere is neutral.

Hence \[ \boxed{ \text{stable: }\Lambda<\Gamma,\qquad \text{unstable: }\Lambda>\Gamma,\qquad \text{neutral: }\Lambda=\Gamma. } \]

3.2. Height reached by an initially warmer parcel

In the linear approximation, \[ T_p(z)\simeq T_{p0}-\Gamma z, \qquad T(z)=T_0-\Lambda z. \] Therefore, \[ T_p(z)-T(z) =(T_{p0}-T_0)-(\Gamma-\Lambda)z. \]

In a stable atmosphere, \(\Gamma-\Lambda>0\), so the parcel’s initial temperature excess decreases with height. The neutral-buoyancy level is obtained by setting this difference to zero: \[ \boxed{ z_{\max}\simeq \frac{T_{p0}-T_0}{\Gamma-\Lambda} }. \] Here the notation \(z_{\max}\) follows the problem’s intended mixing-height model, which neglects inertial overshoot.

For completeness, the neutral-buoyancy height can also be found without the approximation of Part 2.4. Setting the exact parcel temperature equal to the atmospheric temperature gives \[ T_{p0} \left(1-\frac{\Lambda z}{T_0}\right)^{\Gamma/\Lambda} =T_0\left(1-\frac{\Lambda z}{T_0}\right). \] Thus, for \(\Lambda\ne0\), \[ \boxed{ z_{\mathrm n} =\frac{T_0}{\Lambda} \left[ 1-\left(\frac{T_0}{T_{p0}}\right)^{\Lambda/(\Gamma-\Lambda)} \right]. } \] For \(\Lambda=0\), its limit is \[ \boxed{ z_{\mathrm n}=\frac{T_0}{\Gamma} \ln\frac{T_{p0}}{T_0}. } \] Both reduce to the stated linear result when \(T_{p0}-T_0\ll T_0\).

A dynamical qualification is necessary: zero buoyancy means zero acceleration, not necessarily zero velocity. A freely moving parcel can overshoot this level. For example, in the same linear approximation, an undamped parcel released from rest obeys \[ \ddot z\simeq\frac{g}{T_0} \left[(T_{p0}-T_0)-(\Gamma-\Lambda)z\right], \] and its first turning point is \(2z_{\mathrm n}\). The following parts use the intended neutral-buoyancy mixing height.

4.1. Parcel temperatures at 96 m and 119 m

Use straight lines through the measured endpoints of each specified layer. Their lapse rates are

\[ \Lambda_1 =\frac{21.5-20.1}{96-5} =0.0153846\ \mathrm{K\,m^{-1}}, \] \[ \Lambda_2=0, \] and \[ \Lambda_3 =\frac{20.1-22.0}{215-119} =-0.0197917\ \mathrm{K\,m^{-1}}. \]

The first line extrapolates to a ground temperature \[ T(0)\simeq21.5+5\Lambda_1 =21.5769^\circ\mathrm C, \] close to the parcel’s initial \(22^\circ\mathrm C\). All relevant temperature changes are small compared with approximately \(294\ \mathrm K\), so Part 2.4 applies.

Since a kelvin temperature difference equals a Celsius temperature difference, \[ T_p(z)\simeq22^\circ\mathrm C-\Gamma z. \] Consequently, \[ T_p(96) \simeq22-0.00978133(96) =21.06099^\circ\mathrm C, \] and \[ T_p(119) \simeq22-0.00978133(119) =20.83602^\circ\mathrm C. \]

To the precision of the temperature data, \[ \boxed{T_p(96\ \mathrm m)\simeq21.1^\circ\mathrm C}, \qquad \boxed{T_p(119\ \mathrm m)\simeq20.8^\circ\mathrm C}. \]

As a check, applying the exact formula of Part 2.3 separately to the first layer and the isothermal second layer gives \(21.059^\circ\mathrm C\) and \(20.833^\circ\mathrm C\), respectively. These round to the same answers.

4.2. Mixing height

At both 96 m and 119 m, the surrounding temperature is \(20.1^\circ\mathrm C\). The parcel is warmer at both heights, so it continues into the inversion layer above 119 m.

Write the positive temperature gradient there as \[ k=-\Lambda_3 =\frac{22.0-20.1}{215-119} =0.0197917\ \mathrm{K\,m^{-1}}. \] The environmental temperature in that layer is \[ T(z)=20.1^\circ\mathrm C+k(z-119\ \mathrm m). \]

At the mixing height, the parcel and atmosphere have equal temperatures: \[ 22-\Gamma H=20.1+k(H-119), \] with \(H\) in metres and the displayed temperatures in degrees Celsius. Solving, \[ H =\frac{22-20.1+119k}{\Gamma+k} =\frac{4.2552083}{0.029572997} =143.888\ \mathrm m. \] This lies within the assumed inversion layer, as required.

The parcel temperature there is \[ T_p(H) \simeq22-\Gamma H =20.5926^\circ\mathrm C. \] Therefore, \[ \boxed{H\simeq144\ \mathrm m,\qquad T_p(H)\simeq20.6^\circ\mathrm C}. \]

Using the exact parcel-temperature evolution through all three linear atmospheric layers gives \(H=143.78\ \mathrm m\), confirming that the linear parcel approximation is adequate.

5.1. Conservation equation for CO

The uniformly mixed air volume is \[ V=LWH. \] If \(C(t)\) is the CO mass concentration, the total CO mass inside the box is \[ m_{\mathrm{CO}}=LWH\,C. \]

The wind crosses a face of width \(L\) and height \(H\), so its volume flow rate is \[ Q=uLH. \] Because incoming air is clean, the incoming CO flux is zero. The outgoing CO mass rate is \[ QC=uLH\,C. \] Conservation of CO therefore gives \[ \frac{d}{dt}(LWH\,C)=M-uLH\,C. \] The box dimensions are constant, hence \[ \boxed{ \frac{dC}{dt}+\frac{u}{W}C=\frac{M}{LWH} }, \qquad \boxed{C(0)=0}. \]

5.2. Concentration as a function of time

A constant particular solution is \[ C_{\mathrm{ss}}=\frac{M}{uLH}. \] The homogeneous solution is proportional to \(e^{-ut/W}\), so \[ C(t)=\frac{M}{uLH}+A e^{-ut/W}. \] Using \(C(0)=0\), \[ A=-\frac{M}{uLH}. \] Thus \[ \boxed{ C(t)=\frac{M}{uLH}\left(1-e^{-ut/W}\right) }, \qquad 0\le t\le3600\ \mathrm s. \]

The ventilation timescale is \(W/u\). In the limit of no wind, this expression correctly reduces to \(C(t)=Mt/(LWH)\).

5.3. CO concentration at 8:00 am

The total CO emitted during the hour is \[ m_{\mathrm{emitted}} =(8\times10^5)(5\ \mathrm{km}) (12\ \mathrm{g\,km^{-1}}) =4.8\times10^7\ \mathrm g =4.8\times10^4\ \mathrm{kg}. \] Therefore, \[ M=\frac{4.8\times10^4}{3600} =13.3333\ \mathrm{kg\,s^{-1}}. \]

Using \[ L=15000\ \mathrm m,\quad W=8000\ \mathrm m,\quad u=1\ \mathrm{m\,s^{-1}},\quad H\simeq143.888\ \mathrm m, \] the dimensionless elapsed time is \[ \frac{ut}{W} =\frac{1(3600)}{8000} =0.45. \] Hence \[ C(3600) = \frac{13.3333}{(1)(15000)(143.888)} \left(1-e^{-0.45}\right) \ \mathrm{kg\,m^{-3}}, \] giving \[ C(3600) =2.2386\times10^{-6}\ \mathrm{kg\,m^{-3}} =2.2386\ \mathrm{mg\,m^{-3}}. \]

With precision appropriate to the rough input data, \[ \boxed{C(8{:}00\ \mathrm{am})\simeq2.2\ \mathrm{mg\,m^{-3}}}. \]

The solution was checked against the official solution.