Question #298958

Suppose that you are travelling in the Space Shuttle Endeavor on a journey to Mars. The atmosphere (of the vehicle) consists of 20% oxygen and 80% helium by volume. It was noticed before take off from the Kennedy Space Center that there is a small leakage which if unchecked, would lead to a continual loss of gas at a rate of 0.05 atm/day through effusion. If the temperature in the space shuttle is maintained at 22°C and the volume of the vehicle is 15,000 liters. Calculate the amounts of helium and oxygen in grams that must be stored on a 5-day journey to allow for the leakage.

1
Expert's answer
2022-02-18T10:13:01-0500

With loss rate of 0.05 atm/day, in 5 days there will be 0.05×5=0.25 atm0.05\times5=0.25\ atm lost. Since nothing is mentioned about the hypothetical consumption of oxygen by astronauts, the key assumption is that there should be minimum of 0.25 atm of the gas mixture initially. Since pressure and volume are proportional, there are 20% of O2 and 80% of He by pressure, too. Therefore,


PO2=0.25 atm×0.2=0.05 atmP_{O_2}=0.25\ atm\times0.2=0.05\ atm

PHe=0.25 atm×0.8=0.2 atmP_{He}=0.25\ atm\times0.8=0.2\ atm


nO2=PO2VRT=0.05 atm×15000 L0.08206LatmmolK×295 K=31.0 moln_{O_2}=\frac{P_{O_2}V}{RT}=\frac{0.05\ atm\times15000\ L}{0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times295\ K}=31.0\ mol


nHe=PHeVRT=0.2 atm×15000 L0.08206LatmmolK×295 K=124 moln_{He}=\frac{P_{He}V}{RT}=\frac{0.2\ atm\times15000\ L}{0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times295\ K}=124\ mol


m(O2)=31.0 mol×32.00 g1 mol=992 gm(O_2)=31.0\ mol\times\frac{32.00\ g}{1\ mol}=992\ g


m(He)=124 mol×4.003 g1 mol=496 gm(He)=124\ mol\times\frac{4.003\ g}{1\ mol}=496\ g


Answer: 992 g O2; 496 g He


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS