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che359_sp10_exam3_soln

# che359_sp10_exam3_soln - ChE 359 Exam#3 S o lu tio n s p r...

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Page 1 of 4 ChE 359 Exam #3 Solutions prepared by Professor Lo 1. (55 points) Lake Kivu in Rwanda was recently the subject of a BBC Horizon television documentary on “killer lakes”. (In 1986, another killer lake, Lake Nyos in Cameroon, was responsible for suffocating over 1,700 people and over 3,500 livestock within 25 meters of the lake.) In both cases, carbon dioxide (CO 2 ) and methane (CH 4 ) are dissolved in the lake, with the colder layers of water at the bottom of the lake being very dense with these toxic gases, and the warmer layers of water at the top of the lake being less dense but serving to trap the gases at the bottom of the lake. Most of the time, the lake is stable and the gases remain in solution. However, it is feared that volcanic activity would heat the water sufficiently to force CO 2 and CH 4 out of solution and spark an eruption that would suffocate the two million people that live near Lake Kivu. Thus, our role today is to determine the thermodynamic conditions that would spark this eruption and see if future eruptions can be prevented. a. (15 points) Let’s first consider methane (CH 4 ) in Lake Kivu. The lake has a water volume of 560 km 3 , and 54 km 3 (at T = 298.15 K and P = 1 atm ) of methane is dissolved in the water. You can assume that the density of water is 1 g/cm 3 and that the molar mass of water is 18 g/mol. For full credit: 1. State whether methane or water is the volatile component in this mixture, 2. Calculate the number fraction x B of methane in the lake, and 3. Calculate the vapor pressure P B [atm] of methane at T = 298.15 K . Methane (B) is the volatile component. To calculate the number fraction of methane in the lake, we need to know the number of water (A) molecules N A and methane molecules N B . N A = 560 km 3 ( ) × 1000 m 1 km ( ) 3 × 100 cm 1 m ( ) 3 × 1 g cm 3 ( ) × mol 18 g ( ) × N Av = 3.11 × 10 16 × N Av N B = 54 km 3 ( ) × 1000 m 1 km ( ) 3 × 1 atm ( ) 8.20575 × 10 5 m 3 atm mol K ( ) × 298.15 K ( ) × N Av = 2.21 × 10 12 × N Av x B = 7.09 × 10 5 The Henry’s law constant at T = 298.15 K is: k H = 41 × 10 3 atm . Thus, the vapor pressure of methane is: P B = k H x B = 2.91 atm .

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