Question

A one litre vessel contained a gas at 27${}^{0}C$. 6 g of charcoal was introduced into it. The pressure of gas fell down from 700 mm Hg to 400 mm Hg. Calculate the volume of the gas (at NTP) adsorbed per gram of charcoal. Density of charcoal sample used was 1.5 g cm${}^{-3}$.

• A. 60.107 cm${}^3$
• B. 360.642 cm${}^3$
• C. 120.214 cm${}^3$
• D. None of these

Answer 1

The correct option is A 60.107 cm${}^3$

$P_1=700mmHg=700/760atm,V_1=1L,T={27}^{o}C=300K,R=0.0821atm-Lmo{l}^{-1}{K}^{-1}$

Initial number of moles of gas $\left(n_1\right)=\frac{P_1{V}_1}{RT}=\frac{\left(700/760\right)\times 1}{0.0821\times 300}$ $=0.0374mol$

volume occupied by charcoal $=\frac{6}{1.5}=4c{m}^3$

Available volume for the gas $=(1000-4)=996c{m}^3$

$P_2=700mmHg=700/760atm,V_2=0.996L$

Final number of moles $\left(n_2\right)=\frac{\left(400/760\right)\times 0.996}{0.0821\times 300}=0.0213mol$

Number of moles of gas adsorbed $=0.0374-0.0213=0.0161mol$

So, 6 gm of charcoal has adsorbed $=0.0161mol$

Option A is correct.
1 gm of charcoal has adsorbed $=\frac{0.0161}{6}mol$

Volume of gas adsorbed by 1 g charcoal at NTP.

$=\frac{0.0161}{6}\times 22400=60.107c{m}^3$