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A 34.0 dm^(3) cylinder contains 212 g of...

A `34.0 dm^(3)` cylinder contains `212 g` of oxygen gas at `21^(@)C`. What mass of oxygen must be released to reduce the pressure in the cylinder to `1.24` bar ?

Text Solution

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Step I. Calculation of no. of moles of `O_(2)` left in cylinder.
`P = 1.24` bar , `V = 34 dm^(3)`
`T = (21 + 273) = 294 K` , `R = 0.083 dm^(3) "bar" K^(-1) mol^(-1)`
According to ideal gas equation, `PV = nRT`
`n = (PV)/(RT) = ((1.24 "bar") xx (34 dm^(3)))/((0.083 dm^(3) "bar" K^(-1) mol^(-1)) xx (294 K)) = 1.727` mol
Step II. Calculation of mass of hydrogen released
Mass of `O_(2)` left in the cylinder `= n xx M = (1.727 "mol") xx (32 g "mol"^(-1)) = 55.26 g`
Mass of `O_(2)` initially present `= 212 g`
`:.` Mass of `O_(2)` released `= (212 - 55.26) = 156.74 g`
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