Question

A sample of charcoal weighing 6.0 g was brought into contact with a gas contained in a vessel of 1 litre capacity at ${27}^{\circ }\text{C}$. The pressure of the gas falls from 700 to 400 torr. Calculate the volume of the gas (at STP) that is adsorbed per gram of the adsorbent under the condition of the experiment. The density of the charcoal compound is $1.5{\text{g cm}}^{-3}$.

Answer 1

Given :
$P_1=700torr=\frac{700}{760}atm,P_2=400torr=\frac{400}{760}atmT=300K,R=0.0821atm/K/mol/L$
Number of moles of gas before absorption $n_1=\frac{P_{1}V}{RT}$
Number of moles of gas after absorption $n_2=\frac{P_{2}V}{RT}$
Number of moles of gas absorbed $=n_1-n_2=\frac{(P_1-P_2)\times V}{RT}$
$=\frac{(700-400)1}{0.0821\times 300\times 760}$
= 0.016 moles
Volume of gas absorbed(at STP) $=0.016\times 22400=358.4mL$
Volume of gas absorbed per gram $=\frac{358.4mL}{6g}=59.7mL/g=60\left(approx\right)$