A sample of charcoal weighing 6.0 gm was brought into contact with a gas contained in a vessel of 1 litre
capactiy at `27^(@)C`. The pressure of the gus falls from 700 to 400 torr. Calculate the volume of gas (at STP) that
is adsorbed per gm of the adsorbent under the vondition of experiment. the density of charcoal sample was
`1.5 gm 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.`