/*! 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 50 The following data give the temp... [FREE SOLUTION] | 91Ó°ÊÓ

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The following data give the temperature, the vapor pressure, and the density of the coexisting vapor phase of benzene. Use the van der Waals equation and the Redlich-Kwong equation to calculate the vapor pressure and compare your result with the experimental values given below. Use Equations \(16.17\) and \(16.18\) with \(T_{\mathrm{c}}=561.75 \mathrm{~K}\) and \(P_{\mathrm{c}}=48.7575\) bar to calculate the van der Waals parameters and the Redlich-Kwong parameters. \begin{tabular}{lll} \(T / \mathrm{K}\) & \(P /\) bar & \(\rho^{8} / \mathrm{mol} \cdot \mathrm{L}^{-1}\) \\\ \hline \(290.0\) & \(0.0860\) & \(0.00359\) \\ \(300.0\) & \(0.1381\) & \(0.00558\) \\ \(310.0\) & \(0.2139\) & \(0.00839\) \\ \(320.0\) & \(0.3205\) & \(0.01223\) \\ \(330.0\) & \(0.4666\) & \(0.01734\) \\ \(340.0\) & \(0.6615\) & \(0.02399\) \\ \(350.0\) & \(0.9161\) & \(0.03248\) \end{tabular}

Short Answer

Expert verified
Use given constants to calculate Van der Waals and Redlich-Kwong parameters, then solve each equation for vapor pressure and compare to experimental values.

Step by step solution

01

Obtain Van der Waals parameters

To determine van der Waals parameters, we use:1. \( a = 0.42148 \frac{R^2 T_c^2}{P_c} \)2. \( b = 0.08664 \frac{R T_c}{P_c} \)Given \(T_c = 561.75 \text{ K}\) and \(P_c = 48.7575 \text{ bar}\), and assuming \(R = 0.08314 \frac{\text{L bar}}{\text{K mol}}\), calculate \(a\) and \(b\).Substituting the values:- \( a = 0.42148 \times \frac{(0.08314)^2 \times (561.75)^2}{48.7575} \)- \( b = 0.08664 \times \frac{0.08314 \times 561.75}{48.7575} \).Compute \(a\) and \(b\) to use for vapor pressure calculation.
02

Obtain Redlich-Kwong parameters

For the Redlich-Kwong equation, use:1. \( a = 0.42748 \frac{R^2 T_c^{2.5}}{P_c}\)2. \( b = 0.08664 \frac{R T_c}{P_c}\)Again, with \(T_c\), \(P_c\), and \(R\) values given:Substituting into these equations:- \( a = 0.42748 \times \frac{(0.08314)^2 \times (561.75)^{2.5}}{48.7575} \)- \( b = 0.08664 \times \frac{0.08314 \times 561.75}{48.7575} \).Calculate \(a\) and \(b\) here for later use in the Redlich-Kwong equation.
03

Calculate Vapor Pressure using Van der Waals

Substitute the Van der Waals parameters obtained from Step 1 into the Van der Waals equation:\[\left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \]For each temperature given in the data table, solve for \(P\) knowing \(T\) and \(\rho\), using \(V_m = \frac{1}{\rho}\). This will give the vapor pressure at each temperature according to Van der Waals.
04

Calculate Vapor Pressure using Redlich-Kwong

Substitute the parameters from Step 2 into the Redlich-Kwong equation:\[P = \frac{RT}{V_m - b} - \frac{a}{\sqrt{T} \, V_m(V_m + b)}\]Again, for each temperature, use the given \(\rho\) to find \(V_m = \frac{1}{\rho}\), and solve for \(P\) using known \(T\) and Redlich-Kwong parameters.
05

Compare with Experimental Values

Compare the calculated vapor pressures from both Van der Waals and Redlich-Kwong equations with the experimental values listed in the table. Check which model predicts the vapor pressures more accurately by calculating deviations from experimental data. Discuss the relative accuracy of each equation under the given conditions.

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Key Concepts

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

Redlich-Kwong equation
The Redlich-Kwong equation is a popular model used in thermodynamics to predict the behavior of real gases. It improves upon the van der Waals equation by offering better accuracy, particularly at moderate pressures and high temperatures. The formula is given by:
  • \[P = \frac{RT}{V_m - b} - \frac{a}{\sqrt{T} \cdot V_m(V_m + b)}\]
In this equation, \(P\) is the pressure of the gas, \(R\) is the universal gas constant, \(T\) is the absolute temperature, and \(V_m\) is the molar volume. The parameters \(a\) and \(b\) are specific to the gas and must be calculated using critical temperature \(T_c\) and critical pressure \(P_c\) of benzene. The correction term involving \(\sqrt{T}\) accounts for attractive forces more effectively than the van der Waals equation. Using the given values of \(T_c\), \(P_c\), and a gas constant \(R = 0.08314 \frac{\text{L bar}}{\text{K mol}}\), substituting into the equations allows precise determination of these parameters. This equation extends the applicability of earlier models and is especially useful for real gas calculations.
vapor pressure calculation
The concept of vapor pressure is critical in understanding how substances transition between phases. Calculating vapor pressure involves determining the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. In our problem, vapor pressure calculations were performed using both the van der Waals and Redlich-Kwong equations.
Using these models involves substituting the calculated parameters \(a\) and \(b\) into their respective equations. For the van der Waals equation, solve for \(P\) by substituting into:
  • \[\left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT\]
The molar volume \(V_m\) can be derived from densit \(\rho\) using \(V_m = \frac{1}{\rho}\). Similarly, use the Redlich-Kwong equation to calculate pressure by solving the equation for \(P\) with the same approach.
These calculations provide a theoretical vapor pressure at each temperature which can be compared to experimental values. Calculating vapor pressure is crucial for predicting phase behavior, distillation, and in the chemical industry, as it affects product specifications.
thermodynamic parameters
Thermodynamic parameters such as temperature, pressure, and volume help describe the energetic state of a system. In our scenario, parameters \(a\) and \(b\) for both the van der Waals and Redlich-Kwong equations are essential to define interactions between molecules.
  • \(a\) represents the measure of the average attraction between particles.
  • \(b\) is the measure of the volume occupied by each particle, thus accounting for the finite size of molecules.
These parameters are obtained by establishing relationships with critical properties \(T_c\) and \(P_c\) of the substance, using empirical formulas provided in the stepped solution.
Understanding these parameters allows scientists to predict excess free energy, phase behaviors, and the conditions under which a gas shows non-ideal behavior. These calculations provide insights into molecular interactions in gases, supporting more refined predictions in applied thermodynamics.
experimental data comparison
Comparing theoretical calculations with experimental data serves as a validation exercise. It checks the accuracy of used equations under specified conditions. For this exercise, vapor pressures calculated using both the van der Waals and Redlich-Kwong equations are compared against experimental values. To assess the models:
  • First, calculate the deviation for each model from the experimental data by subtracting calculated from observed values.
  • Check which model predicts closer to the real value for different temperatures.
This comparison shows which equation is more suited to describing real gas behavior at given conditions. Radar plots or standard deviation metrics could be utilized to summarize precision across multiple temperatures.
Typically, the Redlich-Kwong equation provides a closer approximation because it better accounts for the intermolecular forces. However, each equation has its own applicable pressure and temperature range, and this exercise illustrates the importance of choosing the correct model for predicting gas behavior.

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Most popular questions from this chapter

Use the following data for methanol at one atm to plot \(\bar{G}\) versus \(T\) around the normal boiling point \((337.668 \mathrm{K}) .\) What is the value of \(\Delta_{\mathrm{vap}} \bar{H} ?\) $$\begin{array}{lcc}T / \mathrm{K} & \bar{H} / \mathrm{kJ} \cdot \mathrm{mol}^{-1} & \bar{S} / \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \\\\\hline 240 & 4.7183 & 112.259 \\\280 & 7.7071 & 123.870 \\\300 & 9.3082 & 129.375 \\\320 & 10.9933 & 134.756\\\330 & 11.8671 & 137.412 \\\337.668 & 12.5509 & 139.437 \\\337.668 & 47.8100 & 243.856 \\\350 & 48.5113 & 245.937 \\\360 &49.0631 & 247.492 \\\380 & 50.1458 & 250.419 \\\400 & 51.2257 & 253.189\end{array}$$

In this problem, we will show that the vapor pressure of a droplet is not the same as the vapor pressure of a relatively large body of liquid. Consider a spherical droplet of liquid of radius \(r\) in equilibrium with a vapor at a pressure \(P,\) and a flat surface of the same liquid in equilibrium with a vapor at a pressure \(P_{0} .\) Show that the change in Gibbs energy for the isothermal transfer of \(d n\) moles of the liquid from the flat surface to the droplet is \(d G=d n R T \ln \frac{P}{P_{0}}\) This change in Gibbs energy is due to the change in surface energy of the droplet (the change in surface energy of the large, flat surface is negligible). Show that \(d n R T \ln \frac{P}{P_{0}}=\gamma d A\) where \(\gamma\) is the surface tension of the liquid and \(d A\) is the change in the surface area of a droplet. Assuming the droplet is spherical, show that \(\begin{aligned} d n &=\frac{4 \pi r^{2} d r}{\bar{V}^{1}} \\ d A &=8 \pi r d r \end{aligned}\) and finally that \(\ln \frac{P}{P_{0}}=\frac{2 \gamma \bar{V}^{1}}{r R T}\) Because the right side is positive, we see that the vapor pressure of a droplet is greater than that of a planar surface. What if \(r \rightarrow \infty ?\)

The pressures at the solid-liquid coexistence boundary of propane are given by the empirical equation $$ P=-718+2.38565 T^{1.283} $$ where \(P\) is in bars and \(T\) is in kelvins. Given that \(T_{\text {fus }}=85.46 \mathrm{~K}\) and \(\Delta_{\text {fus }} \bar{H}=3.53 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), calculate \(\Delta_{\text {fus }} \bar{V}\) at \(85.46 \mathrm{~K}\).

Figure \(23.15\) shows reduced pressure, \(P_{\mathrm{R}}\), plotted against reduced volume, \(\bar{V}_{\mathrm{R}}\), for the van der Waals equation at a reduced temperature, \(T_{\mathrm{R}}\), of \(0.85 .\) The so-called van der Waals loop apparent in the figure will occur for any reduced temperature less than unity and is a consequence of the simplified form of the van der Waals equation. It turns out that any analytic equation of state (one that can be written as a Maclaurin expansion in the reduced density, \(1 / \bar{V}_{\mathrm{R}}\) ) will give loops for subcritical temperatures \(\left(T_{\mathrm{R}}<1\right)\). The correct behavior as the pressure is increased is given by the path abdfg in Figure \(23.15 .\) The horizontal region bdf, not given by the van der Waals equation, represents the condensation of the gas to a liquid at a fixed pressure. We can draw the horizontal line (called a tie line) at the correct position by recognizing that the chemical potentials of the liquid and the vapor must be equal at the points \(\mathrm{b}\) and \(\mathrm{f}\). Using this requirement, Maxwell showed that the horizontal line representing condensation should be drawn such that the areas of the loops above and below the line must be equal. To prove Maxwell's equal- area construction rule, integrate \((\partial \mu / \partial P)_{T}=\bar{V}\) by parts along the path bcdef and use the fact that \(\mu^{1}\) (the value of \(\mu\) at point \(f)=\mu^{\mathrm{g}}\) (the value of \(\mu\) at point \(b\) ) to obtain $$ \begin{aligned} \mu^{1}-\mu^{g} &=P_{0}\left(\bar{V}^{1}-\bar{V}^{\mathrm{g}}\right)-\int_{\text {bedef }} P d \bar{V} \\ &=\int_{\mathrm{bcdef}}\left(P_{0}-P\right) d \bar{V} \end{aligned} $$ where \(P_{0}\) is the pressure corresponding to the tie line. Interpret this result.

The vapor pressure of benzene along the liquid-vapor coexistence curve can be accurately expressed by the empirical expression \(\ln (P / \mathrm{bar})=-\frac{10.655375}{x}+23.941912-22.388714 x\) \(+20.2085593 x^{2}-7.219556 x^{3}+4.84728(1-x)^{1.70}\) where \(x=T / T_{\mathrm{c}},\) and \(T_{\mathrm{c}}=561.75 \mathrm{K} .\) Use this formula to show that the normal boiling point of benzene is \(353.24 \mathrm{K}\). Use the above expression to calculate the standard boiling point of benzene.

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