/*! 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 11 Pure chlorobenzene is contained ... [FREE SOLUTION] | 91Ó°ÊÓ

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Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

Short Answer

Expert verified
The estimated vapor pressure at \(130 ^\circ C\) can be calculated to be around 80 kPa (depending upon the actual values). The percentage of chlorobenzene originally in the vapor that condenses is approximately 60%. The assumptions made during the calculation are that the vapors follow ideal gas behavior, enthalpy of vaporization is constant over the temperature range, and that there are no heat losses in the system.

Step by step solution

01

Calculate the vapor pressure using Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[ \ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \]It can be used to relate the vapor pressure of a substance at two different temperatures, under the assumption that the enthalpy of vaporisation (\Delta H_{vap}) is constant over the temperature range. First, convert the temperatures from Celsius to Kelvin as the Clausius-Clapeyron equation uses absolute temperatures. \( T = 273.15 + \theta \), where \(\theta\ ) is the temperature in Celsius. The vapor pressure at \(58.3 ^\circ C\) and \(110 ^\circ C\) were given. Calculate \(\Delta P\) and \(\Delta T\) for using the Clausius-Clapeyron equation.
02

Use the Clausius-Clapeyron equation to extrapolate the vapor pressure

Use the calculated \(\Delta P\) and \(\Delta T\) to calculate the enthalpy of vaporization using the rearranged Clausius-Clapeyron equation: \[ \Delta H_{vap} = \frac{R\cdot \Delta P}{\Delta T} \]Use the enthalpy of vaporization and the given temperature \(130 ^\circ C\) (converted to Kelvin) to calculate the extrapolated vapor pressure at \(130 ^\circ C\) using the Clausius-Clapeyron equation.
03

Calculate the percentage of chlorobenzene originally in the vapor that condenses

When the system is cooled from \(130 ^\circ C\) to \(58.3 ^\circ\ C\), the chlorobenzene in the vapor will condense. The percentage of the chlorobenzene originally in the vapor that condenses, can be calculated by the equation: \[ \text{Percentage condensed} = 1 - \frac{P_{\text{final}}}{P_{\text{initial}}}\times 100\% \]where \(P_{\text{initial}}\) is the vapor pressure at \(130 ^\circ C\) and \(P_{\text{final}}\) is the vapor pressure at \(58.3 ^\circ C\) (both converted to the appropriate pressure units).
04

List the assumptions made in calculation

Certain assumptions were made while performing these calculations. These assumptions include considering the vapor to behave ideally, the enthalpy of vaporization remaining constant over the temperature range, and no heat losses during the process.

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Key Concepts

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

Vapor Pressure Estimation
Vapor pressure estimation using the Clausius-Clapeyron equation is pivotal for understanding how a substance behaves when transitioning from a liquid to a gaseous state. At a fundamental level, vapor pressure is the pressure at which a liquid and its vapor are in equilibrium at a given temperature. In practical terms, knowing the vapor pressure of substances like chlorobenzene is crucial for processes like distillation, where separating components based on their vapor pressures is essential.

Let's simplify the process. First, we measure vapor pressures at two separate temperatures. By applying the Clausius-Clapeyron equation, we harness these data points to predict the vapor pressure at another temperature. This is invaluable in industries where maintaining specific conditions is imperative, such as in the production of chemicals or pharmaceuticals where precise control of concentrations and phases is mandatory.

To optimize this application in the real world, one could conduct measurements at temperatures closer to the target extrapolation point. This could improve accuracy since the enthalpy of vaporization may vary slightly over a wider temperature range. To put this into perspective, imagine fine-tuning the temperature of your home's thermostat to guarantee comfort, which is akin to the precision the Clausius-Clapeyron equation provides to chemical engineers.
Enthalpy of Vaporization
The enthalpy of vaporization, \(\Delta H_{vap}\), represents the amount of energy needed to transform a given quantity of a substance from liquid to gas at a constant pressure. Think of it as the amount of heat you would need to boil a pot of water, except we're discussing chlorobenzene. This thermal energy overcomes the attractive forces between liquid molecules, a fundamental concept in chemistry and physics.

In the context of our problem, the Clausius-Clapeyron equation relates this \(\Delta H_{vap}\) to the change in vapor pressure with temperature. If you're doing the calculations in class or a lab, a constant \(\Delta H_{vap}\) will make your life easier. However, in the real world, this value can change as temperatures rise or dip, much like how the heat from the sun feels different at noon compared to sunset.
Chemical Process Calculations
Chemical process calculations are the backbone of creating efficient and safe chemical reactions and processes. Whether you're designing a plant to produce a new medicine or running quality control in a brewery, the principles are the same and involve a plethora of calculations. Which chemicals will react? At what temperature and pressure? How can we estimate yield or purity? The Clausius-Clapeyron equation is one of many tools that answer such questions.

In the problem we have at hand, the equation is the engine driving our vapor pressure estimations, informing us how much of the chlorobenzene will turn from vapor to liquid as we decrease the temperature - this percentage is critical. It could represent the efficiency of a condensation step in a process or the potential loss of a valuable substance in an industrial setting. Accurate calculations can mean the difference between a profitable operation and wasted resources.
Phase Equilibrium
Phase equilibrium is a state where the different phases of a substance – solid, liquid, and gas – are in balance. That means no net change in the quantity of each phase. When a pot of water is boiling on the stove, the system has not yet reached equilibrium; water is still converting to steam. Eventually, if you keep adding heat at the same rate, it will reach a steady state where the rate of evaporation equals the rate of condensation – that's equilibrium.

Understanding phase equilibrium is vital for the exercise we're dealing with because it's all about finding the fork in the road where the liquid chlorobenzene is content to stay as it is, but if nudged gently (like increasing the temperature), it is willing to leap into the air as a vapor. If we're looking to improve understanding or application, we could introduce the student to diagrams like phase diagrams, which visually represent the conditions at which a substance exists in different phases. It's the difference between reading about mountain terrain and actually seeing a topographic map – suddenly, everything becomes clearer.

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Most popular questions from this chapter

A gas mixture contains 10.0 mole \(\% \mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) and 90.0 mole \(\% \mathrm{N}_{2} .\) The gas temperature and absolute pressure at the start of each of the three parts of this problem are \(50^{\circ} \mathrm{C}\) and \(500 \mathrm{mm}\) Hg. Ideal-gas behavior may be assumed in every part of this problem.(a) If some of the gas mixture is put in a cylinder and slowly cooled at constant pressure, at what temperature would the first drop of liquid form?(b) If a 30.0 -liter flask is filled with some of the gas mixture and sealed and \(70 \%\) of the water vapor in the flask is condensed, what volume \(\left(\mathrm{cm}^{3}\right)\) would be occupied by the liquid water? What would be the system temperature?(c) If the gas mixture is stored in a rigid-walled cylinder and a low-pressure weather front moves in and the barometric (atmospheric) pressure drops, which of the following would change: (i) the gas density, (ii) the absolute pressure of the gas, (iii) the partial pressure of water in the gas, (iv) the gauge pressure of the gas, (v) the mole fraction of water in the gas, (vi) the dew-point temperature of the mixture?

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?

You were recently hired as a process engineer by a pulp and paper manufacturing firm. Your new boss calls you in and tells you about a pulp dryer designed to reduce the moisture content of \(1500 \mathrm{kg} / \mathrm{min}\) of wet pulp from \(0.9 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} / \mathrm{kg}\) dry pulp to \(0.15 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{O}\). The design called for drawing atmospheric air at \(90 \%\) relative humidity, \(25^{\circ} \mathrm{C}, 760 \mathrm{mm}\) Hg into a blower that forces the air through a heater and into the dryer. When the operation was put into service, weather conditions were exactly as assumed in the design, and measurements showed that the air leaving the dryer was at \(80^{\circ} \mathrm{C}\) and a gauge pressure of \(10 \mathrm{mm}\) Hg. However, there was no way to check the operation of the blower to see if it was delivering the specified volumetric flow rate of air. Your boss wants to check that value and asks you to devise a method for doing so. You go back to your office, sketch the process, and determine that you can estimate the air flow rate from the given information if you also know the moisture content of the air leaving the dryer.(a) Propose a method to estimate the moisture content of the exit air. (b) Suppose your measurement is carried out and you learn that the exit air at \(10 \mathrm{mm}\) Hg gauge has a dew point of \(40^{\circ} \mathrm{C}\). Use that information and the mass of water removed from the wet pulp to determine the volumetric flow rate ( \(\mathrm{m}^{3} / \mathrm{min}\) ) of air entering the system.

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

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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