A sample of charcoal weighing 6 g was brought into contact with a gas contained in a vessel of one litre capacity at `27^(@)`C. The pressure of the gas was found to fall from 700 and 400 mm. Calculate the volume of the gas (reduced to STP) that is adsorbed per gram of the adsorbent under the condition of the experiment (density of charcoal sample is `1.5 g cm^(3)`).

Explanation

The adsorption is taking place in a closed vessel hence if pressure falls there is correspondingly increase in volume constant, excess of the volume of the gas would be adsorbed.
`P_(1) V_(1) = P_(2) V_(2)` ltbr `V_(2) = P_(1) (V_(1))/(P_(2)) = 700 xx (1000)/(400) = 1750 mL.`
Actual volume of the flask = 1000 - volume of charcoal `= 1000 - (6.00)/(1.50) = 996 mL.`
Volume of the gas adsorbed = 1750 - 996 = 754 mL.
Volume of the gas adsorbed per gram at STP `("Using " (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))) = (125.67 xx 400 xx 273)/(300 xx 760) = 60.19 mL.`