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oxygen i used at higher-than-normal pressures. how many grams of oxygen are required to fill a 3.00 x 10^3liter hyperbaric chamber at a pressure of 300 cm Hg and 75 degrees F?

75 degrees F – 32 = Celsius (1.8)
43 = Celsius (1.8)
temp Celsius = 23.89 C
plus 273 = 296.9 Kelvin

300 cm Hg @ 1 atm / 76.0 cm Hg = 3.947 atm

find moles of oxygen:
PV = n RT
(3.947 atm)( 3 e3 litres) = n (o.o821 L-atm/mol-K)(296.9 K)
n = 485.8 moles of O2

use molar mass to find grams:
485.8 molO2 @ 32.00 g/mol = 15,546 grams of O2

which likely should be rounded off to 15,500 grams of O2

2 Responses

1. Kevin Says:
   December 31st, 2009 at 4:10 am

   this requires the ideal gas law equation PV=nRT

   we need to know the amount of Oxygen so need to rearrange the formula to:

   n=PV/(RT)

   Temperature is 75F which is 24C 24+273.2 = 297K

   Pressure is 300 cm Hg
   Sea level pressure is 76 cm of Hg

   the pressure in the hyperbaric chamber would be the addition of both 376 cm of Hg now that is equivalent to 3760 mm of Hg or 3760 Torr

   we need a gas constant R in 62.363 67 L mmHg K−1 mol−1

   n=(3000 x 3760)/(62.363 x 297) {you do the math)

   then take the n and convert it into gras (o2 is 32 g/mol)
   References :
   Ph.D. Chemistry

2. Steve O Says:
   December 31st, 2009 at 4:34 am

   75 degrees F – 32 = Celsius (1.8)
   43 = Celsius (1.8)
   temp Celsius = 23.89 C
   plus 273 = 296.9 Kelvin

   300 cm Hg @ 1 atm / 76.0 cm Hg = 3.947 atm

   find moles of oxygen:
   PV = n RT
   (3.947 atm)( 3 e3 litres) = n (o.o821 L-atm/mol-K)(296.9 K)
   n = 485.8 moles of O2

   use molar mass to find grams:
   485.8 molO2 @ 32.00 g/mol = 15,546 grams of O2

   which likely should be rounded off to 15,500 grams of O2
   References :

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