A one-litre vessel contained a gas at 27^(@)C. 6 g of charcoal was introduced into it. The pressure of the gas fell down from 700 mm to 400 mm. Calculate the volume of the gas (at S.T.P.) adsorbed per gram of charcoal. Density of charcoal sample used was 1.5g cm ^(-3).

A one-litre vessel contained a gas at 27^(@)C. 6 g of charcoal was int

1.	A one-litre vessel contained a gas at 27^(@)C. 6 g of charcoal was introduced into it. The pressure of the gas fell down from 700 mm to 400 mm. Calculate the volume of the gas (at S.T.P.) adsorbed per gram of charcoal. Density of charcoal sample used was 1.5g cm ^(-3).
Answer» Solution :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)`