Marissa

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# Algebra 2 help please!?

A popular size of scuba-diving tank holds the amount of compressed air that would occupy 71.2 ft^3 at a normal surface pressure of 1 atm. The air in the tank is at a pressure of about 2250 lb/in^2 so the tank itself can have a volume much less than 71.2 ft^3. How large does the tank need to be to hold 71.2 ft^3? (Hint: Use Boyle’s Law: PV=k Remember that 1 atm = 14.7 lb/in^2.)

I've finished my entire project but this problem i'm struggling with. Can somebody please explain it to me or help me solve it? Thank you!!

### 1 Answer

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- BryceLv 76 years agoBest Answer
71.2ft^3*12^3in^3/ft^3≈123034in^3

P=1, PV=1*123034=k; k=123034lb-in.

PV=k; 2250lb/in²*V=123034lb-in; V≈54.7in^3

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