In an experiment involving adsorption of a gas on a solid at different pressures of the gas, the values of `log(x/m)` obtained are plotted against logp. This gives a straight line inclined at an angle of `45^(@)`. If the value of Freundlich's constant is 10, then what would be the amount of adsorbed gas per gm of adsorbent, when equilibrium pressure is 0.5 atm?

Explanation

According to Freundlich's adsorption isotherm , `x/m =KP^(1//n)` . This can be written as -
`log((x)/(m))=logk + 1/n log p `
So, a plot of `log(x/m)` vs `logp` will give a striaght line whose slope is `1/n` and the intercept made by it on coordinate is logk. It is given that `1/n = tan 45^(@) and log k =log10=1`
Thus, `1/n=1 and n=1 `
These value give `log(x/m)=1+1xxlog0.5=0.6989`,
i.e. `x/m=5`.
Therefore, the amount of gas adsorbed by 1 g of adsorbent is 5 g.