/*! 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 7 The vapor pressure of an organic... [FREE SOLUTION] | 91影视

91影视

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)?

Short Answer

Expert verified
The amount of the solvent in the gas phase is calculated from the Clausius-Clapeyron and Ideal Gas Law equations. If the species was dimerized, the number of moles would be halved.

Step by step solution

01

Determine the Clausius-Clapeyron equation

The Clausius-Clapeyron equation provides a mathematical relationship between vapor pressure and temperature of a substance. It's given by: \[ \ln \left( \frac{{P_2}}{{P_1}} \right) = -\frac{{\Delta H_{vap}}}{{R}} \left( \frac{{1}}{{T_2}} - \frac{{1}}{{T_1}} \right) \] Here, \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) (all temperatures must be in Kelvin). \( \Delta H_{vap} \) is the enthalpy of vaporization, R is the gas constant. We don't have \(\Delta H_{vap}\), so we need to solve this equation with the given pressures and temperatures to find it.
02

Find the enthalpy of vaporization

Convert temperatures to Kelvin (25掳C = 298.15 K, 45掳C = 318.15 K). Then, substitute \(P_1 = 50 \, \text{mm Hg}\), \(T_1 = 298.15 \, \text{K}\), \(P_2 = 200 \, \text{mm Hg}\), and \(T_2 = 318.15 \, \text{K}\) into the Clausius-Clapeyron equation and solve to find \(\Delta H_{vap}\) .
03

Estimate gas-phase moles at 35掳C (308.15 K)

First, find the vapor pressure at 35掳C by re-writing the Clausius-Clapeyron equation to solve for the required \(P\): \[ \ln \left( \frac{{P}}{{50}} \right) = -\frac{{\Delta H_{vap}}}{{R}} \left( \frac{{1}}{{308.15}} - \frac{{1}}{{298.15}} \right) \]Solve this to find \(P\). Now that we have the pressure, calculate the number of moles (\(n\)) using the ideal gas law (\(PV = nRT\)), where \(V = 150 \, \text{mL} = 0.150 \, \text{L}\), \(P\) is the calculated pressure (converted to atmospheres), \(R = 0.0821 \, \text{L路atm/K路mol}\) and \(T = 308.15 \, \text{K}\).
04

Answer the second part of the problem

The assumptions made in this problem come from the ideal gas law and the Clausius-Clapeyron equation. We assume the gas behaves ideally (no intermolecular interactions and occupies no volume). We also assume constant enthalpy of vaporization over the temperature range and that the system is in equilibrium (liquid and vapor phases). If the species dimerized, the number of moles would be halved, because two molecules would combine to form one.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vapor Pressure
Vapor pressure is a fundamental concept in thermodynamics related to phases of matter. It refers to the pressure exerted by a vapor in equilibrium with its liquid form at a given temperature. When a substance has a higher vapor pressure, it means more molecules are escaping from the liquid phase into the vapor phase. This is because the molecules are energetic enough to overcome the attractive forces within the liquid.
Understanding vapor pressure is crucial when considering how substances evaporate and boil. For example, at higher temperatures, molecules have more kinetic energy, leading to increased vapor pressure. This signifies that more molecules have the necessary energy to transition into the vapor state, leading to evaporation.
  • It is temperature-dependent: As temperature increases, vapor pressure also increases.
  • High vapor pressure implies greater volatility of the substance.
  • Equilibrium is achieved when evaporation and condensation occur at the same rate.
It's important to remember that at any given temperature, a phase equilibrium will exist when the vapor pressure matches the atmospheric pressure. This is what happens, for example, at the boiling point of a liquid.
Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is a vital tool in thermodynamics and chemistry, providing a way to describe the relationship between temperature and vapor pressure. It is particularly useful when comparing the vapor pressures of a substance at two different temperatures.
The equation is represented as: \[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where:- \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively, - \(\Delta H_{vap}\) represents the enthalpy of vaporization,- \(R\) is the universal gas constant, and - \(T_1\) and \(T_2\) are temperatures in Kelvin.
To use the equation effectively, temperatures should always be in Kelvin, and it may require solving for \( \Delta H_{vap} \) if it is unknown. By rearranging the equation, we can also estimate the vapor pressure at any temperature given we know the pressures and temperatures at two points.
This is particularly useful in predicting how a system will behave at temperatures that may not be directly measurable.
Ideal Gas Law
The Ideal Gas Law is a cornerstone of chemistry that helps us understand and predict gas behavior under various conditions. It is a simplified equation of state for gases, bridging pressure, volume, temperature, and the number of moles of gas. It is expressed mathematically as:
\[ PV = nRT \] Here, - \(P\) represents the pressure of the gas,- \(V\) is the volume of the gas,- \(n\) is the number of moles of the gas,- \(R\) is the ideal gas constant, which is usually 0.0821 L路atm K鈦宦 mol鈦宦,- \(T\) stands for the temperature in Kelvin.
The Ideal Gas Law assumes that gases are composed of particles that occupy no volume and experience no intermolecular attractions or repulsions. This is an approximation, as real gases exhibit these properties, but for many cases鈥攅specially at high temperatures and low pressures鈥攖his assumption allows us to predict and calculate gas properties effectively.
  • It is used for estimating gas moles from pressure, volume, and temperature.
  • Assumes ideal behavior, which means the equation is best applied when deviations due to intermolecular forces and molecular size are minimal.
  • Commonly applied to estimate moles of gas in closed systems or compare reactions where gases are involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A vapor mixture of \(n\) -butane (B) and \(n\) -hexane (H) contains 50.0 mole\% butane at \(120^{\circ} \mathrm{C}\) and 1.0 atm. A stream of this mixture flowing at a rate of \(150.0 \mathrm{L} / \mathrm{s}\) is cooled and compressed, causing some but not all of the vapor to condense. (Treat this process as a single-unit operation.) Liquid and vapor product streams emerge from the process in equilibrium at \(T\left(^{\circ} \mathrm{C}\right)\) and \(1100 \mathrm{mm} \mathrm{Hg}\). The vapor product contains 60.0 mole\% butane.(a) Draw and label a flowchart. Perform a degree-of-freedom analysis to show that you have enough information to determine the required final temperature ( \(T\) ), the composition of the liquid product (component mole fractions), and the molar flow rates of the liquid and vapor products from the given information and Antoine expressions for the vapor pressures \(p_{\mathrm{B}}^{*}(T)\) and \(p_{\mathrm{H}}^{*}(T) .\) Just identify the equations - for example, mole balance on butane or Raoult's law for hexane-but don't write them yet.(b) Write in order the equations you would use to determine the quantities listed in Part (a) and also the fractional condensation of hexane (mol \(\mathrm{H}\) condensed/mol \(\mathrm{H}\) fed). In each equation, circle the variable for which you would solve. Do no algebra or calculations.(c) Complete the calculations either manually or with an equation-solving program.(d) State three assumptions you made that could lead to errors in the calculated quantities.

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?

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?

In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20,000-liter cistem has been installed during construction of a new building. The cistem collects water from an HVAC (heating, ventilation, and air-conditioning) system designed to provide 2830 cubic meters of air per minute at \(22^{\circ} \mathrm{C}\) and \(50 \%\) relative humidity after converting it from ambient conditions \(\left(31^{\circ} \mathrm{C}, 70 \% \text { relative humidity }\right) .\) The collected condensate serves as the source of water for lawn maintenance. Estimate (a) the rate of intake of air at ambient conditions in cubic feet per minute and (b) the hours of operation required to fill the cistern.

In-Hexane is used to extract oil from soybeans. (See Problem 6.24 .) The solid residue from the extraction unit, which contains 0.78 kg liquid hexane/kg dry solids, is contacted in a dryer with nitrogen that enters at \(85^{\circ} \mathrm{C}\). The solids leave the dryer containing \(0.05 \mathrm{kg}\) liquid hexane/kg dry solids, and the gas leaves the dryer at \(80^{\circ} \mathrm{C}\) and 1.0 atm with a relative saturation of \(70 \% .\) The gas is then fed to a condenser in which it is compressed to 5.0 atm and cooled to \(28^{\circ} \mathrm{C}\), enabling some of the hexane to be recovered as condensate.(a) Calculate the fractional recovery of hexane (kg condensed/kg fed in wet solids). (b) A proposal has been made to split the gas stream leaving the condenser, combining 90\% of it with fresh makeup nitrogen, heating the combined stream to \(85^{\circ} \mathrm{C},\) and recycling the heated stream to the dryer inlet. What fraction of the fresh nitrogen required in the process of Part (a) would be saved by introducing the recycle? What costs would be incurred by introducing the recycle?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.