In a Langmuir‐type adsorption, a solid adsorbs 0.25 mg of a gas when the pressure of the gas is 50 bar and 0.2 mg of the gas at 20 bar pressure. The percentage of surface coverage at 50 bar is close to:

Concept:

• Langmuir adsorption isotherm talks about the monolayer gas adsorption.
• The  gaseous adsorbate molecules are in dynamic equilibrium adsorbent surface.
• Adsorption of gases liberates heat which is equal for all sites, that is, all sites are equally probable.
• According to this model, fractional coverage is proportional to fraction of the vaccant sites and pressure of the gas, 

 \(\theta \propto (1-\theta )P \)

• so, the relation for fractional coverage can be written as:          

  \(\theta =K(1-\theta)P\)

  \(\theta= \frac{KP}{1+KP}\)------------------(1)

Explanation:

We Know that, fractional coverage is:

  \(\theta =\frac{amount\;of\; gas\;adsorbed}{amount\;of\;gas\;for\;complete\;monolayer\;coverage}=\frac{m}{M_0} \)

\(\frac{m}{M_0}=\frac{KP}{1+KP} \)

for different pressures P₁ and P₂ with corresponding adsorbate mass m₁ and m₂ respectively, fractional coverage can be written as:

 \(\frac{m_1}{M_0} =\frac{KP_1}{1+KP_1}\)        ------------(2)

\(\frac{m_2}{M_0} =\frac{KP_2}{1+KP_2}\)-       -------------(3)

dividing equation (2) and (3) gives,

\(\frac{m_1}{m_2}=\frac{P_1}{1+KP_1}\times \frac{1+KP_2}{P_2} \)

For m₁ = 0.25mg, P₁= 50bar and

 m₂= 0.20mg, P₂ = 20bar

equation can be written as:

\(\frac{0.25}{0.20}=\frac{50}{1+50K}\times\frac{1+20K}{20}\)

 \(K=0.1Pa^{-1}\)

Now, put \(K=0.1Pa^{-1}\) and P = 50 bar in following equation(1) for surface coverage gives,

\(\theta=\frac{ 0.1bar^{-1}\times50bar}{1+ 0.1bar^{-1}\times50bar}\)

\(\theta= 0.83 \)

Percentage of fractional coverage = \(0.83\times100 \%\)

= 83%

Conclusion:

The percentage of surface coverage for adsorption of gas at 50 bar is 83%.