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A cylinder of 20.0 L capacity contains 160 g oxygen gas at `25^(@)C`. What mass of oxygen must be released to reduce the pressure of the cylinder to 1.2 atm ? |
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Answer» Number of moles of oxygen gas present initially in the cylinder`=(160g)/(32"g mol"^(-1))=5" moles"` To calculate the number of moles now present, we have P=1.2 atm, T=298 K, V=20.0L Applying the relation, PV=nRT, we have `n=(PV)/(RT)=(1.2" atm "xx20.0" L")/(0.0821" L atm " K^(-1)mol^(-1)xx298" K")=0.98" mol"` `:.` Number of moles of `O_(2)` required to be released =5-0.98=4.02 mol or Mass of `O_(2)` required to be released `=4.02xx32" g"=128.64" g"` |
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