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` to `400` mm. Calculate the volume of the gas (reduced to STP) that is adsorbent under the condition of the experiment `("density of charcoal sample is" 1.5 g cm^(3))`.

Explanation

Volume of carcoal presentin the eessel `=(6g)/(1.5g cm ^(-3))=4 cm`
`therefore` Volume of the gas initial at `27^(@)C,` 700 mm pressure `=100-4=996 cm^(3)`
Volume of the gas at 400 cm and `27^(@)C` is again equal to the volume of the vessel excluding that of chrocol, i.e., `996 cm^(3).` Let us calculate equivalent volume of the gas at 700 mm at the same temperature
`P_(1)V_(1)=P_(2)V_(2)i.e. 400 xx996=700xxV_(2)or V_(2)=569.1c c`
`therefore` Volume of the gas adsorbed at `27^(@)C,` 700 mm pressure `=996-569.1=426.9cm^(3)`
`therefore `Volume adsorbed pr gram of charcoal `=(426.9)/(6)cm^(3)g^(-1)=711cm^(3)g^(-1)`
Converting this volume to S.T.P. we get `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),i.e., (700xx71.8)/(300)=(760xxV_(2))/(273)or V_(2)=59.6cm^(3)`