To determine the amount of acetone adsorbed (in kg per kg of activated carbon) at a partial pressure of 50 mmHg, we can use the Langmuir isotherm. The Langmuir isotherm is given by the equation:
\(q = \frac{{q_m \cdot K \cdot P}}{{1 + K \cdot P}}\)
Where:
- \(q\) is the amount of adsorbate (acetone) adsorbed at a given pressure \(P\).
- \(q_m\) is the maximum adsorption capacity.
- \(K\) is the Langmuir constant related to the energy of adsorption.
- \(P\) is the partial pressure of the adsorbate.
Given data:
- At \(P_1 = 10 \, \text{mmHg}\), \(q_1 = 0.1 \, \text{kg/kg}\).
- At \(P_2 = 100 \, \text{mmHg}\), \(q_2 = 0.4 \, \text{kg/kg}\).
We use the Langmuir isotherm for both cases to set up two equations:
- \(0.1 = \frac{{q_m \cdot K \cdot 10}}{{1 + K \cdot 10}}\)
- \(0.4 = \frac{{q_m \cdot K \cdot 100}}{{1 + K \cdot 100}}\)
Solving these equations simultaneously will give us the values for \(q_m\) and \(K\). Once determined, these values can calculate \(q\) at \(P = 50 \, \text{mmHg}\).
Re-arranging the first equation, we get:
\(0.1 + 1.0K = q_m \cdot K \cdot 10\)
Re-arranging the second equation, we get:
\(0.4 + 40K = q_m \cdot K \cdot 100\)
Dividing the second equation by the first to eliminate \(q_m\):
\(\frac{{0.4 + 40K}}{{0.1 + 10K}} = \frac{{100}}{{10}}\)
\(0.4 + 40K = 4(\0.1 + 10K)\)
Solving the above equation, you find:
- \(\)
- Substituting back to find \(q_m\): <(q_m = 2)>
Using these values, we can calculate \(q\) for \(P = 50 \, \text{mmHg}\):
\(q = \frac{{2 \cdot 0.05 \cdot 50}}{{1 + 0.05 \cdot 50}}\) = \(\0.3 kg/kg\)
Thus, the amount of acetone adsorbed at a partial pressure of 50 mmHg is 0.3 kg/kg. Therefore, the correct option is 0.3.
Important Questions from Adsorption
- A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is $0.4 \text{ grams}$ VOC per gram of activated carbon. The column contains $4 \text{ grams}$ of activated carbon per $\text{cm}^2$ of cross-section. The feed rate into the adsorber column is $0.2 \text{ grams}$ VOC $\text{cm}^{-2}\text{ h}^{-1}$. Breakthrough time is defined as the time at which the concentration at the exit of the bed ($c$) reaches a value of $0.05c_0$, where $c_0$ is the concentration of VOC in the feed. The breakthrough time for the bed is $2.1 \text{ h}$. The area under $c / c_0$ curve between the initial and breakthrough times is $0.1 \text{ h}$. Which one of the following is the fraction of unused bed at breakthrough?
- The correct statement(s) for a surface catalyzed unimolecular gaseous reaction is(are)
- 500 mg of a dry adsorbent is added to a beaker containing 100 mL solution of concentration 100 mg phenol/(L solution). The adsorbent is separated out after 5 h of rigorous mixing. If the residual concentration in the solution after separating the adsorbent is 30 mg phenol/(L solution), the amount of phenol adsorbed (in mg per gram of dry adsorbent) is
- Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater from 0.04 mol/l to 0.008 mol/l. The adsorption isotherm at the operating temperature can be expressed as $q = 0.025C^{1/3}$; where $q$ is the phenol concentration in solid (mol/g solid) and $C$ is the phenol concentration in water (mol/l). The minimum amount of solid (in grams) required per liter of wastewater (up to one decimal place) is ________
At 273 K and 10 bar, the Langmuir adsorption of a gas on a solid surface gave the fraction of surface coverage as 0.01. The Langmuir adsorption isotherm constant is ________ bar$^{-1}$. (Give the answer to the third decimal place)
