SOLUTIONS FOR EART 121 PROBLEM SET #2 1. [8 pts] Gas Concentrations Consider a mass of air at 20 C and sea level pressure. It contains 21% O2 and 2 ppb methane. (a) Determine the total concentration of air in units of kg/m3 and mol/m3. (b) Do the same as in (a) but for only the oxygen fraction. Report in units of kg/m3, mol/m3, and parts per billion. (c) Do the same as in (b) but for only the methane fraction. SOLUTION: Let’s define c = concentration of gas measured in mol/m3. You need to use the ideal gas law. The forms you’ll need are: n/V = c = P/RT and m/V = ρ = PM/RT (both derived in class) Also recall that Mair = 29 g/mol (a) Using only SI units, ρair = PairMair/RT = (101 x 103 Pa) (0.029 kg/mol) / (8.314 J/mol K * 293 K) = 1.20 kg/m3 cair = n/V = Pair/RT = (101 x 103 Pa) / (8.314 J/mol K * 293 K) = 41.4 mol/m3 (b) If air is 21% oxygen by mole, then: cO2 = 21% * cair = 0.21 * 41.4 mol/m3 = 8.69 mol/m3 [Note that because 21% is by mole, then you can’t do 0.21 * 1.20 kg/m3 to get the right answer for the next part.] Converting cO2 to mass: ρCO2 = 8.69 mol/m3 * 0.032 kg/mol = 0.278 kg/m3 To convert to ppb: if O2 is 21% by mole, then the following is true: 21 / 100 = x / 109, where x is the number we want. Solving, x = 2.1 x 108 ppb = cO2 in ppb (c) Follow the same procedure as for (b), but with MCH4 = 16 g/mol. You get the following values: cCH4 = 8.28 x 10-8 mol/m3 ρCH4 = 1.32 x 10-9 kg/m3

2. [15 pts] Pressure structure On a particular day, the temperature of the atmosphere as a function of height

above is given by T(z) = 25°C – βz, where β is 6.5°C/km (β is called the lapse rate of the atmosphere, the rate at which temperature decreases with height). (a) Derive an equation for the vertical profile of pressure as a function of height, P(z), assuming hydrostatic equilibrium. (b) Calculate the atmospheric pressure in Denver, Colorado (the steps of City Hall are at exactly 1 mile above sea level) using the formula from (a). (c) Plot your answer from (a) using a computer program, and compare it to the solution we determined in class assuming a constant T profile. Start with the same surface T in both cases. Note: Make sure you keep your negative signs straight. Don’t go throwing them away arbitrarily! Solution: Start using the hydrostatic equilibrium equation, dP/dz = – ρg Now use the Ideal Gas Law to substitute for ρ: ρ = PM/RT We will use for T the relationship given in a generic form, T = A – Bz. ** NOTE: In the below solutions, A is used instead of T0 to represent the surface temperature. Just substitute the former for the latter to match your own solution. ** Substituting all these into (1): dP/dz = – (Mg/R) * P / (A – Bz) Rearranging to put all P’s with dP, and all z’s with dz: 1/P * dP = – (Mg/R) * dz/(A – Bz) Now we integrate both sides. I’ll do it using definite limits, but the process is similar with indefinite limits (if you have any questions about how to do it the other way, just ask).

∫ ∫ −−=

P

P

z

dzBzAR

MgdP

P0 0

11

which gives:

zP

PBzA

BR

MgP

00)ln(

1ln −⋅

−⋅−=

RBMg

A

BzA

A

BzA

RB

Mg

P

P

−=

−⋅=

lnlnln

0 Taking the exponential of both sides gives:

RBMg

A

BzPzP

−⋅= 1)( 0

� SOLUTION! where A = 25 C = 298 K, B = 6.5 K/km = 0.0065 K/m, M = 29 g/mol = 0.029 kg/mol, and P0 = 1 atm ~ 1 bar NOTE: You MUST convert A (i.e. T0, a.k.a. the surface temperature), from Celsius to Kelvin, otherwise you will get nonsense once the temperature hits zero Celsius (in which case you would divide by zero). (b) For Denver, z = 1 mile = 1 mile * (1.6 km/1 mile) = 1600 m Substituting, P(1600 m) = 0.829 P0 = 0.829 atm (c) The solution from class using constant T profile (instead of the decreasing T profile) is given by:

−⋅= zRT

MgPzP exp)( 0

where T used should be 298 K (since this is the initial T for the decreasing temperature profile given in this problem). See attached spreadsheet and plot for the solution. Note that the constant T solution does a decent job at lower altitudes, but at higher altitudes, the pressure is lower when the T decrease is accounted for, though not by a tremendous amount (~10% error). Notes on the solution:

Some common mistakes: 1. Used 29 g/mol instead of 0.029 kg/mol. The SI units for mass are kg! 2. Instead of B = 6.5 K/km, you used B = 6.5+273 K/km = 279.5 K/km. This is very very very wrong. Why? Because B represents a temperature difference, i.e. T decreases by 6.5 C (which is the same as 6.5 K) for every 1 km increase in altitude. If you can’t convince yourself of that case, calculate T at 1 km using the original equation in Celsius, then do it again in Kelvin. That should clear things up. 3. Some people solved this using an indefinite integral. Here’s the solution if you chose that route: Most of you got here OK (though some with a dropped minus sign):

∫ ∫ −−= dz

BzAR

MgdP

P

11

From here, integrate, making sure to add a constant of integration: Let’s consider the RHS first: We will use the substitution u = A – Bz; this means du/dz = –B, or dz = –du/B. Don’t lose thatminus sign! Making these substitutions, we get:

)ln(ln11

BzARB

Mgu

RB

MgduuRB

Mg

B

du

uR

Mg−==+=

−

− ∫∫

And the RHS is simply ln P. Adding in the constant of integration, then, our full equation becomes:

CBzARB

MgP +−= )ln(ln

Now, we determine C but using the fact that at z = 0 (sea level), P = P0 = 1 atm = 1.01 x 105 Pa Now it’s all algebra from here. If you put the right values in, you’ll find that C is about –18.44 (in SI units), from which you’ll get the right answer. Note, though, that using an integral between limits is more compact.

3. [7 pts] Sensible and latent heat Consider a 1 m3 volume of air that contains cloud drops in it. These drops are made of liquid water, and the total mass of cloud drops is 2.0 g. (a) If all these drops were to evaporate, would the T of the volume of air increase or decrease? (b) What would be the change in temperature of the air if all the drops evaporated. Neglect any changes in pressure and volume that would occur (they’re very small). Solution: (a) In order to evaporate liquid water, energy must be added due to the latent heat required. This energy comes from the sensible heat of the air, and therefore the air decreases in temperature. (b) The amount of energy that needs to be added is: PE = m * Lv = 2 x 10-3 kg * 2.5 x 106 J/kg = 5 x 103 J where m is mass of liquid water and Lv is latent heat of vaporization of water. This energy comes from the sensible heat of the air, which is given by: PE = m * Cp * PT where m in this case is the mass of the air, since this is what is changing temperature, and Cp is the heat capacity of air = 1000 J/kg K (given in class, or you can look it up). But we don’t know the mass of the air at the moment. But the density of air is ~1.2 kg/m3 (we calculated it in Problem #1). Therefore, mair = 1.2 kg/m3 * 1 m3 = 1.2 kg and therefore PT = 5 x 103 J / (1.2 kg * 1.0 x 103 J/kg K) = 4.2 K This tells us that the evaporation of some cloud water leads to a very appreciable cooling of the air!

4. Albedo and Climate [13 pts] In scenarios of future climate change, warming due to CO2 from fossil fuel burning causes melting of ice and snow, which then replaces a high albedo surface, ice, with a much lower albedo surface such as rock or soil or ocean. This positive feedback leads to further warming worldwide, especially in the high latitudes. (a) Using the “perfect greenhouse gas layer” model, derive an equation for the

change in the Earth’s average surface temperature, Te with respect to albedo α, i.e.

derive an expression for dTe/dα. (b) Use this expression to estimate the change in Te if albedo were to be reduced

from the value of 0.30 that we used in class to 0.29, assuming dTe/dα ~ PTe/Pα. SOLUTION See attached.

Problem #1