/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 A \(50.0-\mathrm{L}\) tank conta... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(50.0-\mathrm{L}\) tank contains an air-carbon tetrachloride gas mixture at an absolute pressure of \(1 \mathrm{atm}, \mathrm{a}\) temperature of \(34^{\circ} \mathrm{C},\) and a relative saturation of \(30 \% .\) Activated carbon is added to the tank to remove the \(\mathrm{CCl}_{4}\) from the gas by adsorption and the tank is then sealed. The volume of added activated carbon may be assumed negligible in comparison to the tank volume.(a) Calculate \(p_{\mathrm{CCl}_{4}}\) at the moment the tank is sealed, assuming ideal-gas behavior and neglecting adsorption that occurs prior to sealing. (b) Calculate the total pressure in the tank and the partial pressure of carbon tetrachloride at a point when half of the CCl_ initially in the tank has been adsorbed. Note: It was shown in Example \(6.7-1\) that at \(34^{\circ} \mathrm{C}\).$$X^{*}\left(\frac{\mathrm{g} \mathrm{CCl}_{4} \text { adsorbed }}{\mathrm{g} \text { carbon }}\right)=\frac{0.0762 p_{\mathrm{CCl}_{4}}}{1+0.096 p_{\mathrm{CCl}_{4}}}$$ where \(p_{\mathrm{CCl}_{4}}\) is the partial pressure (in \(\mathrm{mm} \mathrm{Hg}\) ) of carbon tetrachloride in the gas contacting the carbon.(c) How much activated carbon must be added to the tank to reduce the mole fraction of \(\mathrm{CCl}_{4}\) in the gas to 0.001?

Short Answer

Expert verified
The initial partial pressure of carbon tetrachloride, \(p_{\mathrm{CCl}_{4}}\) is \(71.7 \, mmHg\). After half of it is adsorbed, the new partial pressure becomes \(1 \, atm - 0.5 \times p_{\mathrm{CCl}_{4}}\). The mass of activated carbon needed to reduce the mole fraction of \(\mathrm{CCl}_{4}\) to 0.001 can be calculated through rearrangement of the equation provided and setting \(p_{\mathrm{CCl}_{4}}=1 \, mmHg\).

Step by step solution

01

Calculate initial partial pressure of carbon tetrachloride

The partial pressure of the carbon tetrachloride can be computed considering the relative saturation. The saturation vapor pressure of carbon tetrachloride at \(34 \, ^{\circ}C\) is \(239 \, mmHg\). Using the given relative saturation, \(p_{\mathrm{CCl}_{4}} = 0.3 \times 239 = 71.7 \, mmHg\). This represents the initial partial pressure at the moment the tank is sealed.
02

Calculate total pressure and partial pressure after adsorption

Once we consider half of the carbon tetrachloride is adsorbed, the decrease in number of moles, and therefore pressure, will also be to half the initial pressure. The partial pressure will decrease by \(0.5 \times p_{\mathrm{CCl}_{4}}\) while the total pressure in the tank will remain constant at \(1 \, atm\). Subtracting this partial pressure decrease from the total pressure would give us the new partial pressure.
03

Calculate the amount of carbon needed

The formula given in the problem that relates the adsorption of CCl4 to its partial pressure can be used. The volume of CCl4 is required and it can be obtained from the partial pressure and Ideal Gas Law. Rearranging the formula will provide the mass of carbon needed. We set \(p_{\mathrm{CCl}_{4}}=1 \, mmHg\) for the desired end mole fraction and calculate. Remember to apply the proper unit conversion values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure Calculation
Partial pressure is a fundamental concept in chemistry, especially when dealing with gas mixtures. It represents the pressure that a single gas component would exert if it occupied the entire volume of the mixture. Calculation of partial pressure is crucial in many industrial processes, including gas treatment.

To calculate the partial pressure of a gas, one must first understand Dalton's Law of Partial Pressures. This law states that in a mixture of non-reacting gases, the total pressure of the mixture is equal to the sum of the partial pressures of each individual gas. For a component gas in a mixture, its partial pressure is determined by multiplying the total pressure by the mole fraction of that gas.

In the context of the exercise, the initial partial pressure of carbon tetrachloride (\( p_{CCl_4} \) is calculated using the relative saturation of carbon tetrachloride in the mixture. Given that the saturation vapor pressure at a certain temperature is known, the initial partial pressure is simply the product of the relative saturation and the saturation vapor pressure. By grasping this concept, students can move on confidently to address more complex scenarios, including those where gas is removed via adsorption.
Carbon Tetrachloride Removal

Understanding Carbon Tetrachloride Removal

Removal of carbon tetrachloride (\(CCl_4\)) from gaseous mixtures is essential in industrial processes to mitigate environmental impacts and health hazards. Activated carbon is frequently used for this purpose due to its exceptional adsorption properties.

In industrial gas treatment applications, carbon tetrachloride can be removed through physical adsorption on a solid surface, such as activated carbon. This process is influenced by the partial pressure of the carbon tetrachloride and temperature—the factors that dictate how much of the gas is removed. Knowing how to calculate both the total and partial pressure after the adsorption process, as demonstrated in the exercise's steps, equips students with the necessary understanding to evaluate the efficiency of the removal process.

The relationship between the amount of \(CCl_4\) adsorbed and the partial pressure is also dictated by adsorption isotherms, specific to the adsorbent and adsorbate at a given temperature. The exercise presented an example of such an isotherm equation, facilitating a deeper understanding of the adsorption process and how it can be quantified.
Activated Carbon Adsorption

Role of Activated Carbon in Adsorption

Activated carbon is a porous material with a large surface area that makes it highly effective for adsorption processes. It is used widely in gas treatment to remove contaminants like \(CCl_4\) through its extensive network of pores where adsorption occurs.

The effectiveness of activated carbon in adsorbing gaseous substances comes from its capacity to attract and hold gas molecules onto its surface, primarily through physical forces—such as van der Waals forces—and occasionally through chemical interactions as well. Adsorption capacity, which is related to the partial pressure of the adsorbate and temperature, is key to determining the amount of activated carbon needed.

In exercises like the one provided, students learn how to calculate the amount of activated carbon required for a desired level of contaminant removal. They use the adsorption isotherm equation to relate partial pressure to the weight of the adsorbate and the weight of activated carbon. Through this calculation, the broader topic of adsorption is demystified, showcasing how theoretical knowledge can be applied to practical and environmentally critical applications.

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Most popular questions from this chapter

A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid mixture of 40 mole \(\%\) benzene-60 mole\% toluene at 100^'C and 10 atm. Estimate the gas-phase composition (mole fractions) using Raoult's law. State your assumptions. Why would you have confidence in the accuracy of Raoult's law?

Nitric acid is a chemical intermediate primarily used in the synthesis of ammonium nitrate, which is used in the manufacture of fertilizers. The acid also is important in the production of other nitrates and in the separation of metals from ores. Nitric acid may be produced by oxidizing ammonia to nitric oxide over a platinum-rhodium catalyst, then oxidizing the nitric oxide to nitrogen dioxide in a separate unit where it is absorbed in water to form an aqueous solution of nitric acid.The reaction sequence is as follows:$$\begin{aligned} 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \\\4 \mathrm{NO}+2 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}_{2} \\\4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2} & \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq}) \end{aligned}$$.Ammonia vapor produced by vaporizing pure liquid ammonia at 820 kPa absolute is mixed with air, and the combined stream enters the ammonia oxidation unit. Air at \(30^{\circ} \mathrm{C}, 1\) atm absolute, and \(50 \%\) relative humidity is compressed and fed to the process. A fraction of the air is sent to the cooling and hydration units, while the remainder is passed through a heat exchanger and mixed with the ammonia. The total oxygen fed to the process is the amount stoichiometrically required to convert all of the ammonia to HNO \(_{3},\) while the fraction sent to the ammonia oxidizer corresponds to the stoichiometric amount required to convert ammonia to NO.The ammonia reacts completely in the oxidizer, with \(97 \%\) forming NO and the rest forming \(\mathrm{N}_{2}\). Only a negligible amount of \(\mathrm{NO}_{2}\) is formed in the oxidizer. However, the gas leaving the oxidizer is subjected to a series of cooling and hydration steps in which the NO is completely oxidized to \(\mathrm{NO}_{2}\) which in turn combines with water (some of which is present in the gas from the oxidizer and the rest is added) to form a 55 wt\% aqueous solution of nitric acid. The product gas from the process may be taken to contain only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). (a) Taking a basis of \(100 \mathrm{kmol}\) of ammonia fed to the process, calculate (i) the volumes \(\left(\mathrm{m}^{3}\right)\) of the ammonia vapor and air fed to the process using the compressibility-factor equation of state; (ii) the amount (kmol) and composition (in mole fractions) of the gas leaving the oxidation unit; (iii) the required volume of liquid water \(\left(\mathrm{m}^{3}\right)\) that must be fed to the cooling and hydration units; and (iv) the fraction of the air fed to the ammonia oxidizer. (b) Scale the results from Part (a) to a new basis of 100 metric tons per hour of 55\% nitric acid solution.(c) Nitrogen oxides (collectively referred to as \(\mathrm{NO}_{x}\) ) are a category of pollutants that are formed in many ways, including processes like that described in this problem. List the annual emission rates of the three largest sources of \(\mathrm{NO}_{x}\) emissions in your home region. What are the effects of exposure to excessive concentrations of \(\mathrm{NO}_{x} ?\) (d) A platinum-rhodium catalyst is used in ammonia oxidation. Fxplain the function of the catalyst, describe its structure, and explain the relationship of the structure to the function.

A fuel cell is an electrochemical device in which hydrogen reacts with oxygen to produce water and DC electricity. A 1-watt proton-exchange membrane fuel cell (PEMFC) could be used for portable applications such as cellular telephones, and a \(100-\mathrm{kW}\) PEMFC could be used to power an automobile. The following reactions occur inside the PEMFC:Anode: \(\quad \mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Cathode: \(\quad \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}\) Overall: \(\quad \overline{\mathrm{H}}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\) A flowchart of a single cell of a PEMFC is shown below. The complete cell would consist of a stack of such cells in series, such as the one shown in Problem 9.19.The cell consists of two gas channels separated by a membrane sandwiched between two flat carbonpaper electrodes- -the anode and the cathode- -that contain imbedded platinum particles. Hydrogen flows into the anode chamber and contacts the anode, where \(\mathrm{H}_{2}\) molecules are catalyzed by the platinum to dissociate and ionize to form hydrogen ions (protons) and electrons. The electrons are conducted throughthe carbon fibers of the anode to an extemal circuit, where they pass to the cathode of the next cell in the stack. The hydrogen ions permeate from the anode through the membrane to the cathode.Humid air is fed into the cathode chamber, and at the cathode \(\mathrm{O}_{2}\) molecules are catalytically split to form oxygen atoms, which combine with the hydrogen ions coming through the membrane and electrons coming from the external circuit to form water. The water desorbs into the cathode gas and is carried out of the cell. The membrane material is a hydrophilic polymer that absorbs water molecules and facilitates the transport of the hydrogen ions from the anode to the cathode. Electrons come from the anode of the cell at one end of the stack and flow through an extemal circuit to drive the device that the fuel cell is powering, while the electrons coming from the device flow back to the cathode at the opposite end of the stack to complete the circuit. is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance. 400-sell 300-yolt PEMFS anerates at stady state witha nonwer outnul of 36 k W, The air fod to It is important to keep the water content of the cathode gas between upper and lower limits. If the content reaches a value for which the relative humidity would exceed \(100 \%,\) condensation occurs at the cathode (flooding), and the entering oxygen must diffuse through a liquid water film before it can react. The rate of this diffusion is much lower than the rate of diffusion through the gas film normally adjacent to the cathode, and so the performance of the fuel cell deteriorates. On the other hand, if there is not enough water in the cathode gas (less than \(85 \%\) relative humidity), the membrane dries out and cannot transport hydrogen efficiently, which also leads to reduced performance.A 400-cell 300-volt PEMFC operates at steady state with a power output of 36 kW. The air fed to the cathode side is at \(20.0^{\circ} \mathrm{C}\) and roughly 1.0 atm (absolute) with a relative humidity of \(70.0 \%\) and a volumetric flow rate of \(4.00 \times 10^{3}\) SLPM (standard liters per minute). The gas exits at \(60^{\circ} \mathrm{C}\). (a) Explain in your own words what happens in a single cell of a PEMFC. (b) The stoichiometric hydrogen requirement for a PEMFC is given by \(\left(n_{\mathrm{Hz}}\right)_{\text {conanmad }}=I N / 2 F,\) where \(I\) is the current in amperes (coulomb/s), \(N\) is the number of single cells in the fuel cell stack, and \(F\) is the Faraday constant, 96,485 coulombs of charge per mol of electrons. Derive this expression. (Hint: Recall that since the cells are stacked in series the same current flows through each one, and the same quantity of hydrogen must be consumed in each single cell to produce that current at each anode.) (c) Use the expression of Part (b) to determine the molar rates of oxygen consumed and water generated in the unit with the given specifications, both in units of mol/min. (Remember that power = voltage \(\times\) current.) Then determine the relative humidity of the cathode exit stream, \(h_{\mathrm{r} \text { rout. }}\) (d) Determine the minimum cathode inlet flow rate in SLPM to prevent the fuel cell from flooding ( \(h_{\mathrm{r}, \text { out }}=100 \%\) ) and the maximum flow rate to prevent it from drying \(\left(h_{\mathrm{r}, \text { out }}=85 \%\right)\) .

An important parameter in the design of gas absorbers is the ratio of the flow rate of the feed liquid to that of the feed gas. The lower the value of this ratio, the lower the cost of the solvent required to process a given quantity of gas but the taller the absorber must be to achieve a specified separation.Propane is recovered from a 7 mole \(\%\) propane \(-93 \%\) nitrogen mixture by contacting the mixture with liquid \(n\) -decane. An insignificant amount of decane is vaporized in the process, and \(98.5 \%\) of the propane entering the unit is absorbed.(a) The highest possible propane mole fraction in the exiting liquid is that in equilibrium with the propane mole fraction in the feed gas (a condition requiring an infinitely tall column). Using Raoult's law to relate the mole fractions of propane in the feed gas and liquid, calculate the ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) corresponding to this limiting condition.(b) Suppose the actual feed ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) is 1.2 times the value calculated in Part (a) and the percentage of the entering propane absorbed is the same (98.5\%). Calculate the mole fraction of propane in the exiting liquid.(c) What are the costs and benefits associated with increasing \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) from its minimum value [the value calculated in Part (a)]? What would you have to know to determine the most cost-effective value of this ratio?

When air ( \(\approx 21\) mole\% \(\mathrm{O}_{2}, 79 \% \mathrm{N}_{2}\) ) is placed in contact with \(1000 \mathrm{cm}^{3}\) of liquid water at body temperature, \(36.9^{\circ} \mathrm{C},\) and 1 atm absolute, approximately 14.1 standard cubic centimeters \(\left[\mathrm{cm}^{3}(\mathrm{STP})\right]\) of gas are absorbed in the water at equilibrium. Subsequent analysis of the liquid reveals that 33.4 mole\% of the dissolved gas is oxygen and the balance is nitrogen.(a) Estimate the Henry's law coefficients (atm/mole fraction) of oxygen and nitrogen at \(36.9^{\circ} \mathrm{C}\). (b) An adult absorbs approximately \(0.4 \mathrm{g} \mathrm{O}_{2} / \mathrm{min}\) in the blood flowing though the lungs. Assuming that blood behaves like water and that it enters the lungs free of oxygen, estimate the flow rate of blood into the lungs in L/min. (c) The actual flow rate of blood into the lungs is roughly 5 L/min. Identify the assumptions made in the calculation of Part (b) that are likely causes of the discrepancy between the calculated and actual blood flows.

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