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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?

Short Answer

Expert verified
The mole fractions of each component in the gas phase can be calculated using Raoult's law. This result is highly confident due to the similar intermolecular interactions between benzene and toluene.

Step by step solution

01

Understanding Raoult’s law

Raoult's law states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. The formula can be written as, \( P_{i} \) = \( X_{i} \) * \( P_{i}^{*} \), where \( P_{i} \) is the partial pressure of component i, \( X_{i} \) is the mole fraction of component i in the solution, and \( P_{i}^{*} \) is the vapor pressure of pure component i at the given temperature.
02

Identifying the mole fractions

From the problem, we know that the liquid mixture contains 40% benzene and 60% toluene. Thus, the mole fractions of benzene (\( X_{B} \)) and toluene (\( X_{T} \)) in the liquid phase are 0.4 and 0.6 respectively.
03

Calculating the partial pressures

Assuming that the vapor pressures of pure benzene and toluene at 100°C are known (let's denote them as \( P_{B}^{*} \) and \( P_{T}^{*} \)), we can calculate the partial pressures of benzene and toluene in the gas phase using Raoult's law. For benzene: \( P_{B} \) = \( X_{B} \) * \( P_{B}^{*} \), and for toluene: \( P_{T} \) = \( X_{T} \) * \( P_{T}^{*} \)
04

Calculating the mole fractions in the gas phase

The total pressure of the gas mixture is the sum of the partial pressures of all three components. As nitrogen is an inert gas, it does not interact with benzene or toluene, so its partial pressure equals the total pressure minus the sum of the partial pressures of benzene and toluene. Thus,\( P_{N} \) = \( P_{total} \) - \( P_{B} \) - \( P_{T} \). Then, the mole fractions of each component in the gas phase can be calculated by dividing its partial pressure by the total pressure. For nitrogen: \( Y_{N} \)= \( P_{N} / P_{total} \), for benzene: \( Y_{B} \)= \( P_{B} / P_{total} \), and for toluene: \( Y_{T} \) = \( P_{T} / P_{total} \).
05

Confidence in the accuracy of Raoult's law

Raoult’s law is accurate for ideal solutions where interactions between molecules of different components are similar to those between molecules of the same component. Since benzene and toluene are both hydrocarbon compounds, they are likely to exhibit similar intermolecular interactions, which give us confidence in the accuracy of Raoult's law for this specific case

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Key Concepts

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

Vapor-Liquid Equilibrium
In a solution, **vapor-liquid equilibrium** occurs when the rate of evaporation of the liquid matches the rate at which the gas condenses back into the liquid. This balance between phases at a given temperature and pressure is crucial for explaining the behavior of mixtures, particularly when applying Raoult's law. In our case, the mixture of benzene and toluene reaches a state of equilibrium. This allows us to predict the composition of the gas phase using known properties of the liquid phase. The equilibrium ensures that each component distributes between the liquid and vapor phases according to its own partial pressure over its pure form, multiplied by its mole fraction in the solution.
Understanding vapor-liquid equilibrium helps in determining how components distribute themselves between the liquid and gas phases, providing a foundation for further calculations using Raoult's Law.
Partial Pressure Calculation
**Partial pressure calculation** is a key step in analyzing vapor-liquid equilibrium. Each component in a mixture contributes to the total pressure within the gas phase. Raoult's law facilitates the computation of partial pressures by relating them to mole fractions and the vapor pressures of the pure components.
- For benzene and toluene, you need the vapor pressures at the given temperature. Let's use hypothetical values as an example. If benzene's vapor pressure at 100°C is, let's say, 100 kPa and toluene's is 200 kPa, the partial pressures would be calculated as follows:
- Benzene's partial pressure: \[ P_{B} = X_{B} \times P_{B}^{*} \] where \( X_{B} \) is benzene's mole fraction and \( P_{B}^{*} \) is benzene's vapor pressure.
- Similarly, for toluene: \[ P_{T} = X_{T} \times P_{T}^{*} \] where \( X_{T} \) is toluene's mole fraction and \( P_{T}^{*} \) is toluene's vapor pressure.
The sum of these partial pressures provides insight into how each component contributes to the overall behavior of the gas mixture, crucial for determining gas-phase compositions.
Ideal Solutions
**Ideal solutions** are those in which the interactions between different molecules are similar to those within the same component. This concept is essential when applying Raoult's law. In an ideal solution, each component's contribution to the total vapor pressure directly correlates to its concentration in the liquid phase and its pure component vapor pressure.
It's assumed that the molecules do not experience additional attractions or repulsions when mixed together, making calculations straightforward. Benzene and toluene, both being hydrocarbons, tend to follow ideal behavior fairly closely. This is why the accuracy of Raoult’s law in predicting their vapor pressures and partial pressures is enhanced. Using Raoult’s law under these conditions generally yields reliable results, as intermolecular forces don't significantly deviate from ideal predictions.

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Most popular questions from this chapter

A gas containing nitrogen, benzene, and toluene is in equilibrium with a liquid consisting of 35 mole\% benzene and 65 mole \(\%\) toluene at \(85^{\circ} \mathrm{C}\) and 10 atm. Estimate the gas composition (mole fractions) using Raoult's law and assuming ideal-gas behavior.

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

A gas mixture containing 85.0 mole \(\% \mathrm{N}_{2}\) and the balance \(n\) -hexane flows through a pipe at a rate of \(100.0 \mathrm{m}^{3} / \mathrm{h} .\) The pressure is 2.00 atm absolute and the temperature is \(100^{\circ} \mathrm{C}\). (a) What is the molar flow rate of the gas in \(\mathrm{kmol} / \mathrm{h}\) ? (b) Is the gas saturated? If not, to what temperature ( \(^{C} C\) ) would it have to be cooled at constant pressure in order to begin condensing hexane? (c) To what temperature ( \(C\) ) would the gas have to be cooled at constant pressure in order to condense \(80 \%\) of the hexane?

The vapor leaving the top of a distillation column goes to a condenser in which either total or partial condensation takes place. If a total condenser is used, a portion of the condensate is returned to the top of the column as \(r e f l u x\) and the remaining liquid is taken off as the overhead product (or distillate). (See Problem 6.63.) If a partial condenser is used, the liquid condensate is returned as reflux and the uncondensed vapor is taken off as the overhead product.The overhead product from an \(n\) -butane- \(n\) -pentane distillation column is 96 mole \(\%\) butane. The temperature of the cooling fluid limits the condenser temperature to \(40^{\circ} \mathrm{C}\) or higher.(a) Using Raoult's law, estimate the minimum pressure at which the condenser can operate as a partial condenser (i.e., at which it can produce liquid for reflux) and the minimum pressure at which it can operate as a total condenser. In terms of dew point and bubble point, what do each of these pressures represent for the given temperature?(b) Suppose the condenser operates as a total condenser at \(40^{\circ} \mathrm{C}\), the production rate of overhead product is \(75 \mathrm{kmol} / \mathrm{h}\), and the mole ratio of reflux to overhead product is \(1.5: 1 .\) Calculate the molar flow rates and compositions of the reflux stream and the vapor feed to the condenser.(c) Suppose now that a partial condenser is used, with the reflux and overhead product in equilibrium at \(40^{\circ} \mathrm{C}\) and the overhead product flow rate and reflux-to-overhead product ratio having the values given in Part (b). Calculate the operating pressure of the condenser and the compositions of the reflux and vapor feed to the condenser.

Recovery and processing of various oils are important elements of the agricultural and food industries. For example, soybean hulls are removed from the beans, which are then flaked and contacted with hexane. The hexane extracts soybean oil and leaves very little oil in the residual solids. The solids are dried at an elevated temperature, and the dried solids are used to feed livestock or further processed to extract soy protein. The gas stream leaving the dryer is at \(80^{\circ} \mathrm{C}\) 1 atm absolute, and 50\% relative saturation.(a) To recover hexane, the gas leaving the dryer is fed to a condenser, which operates at 1 atm absolute. The gas leaving the condenser contains 5.00 mole \(\%\) hexane, and the hexane condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). (b) In an altemative arrangement, the gas leaving the dryer is compressed to 10.0 atm and the temperature simultancously is increased so that the relative saturation remains at \(50 \% .\) The gas then is cooled at constant pressure to produce a stream containing 5.00 mole \(\%\) hexane. Calculate the final gas temperature and the ratio of volumetric flow rates of the gas streams leaving and entering the condenser. State any assumptions you make.(c) What would you need to know to determine which of processes (a) and (b) is more cost- effective?

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