CHEMISTRY 253

Spring, 2015 - Dixon

Homework Set 3.1 Solutions

For Problems to be Turned In

Ch. 5: Problems: 1, 6

1.  Calculate the ratio of the rates of energy release by two otherwise-identical blackbodies, one of which is at 0ºC and the other at 17ºC.  At what temperature is the rate of release twice that at 0ºC?

E₁/E₂ = T₁⁴/T₂⁴ = (273.15K/290.15)⁴ = 0.785

T₂⁴ = (E₂/E₁)T₁⁴ or T₂ = (E₂/E₁)^0.25T₁ = (2)^0.25(273.15K) = 324.8 K = 51.7ºC

6.  (a) Given that the atmospheric burden of carbon (as CO₂) increases by about 4.1 Gt annually, calculate the increase in the ppm concentration of carbon dioxide that this brings about.

(b) Given that its total concentration was 390 ppm in 2010, calculate the total mass of CO₂ that was present in the air.  After converting to mass of carbon, does your answer agree with the value listed in Figure 5-9?  Note that the mass of the atmosphere = 5.1 x 10²¹ g, and that air’s average molar mass = 29.0 g mol^-1.

a)  D<CO₂ mixing ratio>/Dt

= [(4.1 x 10⁹ t C)(10⁶ g/1 t)/(12.0 g/mol C)/1 year]/[(5.1 x 10²¹ g air)/(29.0 g mol^-1)]

= (3.417 x 10¹⁴ mol/1.759 x 10²⁰ mol)(10⁶) = 1.9 ppm/year

b) moles CO₂ =  (390 ppm)(5.1 x 10²¹ g air)/(29.0 g mol^-1)/10⁶ = 6.86 x 10¹⁶ mol CO₂

mass CO₂ = (6.86 x 10¹⁶ mol CO₂)(44.0 g/mol) = 3.02 x 10¹⁸ g

To compare this to figure 5-9, we convert to g C (x 12/44) and then to Gt (x1 Gt/10¹⁵ g) = 820 Gt vs. ~800 Gt in figure.

Additional Problems: 3, 6, 8

3.  Anthropogenic carbon dioxide emissions into the atmosphere amounted to 178 Gt from Jan. 1990 to Dec. 1997.  Calculate the fraction of this emitted carbon dioxide that remained in the air, given that in  that same eight-year  period, the carbon dioxide concentration in air rose by 11.1 ppm.  Note that the molar masses of C, O, and air, respectively, are 12.0, 16.0, and 29.0 g, that the mass of the atmosphere is 5.1 x 10²¹ g, and that 1 Gt is 10¹⁵ g.

Mass gain in CO₂ = (mole gain in CO₂)(44.0 g/mol) where mole gain in CO₂ =

(11.1/10⁶)(5.1 x 10²¹ g air)/(29.0 g mol^-1) = 1.95 x 10¹⁵ mol

Mass gain in CO₂ = (1.95 x 10¹⁵ mol)(44.0 g/mol) = (8.59 x 10¹⁶ g)(1 Gt/10¹⁵ g) = 85.9 Gt

% left in atmosphere = 85.9*100/178 = 48%

6.  The vapor pressure P of a liquid rises exponentially when it is heated according to the equation ln(P₂/P₁) = -DH/R(1/T₂ – 1/T₁)

Here P₁ and P₂ are the vapor pressures of the liquid at the Kelvin temperatures T₁ and T₂ after and before the temperature increase, R is the gas constant, and DH is the liquid’s enthalpy of vaporization, which for water is 44 kJ/mol.  Calculate the percentage increase in the vapor pressure of water that occurs if the temperature is raised from 15ºC to 18ºC.  Give several reasons why  the amount of outgoing thermal infrared in water’s absorption bands may not be increased by exactly the percentage you calculate if the average air/surface temperature is increased to 18ºC.

ln(P₂/P₁) = -(44,000 J/mol)/(8.314 J/(mol K))(1/291.15 – 1/288.15) = 0.189

P₂/P₁ = e^0.189 = 1.21 or 21 % increase in water vapor pressure.  A smaller than expected increase in expected outgoing thermal IR may occur because absorption is not linearly related to concentration at high concentrations and because a greater lapse rate (change in temperature with altitude) may lead to less total humidification of the atmosphere.

8.  Given that the concentration of CH₄ in the atmosphere is 1.8 ppm, calculate the total mass of this gas that is present in the atmosphere. Note that the mass of the atmosphere = 5.1 x 10²¹ g, and that air’s average molar mass = 29.0 g mol^-1.

moles CH₄ in the atmosphere = (1.8/10⁶)(5.1 x 10²¹ g air)/(29.0 g mol^-1) = 3.17 x 10¹⁴ moles

mass CH₄ in the atmosphere = (3.17 x 10¹⁴ moles)(16.04 g/mol) = 5.09 x 10¹⁵ g

mass CH₄ in the atmosphere = (5.09 x 10¹⁵ g)(1 Gt/10¹⁵ g) = 5.09 Gt