Question 3 [15 points]. The average atmospheric carbon dioxide mole fraction mea

Question

Question 3 [15 points]. The average atmospheric carbon dioxide mole fraction measured in December 2017 at Mauna Loa in Hawaii wag 406.82 parts per million (ppm). The average molar mass of air i 28.97 g/mo a) What was the density of the air in December, assuming an air pressure of 1.00 atm and a typical Hawaiian air temperature of 82? b) What was the (mass) concentration of carbon dioxide measured at Mauna Loa? c) During a rain storm, air pressure decreased by 3% and absolute temperature decreased by 1% but the mole fraction of carbon dioxide and the molar mass of air did not change. How much did the mass concentration of carbon dioxide increase or decrease? (Your answer can be either a concentration change or a % change)

Explanation / Answer

Ans 3

Pressure P = 1 atm = 101325 Pa

Temperature T(K) = (T(°F) + 459.67)× 5/9

= (82 + 459.67) *5/9 = 300.92 K = 301 K

Part a

From the ideal gas equation

PM = density x RT

Density = (101325 Pa x 28.97 x 10^-3 kg/mol) / (8.314 J/mol·K x 301 K)

= 1.1729 kg/m3

Part b

V = RT/P

= 0.0821 L-atm/mol-K x 301 K / 1 atm

= 24.71 L/mol

Mass concentration in mg/m3

= ppm concentration x Molecular weight /V

= 406.82 x 44/24.71

= 724.4 mg/m3

Part C

Pressure decreases by 3%

Pressure P = 1 - 0.03 = 0.97 atm

Temperature T = 301 - (301*0.01) = 297.99 K = 298 K

V = RT /P

= 0.0821 L-atm/mol-K x 298 K / 0.97 atm

= 25.222 L/mol

Mass concentration in mg/m3

= ppm concentration x Molecular weight /V

= 406.82 x 44/25.222

= 709.7 mg/m3

Concentration of CO2 decreased from 724.4 to 709.7 mg/m3

Concentration change = 724.4 - 709.7 = 14.7 mg/m3

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