Sealab II, a habitat used for underwater research, had a volume of 2.2×104 L, and in...
Sealab II, a habitat used for underwater research, had a volume of 2.2×104 L, and in use it was held at an internal pressure of 5.4×103 mmHg with a mixture of nitrogen, oxygen, and helium.
1.) What volume in liters of this mixture would be needed to fill Sealab II at 0.948 atm?
Express your answer to two significant figures.
Homework Answers
Given:
Vi = 2.2*10^4 L
Pi = 5.4*10^3 mm Hg
= (5.4*10^3/760) atm
= 7.1053 atm
Pf = 0.948 atm
use:
Pi*Vi = Pf*Vf
7.1053 atm * 2.2*10^4 L = 0.948 atm * Vf
Vf = 1.6*10^5 L
Answer: 1.6*10^5 L
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