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Let us do some conversions first, then use the ideal gas law.

PV = nRT

1 Bar = 1.01325 atmospheres

80 Bar (1.01325 ATM/1 Bar) = 81.06 ATM

20 C = 293.15 K

new R = 0.08206 L*ATM/mol*K

(81.06 ATM)(100 L) = nI0.08206 L*ATM/mol*K)(293.15 K)

moles O2 = 337 moles O2 * 32 grams

= 10784 grams

about 24 pounds of oxygen

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Q: Find the mass of oxygen in a cylinder having a capacity of 100 liters at a temperature of 20 degree Celsius The pressure in the cylinder is maintained at 80 bar R equals 0.259859?
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