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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 ? |
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Answer» 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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