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InterviewSolution
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An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm. And a temperature of `27^(@)C`. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm. And its temperature drops to `17^(@)C`. Estimate the mass of oxygen taken out of the cylinder. `(R = 8.1 J "mole"^(-1) K^(-1)`, molecular mass of `O_(2) = 32 u)`. |
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Answer» Initially in the oxygen cylinder, `V_(1) = 30 litre = 30 xx 10^(-3) m^(3)`, `P_(1) = 15 atm = 15 xx 1.01 xx 10^(5) Pa, T_(1) = 27 +273 = 300 K` If the cylinder contains `n_(1)` mole of oxygen gas, then `P_(1) V_(1) = n_(1)RT_(1)` or `n_(1) =(P_(1)V_(1))/(RT_(1)) = ((15 xx 1.01 xx 10^(5))xx(30xx10^(-3)))/(8.3 xx 300) = 18.253` For oxygen moleculaer weight, `M= 32 g` Initial mass of oxygen in the cylinder cylinder, `m_(1) = n_(1) M= 18.253 xx 32 = 548.1g` Finally in the oxygen gas in the cylinder, let `n_(2)` moles of oxygen be left, Here, `V_(2) = 30 xx 10^(-3) m^(3) , P_(2) = 11 xx 1.01 xx 10^(5)Pa , T_(2) = 17 + 273 = 290 K` Now, `n_(2) = (P_(2)V_(2))/(RT_(2)) = ((11xx1.01 xx 10^(5))xx(30xx10^(-3)))/(8.3 xx 290) = 13.847` `:.` Final mass of oxygen gas in the cylinder , `m_(2) = 13.847 xx 32 = 453.1 g` `:.` Mass of the oxygen gas withdrawn = `m_(1)-m_(2) = 584.1 - 453.1 = 131.0 g`. |
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