Data for Titan, a moon of Saturn, is given below, and may be used to answer problems 1 and 2.

Transcription

1 CHM 5423 Atmospheric Chemistry Problem Set 1 Due date: Thursday, September 10 th. Do the following problems. Show your work. Data for Titan, a moon of Saturn, is given below, and may be used to answer problems 1 and 2. M = 1.35 x kg r = km P s = 1.45 bar T s = 95. K T ex = 150. K (mass of Titan) (radius of Titan) (surface pressure of Titan) (surface temperature of Titan) (temperature of exophere of Titan) Because of the presence of liquid methane (CH 4) at the surface of Titan, the percentage of methane in the atmosphere varies from %, in a manner similar to the variation of water in the atmosphere of the Earth. Virtually all of the atmosphere of Titan is molecular nitrogen (N 2), with a small percentage (0.10 %) of molecular hydrogen (H 2) and trace amounts of other gases. 1) Using the above information, find the amount of molecular hydrogen (H 2) present in the atmosphere of Titan at surface level. Give your answer in: a) Partial pressure of H 2 (in units of bar) b) Number density of hydrogen molecules (in units of molecules/cm 3 ) c) Density of hydrogen (in units of g/l) 2) The following question concerns the escape of molecules from the atmosphere of Titan. a) Find v esc and v rms for the following molecules: H 2, CH 4, and N 2. Use T = 150. K (the current average temperature in the exosphere of Titan) in your calculations. b) Using your answers in a, find the value for the ratio v esc/v rms for H 2, CH 4, and N 2. Based on the results, explain why the atmosphere of Titan contains relatively little H 2, but a lot of N 2. 3) One indirect piece of evidence indicating that free molecular oxygen was uncommon in the early atmosphere of the Earth is the presence of pyrites (metal sulfides) in mineral samples that have been dated several billion years old. Such compounds are thermodynamically unstable in the presence of oxygen. More recent mineral samples have iron oxides present, but usually do not contain iron pyrite. a) Using the information below, find the equilibrium constant for the process 2 FeS 2(s) + O 2(g) 2 FeO(s) + 4 S(s) (3.1) b) Based on your answer in a, estimate the partial pressure of O 2(g) in the atmosphere at which iron pyrite (FeS 2) becomes thermodynamically unstable relative to iron II oxide (FeO). What does this imply about the content of oxygen in the early atmosphere of the Earth? (Note that there are other factors involved, including the rate of reaction 3.1 and the temperature dependence of the equilibrium constant, but the argument based on the thermodynamics of the reaction remains valid). Substance H (kj/mol) G (kj/mol) S (J/mol K) FeO(s) FeS 2(s) O 2(g) S(s)

2 4) At what altitude (in km) is the pressure of the Earth's atmosphere equal to the surface pressure on Mars ( bar)? Use H = 7.4 km for the scale height for Earth. 5) Simple calculations can often tell us interesting information about planetary atmospheres. a) Based on the information in the Chapter 1 handout, find the total mass of the atmosphere of Earth and Venus. Compare the result obtained for the Earth to the value given in the Chapter 1 notes. (HINT: The force exerted by a mass m in the gravitational field of a planet is F = mg s, where m is mass and G s is the gravitational constant for the planet. Force is also related to pressure by the relationship p = F/A, where A is area.) b) Use your result from a and the information on atmospheric composition to find the total mass of argon in the atmosphere of the Earth and Venus. Comment on your results. 6) For an adiabatic expansion of an ideal gas, it may be shown that (T i/t f) = (p i/p f) -1 (6.1) where = C p/c v is the ratio of the constant pressure heat capacity to the constant volume heat capacity of the gas. Eq 6.1 assumes that over the temperature range of the expansion is constant. a) According to the barometric equation p z = p 0 exp(-z/h) (6.2) where p z is atmospheric pressure at altitude z and p 0 is atmospheric pressure at some reference altitude (taken to be sea level for the Earth). By combining eq 6.1 and 6.2 find an expression for T z, the temperature for an air parcel initially at sea level and temperature T 0 after it rises to an altitude z, and assuming that it expands adiabatically as it rises. Give your expression in terms of p 0, T 0,, H, z, and/or other constants. (HINT: Let T 0 and p 0 be the temperature and pressure of the air at sea level, where z = 0. Then T f and p f are equivalent to T z and p z, the temperature and pressure at altitude z.) b) Using the following values for the terms in your expression for T z, find the value for T z - T 0 for the case z = 1.0 km. Note that this value for T z - T 0 corresponds approximately to the adiabatic lapse rate for dry air in the Earth's atmosphere. = 1.40 H = 7.4 km T 0 = 288. K p 0 = 1.00 atm

3 Solutions 1) a) p(h 2) = (0.001) (1.45 bar) = bar ( = 145. Pa) b) pv = nrt, and so (n/v) = p/rt Multiplying by N A (Avogadro s number) gives nn A/V = N/V = pn A/RT So N = ( bar) (6.022 x molecule/mol) 1. L = 1.11 x molecule/cm 3 V ( L bar/mol K) (95. K) cm 3 c) D = 1.11 x molecule cm 3 1 mol 2.02 g 10 6 g = 3.71 x 10-4 g/l cm 3 1. L x molecule mol g 2) a+b) H 2, M = 2.02 g/mol CH 4, M = g/mol N 2, M = g/mol = 371. g/l v esc = (2M pg/r p) 1/2 = [2 (1.35 x kg) (6.67 x N. m 2 /kg 2 ) / x 10 3 m)] 1/2 = m/s v rms = (3RT/M) 1/2 R = J/mol. K T = 320. K f = v esc/v rms gas v rms (m/s) v esc/v rms H CH N The value of v esc/v rms for H 2 is large enough that H 2 in the atmosphere of Titan quickly escapes into space. Any hydrogen found there is likely of recent origin (perhaps formed photochemically or from venting of underground sources of H 2). The value of v esc/v rms for N 2 is small enough that most of the N 2 originally present in the atmosphere of Titan is likely still there. The value of v esc/v rms for CH 4 is such that some of the methane originally present is likely still in the atmosphere. The presence of liquid methane at the surface of Titan would also slow the escape of methane from the atmosphere. Volcanos are also believed to inject new methane into the atmosphere. 3) a) ln K = - G rxn/rt G rxn = [2 G f(feo(s)) + 4 G f(s(s))] - [2 G f(fes 2(s)) + G f(o 2(g))] [ 2( kj/mol) ] - [ 2( kj/mol) ] = kj/mol ln K = ( J/mol) = K = e = 2.5 x (8.314 J/mol K)(298. K) b) Since everything in the reaction other than O 2 is a solid K = 1/p(O 2) p(o 2) = 1/K = 1/2.5 x = 4. x bar

4 This applies at the temperature of the thermochemical data (298. K). While the value of K might change with temperature, and there is no information on the kinetics of the reaction, thermodynamically FeO(s) is far more stable than FeS 2(s), and would be expected to form if there was any significant partial pressure of O 2(g) in the atmosphere. 4) p z = p 0 exp(-z/h) and so (p z/p 0) = exp(-z/h) ln(p z/p 0) = - z/h z = - H ln(p z/p 0) = H ln(p 0/p z) For p z = bar, p 0 = bar z = (7.4 km) ln(1.0/0.0060) = 38. km 5) a) F = mg s where m = mass G s = surface gravity of planet But p = F/A where p = pressure A = area So pa = mg s m = pa/g s Now A for a sphere is A = 4 r 2, and so m = 4 r p 2 p/g s For Earth m = 4 (6380. x 10 3 m) 2 (1.0 x 10 5 N/m 2 ) = 5.23 x kg (vs 5.14 x kg in Table 1.1) (9.78 m/s 2 ) For Venus m = 4 (6050. x 10 3 m) 2 (92. x 10 5 N/m 2 ) = 4.77 x kg (8.87 m/s 2 ) b) The mass of argon in a planetary atmosphere is m(ar) = fraction of Ar (by number) M(Ar) m(atmosphere) M ave where Mave is the average molecular mass of the atmosphere. Note we need the second factor above because we usually give composition by percent (or parts per million) by number, not by mass. M(Ar) = g/mol M ave(earth) = g/mol (Table 1.1, Chapter 1) M ave(venus) = g/mol (using 96.5 % CO 2, 3.5 % N 2) So mass argon for Earth = (0.0093) (39.95/28.96)(5.23 x kg) = 6.71 x kg So mass argon for Venus = (70. x 10-6 ) (39.95/43.45)(4.77 x kg) = 3.07 x kg So the Earth and Venus have roughly equal masses of argon. Since argon is not expected to escape from either atmosphere, that suggests the argon that is present has a similar origin.

5 6) T i = T 0 T f = T z p i = p 0 p f = p z = p 0 exp(-z/h) And so (T 0/T i) = (p 0/p z) -1 Invert (T z/t 0) = (p z/p 0) -1 = (e -z/h ) -1 T z = T 0 (e -z/h ) -1 or T z = T 0 (e -z/h ) -1/ For z = 1.0 km T 0 = 288. K H = 7.4 km = 1.40 T z = (288. K) (e -1.0/7.4 ) 0.4/1.4 = K T = K K = K So the adiabatic lapse rate for dry air at sea level is approximately 11 K/km.