Question

Question 1 :

Carbon dioxide scrubbers containing lithium hydroxide continually work to remove exhaled CO_2​2​​(g) from the breath of astronauts on board the international space station:

2LiOH(s) + CO_2​2​​(g) Li_2​2​​CO_3​3​​(s) + H_2​2​​O(l)

If there are three astronauts aboard the space station, and each astronaut requires 2150 Cal from food per day, what mass (kg) of LiOH is required to remove 100% of the CO_2​2​​ produced by the three astronauts in seven days? (Assume that the astronauts' energy requirements are met exclusively by the combustion of glucose, ΔH of glucose = -1273.3 kJ/mol. The metabolism of glucose is only 66% efficient).

Question 2:

An engine whose piston moves vertically does more work than an engine whose prison moves horizontally due to its orientation in a gravitational field. Two such engines are compared. In each engine, a 235 g piston is pushed through a total distance of 145 cm against an external pressure of 1.50 atm. The radius of each piston is 4.418 cm. How much extra work (%) must the engine do whose piston moves vertically? (acceleration due to gravity is 9.8 m/s

^2​2​​).

Answer 1

We have glucose oxidation in body to get energy:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + Energy (-1273kJ)

total energy needed per day per person = 2150 cal

for 3 person = 2150 cal x 3 = 6450 cal

for 7 days = 6450 x 7 cal = 45150 cal

1 cal = 4.184 J

45150 cal = 188907.6 J = 188.907 kJ

it is 66% efficient and hence the new total energy needed 188.907kJ/T x 100 = 66

Total energy needed for intake = 286.224 kJ

1 mol C6H12O6 gives 1273.3 kJ

286.224 kJ will be given by 286.224/1273.3 kJ = 0.2248 mol

1 mol C6H12O6 produces 6 mol CO2

0.2248 glucose produces = 6 x 0.2248 = 1.35 mol CO2

we have carbon dioxide fixation as :

2LiOH(s) + CO₂​​​(g) --------> Li₂(CO₃​)(s) + H_​2​​O(l)

2 mol LiOH fixes 1 mol CO2

1.35 mol CO2 will be fixed by 2 x 1.35 = 2.70 mol LiOH

molar mass of LiOH = 23.95 g/mol

mass of LiOH = moles x molar mass = 23.95 g/mol x 2.70 = 64.62g