/*! 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 53 The solubility coefficient of a ... [FREE SOLUTION] | 91Ó°ÊÓ

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The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in \(1 \mathrm{cm}^{3}\) of a solvent under a partial pressure of 1 atm. The solubility coefficient of \(\mathrm{CO}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(0.0901 \mathrm{cm}^{3} \mathrm{CO}_{2}(\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) Calculate the Henry's law constant in atm/mole fraction for \(\mathrm{CO}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) from the given solubility coefficient. (b) How many grams of \(\mathrm{CO}_{2}\) can be dissolved in a \(12-\mathrm{oz}\) bottle of soda at \(20^{\circ} \mathrm{C}\) if the gas above the soda is pure \(\mathrm{CO}_{2}\) at a gauge pressure of 2.5 atm ( 1 liter \(=33.8\) fluid ounces)? Assume the liquid properties are those of water. (c) What volume would the dissolved \(C O_{2}\) occupy if it were released from solution at body temperature and pressure \(-37^{\circ} \mathrm{C}\) and 1 atm?

Short Answer

Expert verified
a) The Henry's Law constant will be calculated as per the conversion explained above. b) To determine the amount of \(CO_2\) dissolved, moles of \(CO_2\) will be calculated using Henry's law and then will be converted into mass with the molar mass. c) Finally, the volume that the dissolved \(CO_2\) would occupy if it were released, is to be calculated using the ideal gas law.

Step by step solution

01

Convert Solubility Coefficient to Henry's Constant

To convert the solubility coefficient \(0.0901 \, \mathrm{cm}^{3} \mathrm{CO}_{2} \, (\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O} (\mathrm{l})\) to Henry's constant in atm/mole fraction we first need to convert the volume of CO2 to moles at STP. At STP, 1 mole of a gas occupies 22.4 liters. \nSo, \(1 \, \mathrm{cm}^{3}=10^{-3} \, \mathrm{liters}\), therefore, \(0.0901 \, \mathrm{cm}^{3} = 0.0901 \times 10^{-3} \, \mathrm{moles}\) \nThen, Henry's law constant \(K_{H} = \frac{P}{X}\), where P is the partial pressure of the gas and X is the mole fraction. Here P is 1 atm and hence the mole fraction X becomes \(X= \frac{n_{\mathrm{CO2}}}{n_{\mathrm{CO2}} + n_{\mathrm{H2O}}}\) . Since, n_{\mathrm{H2O}} >> n_{\mathrm{CO2}} , X ~= n_{\mathrm{CO2}} and thus, \(K_{H} = 1 \, \mathrm{atm} / 0.0901 \times 10^{-3} \, \mathrm{moles}\)
02

Calculate the amount of dissolved CO2

For a 12-ounce bottle, the volume of water = 12/33.8 liters. At 2.5 atm, we can calculate the number of moles of CO2. Since the Henry's Law states that the amount of dissolved gas is directly proportional to its partial pressure in the gas phase, the amount of CO2 dissolved would be: \n\(n_{\mathrm{CO2}} = P/K_{H} = 2.5 \, \mathrm{atm} / K_{H}\) and the mass of CO2 would be \(m_{\mathrm{CO2}} = n_{\mathrm{CO2}} \times M_{\mathrm{CO2}}\) where \(M_{\mathrm{CO2}} = 44 \, \mathrm{g}\, / \, \mathrm{mole}\)
03

Calculate the volume of released CO2

To find the volume that the CO2 would occupy when released, we’d use the Ideal Gas Law equation \(PV = nRT\). We need to convert the temperature to Kelvin by adding 273 to the Celsius temperature. Thus, the volume is : \(V = nRT / P = n_{\mathrm{CO2}} \times R \times (37 + 273) \, K / 1 \, atm\), where R = 0.08206 L-atm/mol-K.

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Key Concepts

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

Solubility Coefficient
The solubility coefficient tells us how much gas can dissolve in a liquid at a specific pressure. It represents the volume of gas (in cubic centimeters, STP) that dissolves in one cubic centimeter of solvent at a partial pressure of 1 atm. In simpler terms, it measures how easily a gas mixes with a liquid. For example, at 20°C, the solubility coefficient of CO₂ in water is 0.0901 cm³ of CO₂ (STP) per cm³ of water.
Understanding this concept is important because it helps us predict the behavior of gases in liquids. When a gas has a high solubility coefficient, it means the gas can dissolve significantly in the liquid. This concept is used frequently in fields like chemistry and environmental science.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates pressure, volume, temperature, and moles of a gas. It is expressed as: \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant (0.08206 L-atm/mol-K), and \(T\) is the temperature in Kelvin.
This equation helps predict how a gas will behave under different conditions or how it will change when any of these variables are altered. For instance, when calculating the volume that dissolved COâ‚‚ would occupy if released, the Ideal Gas Law allows us to account for the new conditions at body temperature by adjusting these variables to reflect the properties of gases.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the moles of one component to the total moles in the mixture. For a gas such as COâ‚‚ dissolving in water, the mole fraction \(X\) can be calculated using the formula: \(X = \frac{n_{\mathrm{CO2}}}{n_{\mathrm{CO2}} + n_{\mathrm{H2O}}}\).
Here, \(n_{\mathrm{CO2}}\) represents the moles of CO₂ and \(n_{\mathrm{H2O}}\) represents the moles of water. In most cases, because the moles of water are much larger compared to the moles of dissolved gas, the mole fraction simplifies to \(X \approx n_{\mathrm{CO2}}\). Understanding the mole fraction is crucial in calculating Henry’s law constant and predicting how much gas will dissolve under certain conditions.

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Most popular questions from this chapter

A stage of a separation process is defined as an operation in which components of one or more feed streams divide themselves between two phases, and the phases are taken off separately. In an ideal stage or equilibrium stage, the effluent (exit) streams are in equilibrium with each other.Distillation columns often consist of a series of vertically distributed stages. Vapor flows upward and liquid flows downward between adjacent stages; some of the liquid fed to each stage vaporizes,and some of the vapor fed to each stage condenses. A representation of a section of a distillation column is shown below. (See Problem 4.42 for a more realistic representation.) Consider a distillation column operating at 0.4 atm absolute in which benzene and styrene are being separated. A vapor stream containing 65 mole\% benzene and 35 mole\% styrene enters stage 1 at a rate of \(200 \mathrm{mol} / \mathrm{h}\), and liquid containing 55 mole\% benzene and 45 mole\% styrene leaves this stage at a rate of 150 mol/h. You may assume (1) the stages are ideal, (2) Raoult's law can be used to relate the compositions of the streams leaving each stage, and (3) the total vapor and liquid molar flow rates do not change by a significant amount from one stage to the next.(a) How would you expect the mole fraction of benzene in the liquid to vary from one stage to another, beginning with stage 1 and moving up the column? In light of your answer and considering that the pressure remains essentially constant from one stage to another, how would you then expect the temperature to vary at progressively higher stages? Briefly explain. (b) Estimate the temperature at stage 1 and the compositions of the vapor stream leaving this stage and the liquid stream entering it. Then repeat these calculations for stage 2 . (c) Describe how you would calculate the number of ideal stages required to reduce the styrene content of the vapor to less than 5 mole\%.

Sodium bicarbonate is synthesized by reacting sodium carbonate with carbon dioxide and water at \(70^{\circ} \mathrm{C}\) and \(2.0 \mathrm{atm}\) gauge pressure: $$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaHCO}_{3}$$ An aqueous solution containing 7.00 wt\% sodium carbonate and a gas stream containing 70.0 mole\% \(\mathrm{CO}_{2}\) and the balance air are fed to the reactor. All of the sodium carbonate and some of the carbon dioxide in the feed react. The gas leaving the reactor, which contains the air and unreacted \(\mathrm{CO}_{2},\) is saturated with water vapor at the reactor conditions. A liquid-solid slurry of sodium bicarbonate crystals in a saturated aqueous solution containing \(2.4 \mathrm{wt} \%\) dissolved sodium bicarbonate and a negligible amount of dissolved \(\mathrm{CO}_{2}\) leaves the reactor and is pumped to a filter. The wet filter cake contains 86 wt\% sodium bicarbonate crystals and the balance saturated solution, and the filtrate also is saturated solution. The production rate of solid crystals is \(500 \mathrm{kg} / \mathrm{h}\).Suggestion: Although the problems to be given can be solved in terms of the product flow rate of \(500 \mathrm{kg} \mathrm{NaHCO}_{3}(\mathrm{s}) / \mathrm{h},\) it might be easier to assume a different basis and then scale the process to the desired production rate of crystals.(a) Calculate the composition (component mole fractions) and volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of the gas stream leaving the reactor. (b) Calculate the feed rate of gas to the process in standard cubic meters/min \(\left[\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{min}\right]\) (c) Calculate the flow rate \((\mathrm{kg} / \mathrm{h})\) of the liquid feed to the process. What more would you need to know to calculate the volumetric flow rate of this stream? (d) The filtrate was assumed to leave the filter as a saturated solution at \(70^{\circ} \mathrm{C}\). What would be the effect on your calculations if the temperature of the filtrate actually dropped to \(50^{\circ} \mathrm{C}\) as it passed through the filter? (e) The reactor pressure of 2 atm gauge was arrived at in an optimization study. What benefit do you suppose would result from increasing the pressure? What penalty would be associated with this increase? The term "Henry's law" should appear in your explanation. (Hint: The reaction occurs in the liquid phase and the \(\mathrm{CO}_{2}\) enters the reactor as a gas. What step must precede the reaction?)

The following diagram shows a staged absorption column in which \(n\) -hexane (H) is absorbed from a gas into a heavy oil.A gas feed stream containing 5.0 mole \(\%\) hexane vapor and the balance nitrogen enters at the bottom of an absorption column at a basis rate of \(100 \mathrm{mol} / \mathrm{s}\), and a nonvolatile oil enters the top of the column in a ratio 2 mol oil fed/mol gas fed. The absorber consists of a series of ideal stages (see Problem 6.66), arranged so that gas flows upward and liquid flows downward. The liquid and gas streams leaving each stage are in equilibrium with each other (by the definition of an ideal stage), with compositions related by Raoult's law. The absorber operates at an approximately constant temperature \(T\left(^{\circ} \mathrm{C}\right)\) and \(760 \mathrm{mm}\) Hg. Of the hexane entering the column, \(99.5 \%\) is absorbed and leaves in the liquid column effluent. At the given conditions it may be assumed that \(\mathrm{N}_{2}\) is insoluble in the oil and that none of the oil vaporizes.(a) Calculate the molar flow rates and mole fractions of hexane in the gas and liquid streams leaving the column. Then calculate the average values of the liquid and gas molar flow rates in the column, \(\dot{n}_{\mathrm{L}}(\mathrm{mol} / \mathrm{s})\) and \(\dot{n}_{\mathrm{G}}(\mathrm{mol} / \mathrm{s}) .\) For simplicity, in subsequent calculations use the average values for liquid and gas molar flow rates within the column, but the actual values for the corresponding flow rates entering and leaving the column.(b) Considering the bottom stage to be ideal, estimate the mole fractions of hexane in the gas leaving that stage \(\left(y_{N}\right)\) and in the liquid entering it \(\left(x_{N-1}\right)\) if the column temperature is \(50^{\circ} \mathrm{C}\). (c) Suppose that \(x_{i}\) and \(y_{i}\) are the mole fractions of hexane in the liquid and gas streams leaving stage \(i\) Derive the following equations from an equilibrium relationship and a mass balance around a section of the column encompassing stage \(i\) and the bottom of the column:$$\begin{array}{c}y_{i}=x_{i} p_{i}^{*}(T) / P \\ x_{i-1}=\left(x_{N} n_{L, N}+y_{i} \dot{n}_{\mathrm{G}}-y_{N+1} \dot{n}_{\mathrm{G}+1}\right) / \dot{n}_{\mathrm{L}}\end{array}$$Verify that these equations yield the answers you calculated in Part (b). (d) Examine the effect of operating temperature on the column by estimating the number of ideal stages necessary to achieve the desired separation. In the calculations, which will be done using a spreadsheet, take the pressure in the column to be constant at 760 torr, but consider three different operating temperatures: \(30^{\circ} \mathrm{C}\) \(50^{\circ} \mathrm{C},\) and \(70^{\circ} \mathrm{C} .\) The calculations will follow a stage-to-stage strategy beginning at the bottom of the column and repeatedly applying Equations (1) and (2) until the mole fraction of hexane in the vapor leaving the column is less than or equal to that calculated in Part (a). You may use APEx or the Antoine equation and Table B.4 to estimate the hexane vapor pressure. The calculations for the case of \(T=30^{\circ} \mathrm{C}\) illustrate how to proceed; for this case, \(y_{N-1} < y_{1}=0.00263\) after only two stages.(e) You can see that the number of stages required increases as the column temperature increases. In fact, there is a maximum temperature beyond which the required separation cannot be achieved. At that temperature, the entering gas and leaving liquid are approximately in equilibrium, so that \(x_{N} p^{*}(T)=y_{N+1} P .\) Use either APEx or the Antoine equation to estimate the maximum temperature at which the separation can be achieved.

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?

Dehydration of natural gas is necessary to prevent the formation of gas hydrates, which can plug valves and other components of a gas pipeline, and also to reduce potential corrosion problems. Water removal can be accomplished as shown in the following schematic diagram: Natural gas containing \(80 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas \(\left[\mathrm{SCF}=\mathrm{ft}^{3}(\mathrm{STP})\right]\) enters the bottom of an absorber at a rate of \(4.0 \times 10^{6}\) SCF/day. A liquid stream containing triethylene glycol (TEG, molecular weight \(=150.2\) ) and a small amount of water is fed to the top of the absorber. The absorber operates at 500 psia and \(90^{\circ} \mathrm{F}\). The dried gas leaving the absorber contains \(10 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O} / 10^{6} \mathrm{SCF}\) gas. The solvent leaving the absorber, which contains all the TEG-water mixture fed to the column plus all the water absorbed from the natural gas, goes to a distillation column. The overhead product stream from the distillation column contains only liquid water. The bottoms product stream, which contains TEG and water, is the stream recycled to the absorber.(a) Draw and completely label a flowchart of this process. Calculate the mass flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) and volumetric flow rate (ft \(^{3}\) /day) of the overhead product from the distillation column. (b) The greatest possible amount of dehydration is achieved if the gas leaving the absorption column is in equilibrium with the solvent entering the column. If the Henry's law constant for water in TEG at \(90^{\circ} \mathrm{F}\) is \(0.398 \mathrm{psia} / \mathrm{mol}\) fraction, what is the maximum allowable mole fraction of water in the solvent fed to the absorber?(c) A column of infinite height would be required to achieve equilibrium between the gas and liquid at the top of the absorber. For the desired separation to be achieved in practice, the mole fraction of water in the entering solvent must be less than the value calculated in Part (b). Suppose it is \(80 \%\) of that value and the flow rate of TEG in the recirculating solvent is 37 Ib \(_{\mathrm{m}}\) TEG/lb \(_{\mathrm{m}}\) water absorbed in the column. Calculate the flow rate ( \(\left(\mathrm{b}_{\mathrm{m}} / \mathrm{day}\right)\) of the solvent stream entering the absorber and the mole fraction of water in the solvent stream leaving the absorber. (d) What is the purpose of the distillation column in the process? (Hint: Think about how the process would operate without it.)

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