/*! 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 37 The vapor pressures of 1-chlorot... [FREE SOLUTION] | 91Ó°ÊÓ

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The vapor pressures of 1-chlorotetradecane at several temperatures are tabulated here. $$\begin{array}{|c|c|c|c|c|c|c|} \hline T\left(^{\circ} \mathrm{C}\right) & 98.5 & 131.8 & 148.2 & 166.2 & 199.8 & 215.5 \\\\\hline p^{*}(\mathrm{mm} \mathrm{Hg}) & 1 & 5 & 10 & 20 & 60 & 100 \\\\\hline\end{array}$$ (a) Use two-point linear interpolation to estimate the value of \(p^{*}\) at \(T=185^{\circ} \mathrm{C}\). (b) Suppose you only know the data at \(98.5^{\circ} \mathrm{C}\) and \(215.5^{\circ} \mathrm{C}\). Use two-point linear interpolation to estimate the vapor pressure at \(148.2^{\circ} \mathrm{C}\). Assume the measured vapor pressure in the table is the true value, and calculate the percentage error in your interpolated value. Why would you expect the error associated with the estimate in Part (a) to be significantly less than that of Part (b)?

Short Answer

Expert verified
For part (a), the estimated value of vapor pressure at 185 °C is 43.6 mmHg using two-point linear interpolation. For part (b), the estimated value of vapor pressure at 148.2 °C is 28.8 mmHg with a percentage error of -188%.

Step by step solution

01

Understand Two-Point Linear Interpolation

The two-point linear interpolation formula is \(p^{*} = p_1 + ((T - T_1) / (T_2 - T_1)) * (p_2 - p_1)\), where \(p^{*}\) is the estimated pressure at temperature \(T\), \(T_1\) and \(T_2\) are the given temperatures that \(T\) lies between, and \(p_1\) and \(p_2\) are the corresponding pressures to temperatures \(T_1\) and \(T_2\).
02

Apply Interpolation for Part (a)

To estimate the value of \(p^{*}\) at \(T=185^{\circ} \mathrm{C}\), we use the given temperatures and pressures at \(T=166.2^{\circ} \mathrm{C}\) (\(p = 20\)) and \(T=199.8^{\circ} \mathrm{C}\) (\(p = 60\)). Inserting these values into the interpolation formula gives us: \[p^{*} = 20 + ((185 - 166.2) / (199.8 - 166.2)) * (60 - 20)\].
03

Calculate the Result for Part (a)

By simplifying the above equation, we get the estimate of \(p^{*}\) at \(T=185^{\circ} \mathrm{C}\), which should be around \(43.6\).
04

Apply Interpolation for Part (b)

To estimate the vapor pressure at \(148.2^{\circ} \mathrm{C}\), we use the given temperatures and pressures at \(T=98.5^{\circ} \mathrm{C}\) (\(p = 1\)) and \(T=215.5^{\circ} \mathrm{C}\) (\(p = 100\)). We insert these values into the interpolation formula: \[p^{*} = 1 + ((148.2 - 98.5) / (215.5 - 98.5)) * (100 - 1)\].
05

Calculate the Result for Part (b)

By simplifying the above equation, we get the estimate of \(p^{*}\) at \(T=148.2^{\circ} \mathrm{C}\), which should be around \(28.8\).
06

Calculate the Percentage Error

The percentage error is calculated as \[Error = ((p_{actual} - p_{estimated}) / p_{actual}) * 100\%.\] Here, \(p_{actual} = 10\), hence, the percentage error is \[((10 - 28.8) / 10) * 100 = -188\%.\]
07

Discuss the Error Differences

The error in part (a) should be less than in part (b) due to the selection of points for interpolation. In part (a), the selected points are closer to the point of interest, making the linear approximation more accurate. In part (b), the points are farther apart, increasing the chance for error due to nonlinear behavior between the points.

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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 is a critical concept in thermodynamics, particularly when predicting the boiling point of substances or understanding their phase behavior under different conditions. The vapor pressure of a substance is the pressure exerted by its vapor when the vapor and the liquid (or solid) phase are in equilibrium at a given temperature.When exact vapor pressure values are not available for the specific temperatures under study, methods such as two-point linear interpolation can provide estimations. This method assumes a linear relationship between adjacent data points and is particularly useful for predicting values within a range bounded by known points. Taking two temperature-pressure pairs from a dataset, linear interpolation uses the formula \( p^{*} = p_1 + ((T - T_1) / (T_2 - T_1)) \times (p_2 - p_1) \) to estimate the pressure (\( p^{*} \) at an intermediate temperature \( T \) lying between \( T_1 \) and \( T_2 \) on a linear scale. The resulting figure reflects the vapor pressure at the desired temperature based on the trend from the given data.It is important to note that the accuracy of this estimation relies heavily on the assumption that the changes in vapor pressure occur linearly between the given data points. This is often but not always, the case, as real-world thermodynamic behavior can exhibit pronounced non-linear characteristics, especially over greater temperature ranges.
Percentage Error Calculation
The calculation of percentage error is essential in assessing the accuracy of an estimated value compared to the actual, or true, value. The percentage error quantifies the deviation of the estimated value from the actual value and is expressed as a percentage. This measure is invaluable for validating experimental results and ensuring estimates fall within acceptable ranges.For vapor pressure estimation, when an interpolated value (\( p_{estimated} \) is determined, it can be compared to the actual vapor pressure (\( p_{actual} \) from the data using the formula \( \text{Error} = ((p_{actual} - p_{estimated}) / p_{actual}) \times 100\% \). A positive result indicates an underestimation, while a negative value indicates an overestimation.Understanding and calculating the percentage error aids students in critically evaluating their approaches and the reliability of their interpolated values. The smaller the percentage error, the closer the estimation is to the true value, which implies a higher level of accuracy in the estimation method used.
Interpolation Accuracy
Interpolation accuracy refers to how closely an interpolated value approximates the true, unknown value between known data points. When using two-point linear interpolation, the selection of points significantly influences the accuracy of the estimate.For example, when temperature points are closely spaced and encompass the temperature of interest, the linear approximation may be very accurate, as is postulated for part (a) of the exercise. However, if the points are far apart, as in part (b), the linearity assumption may not hold well and can lead to larger errors. This is because thermodynamic properties like vapor pressure tend not to change linearly over a wide range of temperatures, and interpolating over a large interval disregards any potential curvature in the true relationship between temperature and pressure.Incorporating closer data points around the temperature of interest typically yields a more accurate result and decreases the likelihood of significant error. However, when only extreme values are available, the introduction of midpoint data or using higher-order interpolation methods may improve the accuracy of the estimation. Interpolation accuracy is critical in all scientific and engineering fields; thus, a deeper understanding of when and how to apply linear interpolation is paramount for students.

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Most popular questions from this chapter

The following table is a summary of data taken on the growth of yeast cells in a bioreactor: $$\begin{array}{|c|c|}\hline \text { Time, } t(\mathrm{h}) & \text { Yeast Concentration, } X(\mathrm{g} / \mathrm{L}) \\\\\hline 0 & 0.010 \\\\\hline 4 & 0.048 \\\\\hline 8 & 0.152 \\\\\hline 12 & 0.733 \\\\\hline 16 & 2.457 \\ \hline\end{array}$$ The data can be fit with the function \(X=X_{0} \exp (\mu t)\) where \(X\) is the concentration of cells at any time \(t, X_{0}\) is the starting concentration of cells, and \(\mu\) is the specific growth rate. (a) Based on the data in the table, what are the units of the specific growth rate? (b) Give two ways to plot the data so as to obtain a straight line. Each of your responses should be of the form "plot ______ Versus ________ on _______ axes." (c) Plot the data in one of the ways suggested in Part (b) and determine \(\mu\) from the plot. (d) How much time is required for the yeast population to double?

The following reactions take place in a batch reactor: \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) (desired product) \(\mathrm{B}+\mathrm{C} \rightarrow \mathrm{D}\) (hazardous product) As the reaction proceeds, D builds up in the reactor and could cause an explosion if its concentration exceeds 15 mol/L. To ensure the safety of the plant personnel, the reaction is quenched (e.g., by cooling the reactor contents to a low temperature) and the products are extracted when the concentration of \(D\) reaches \(10 \mathrm{mol} / \mathrm{L}\). The concentration of \(C\) is measured in real-time, and samples are periodically taken and analyzed to determine the concentration of D. The data are shown below: $$\begin{array}{|c|c|}\hline C_{\mathrm{C}}(\mathrm{mol} / \mathrm{L}) & C_{\mathrm{D}}(\mathrm{mol} / \mathrm{L}) \\ \hline 2.8 & 1.4 \\\\\hline 10 & 2.27 \\\\\hline 20 & 2.95 \\\\\hline 40 & 3.84 \\\\\hline 70 & 4.74 \\\\\hline 110 & 5.63 \\ \hline 160 & 6.49 \\\\\hline 220 & 7.32 \\\\\hline\end{array}$$ (a) What would be the general form of an expression for \(C_{\mathrm{D}}\) as a function of \(C_{\mathrm{C}} ?\) (b) Derive the expression. (c) At what concentration of \(C\) is the reactor stopped? (d) Someone proposed not stopping the reaction until \(C_{\mathrm{D}}=13 \mathrm{mol} / \mathrm{L},\) and someone else strongly objected. What would be the major arguments for and against that proposal?

The following empirical equation correlates the values of variables in a system in which solid particles are suspended in a flowing gas: $$\frac{k_{g} d_{p} y}{D}=2.00+0.600\left(\frac{\mu}{\rho D}\right)^{1 / 3}\left(\frac{d_{p} u \rho}{\mu}\right)^{1 / 2}$$ Both \((\mu / \rho D)\) and \(\left(d_{p} u \rho / \mu\right)\) are dimensionless groups; \(k_{g}\) is a coefficient that expresses the rate at which a particular species transfers from the gas to the solid particles; and the coefficients 2.00 and 0.600 are dimensionless constants obtained by fitting experimental data covering a wide range of values of the equation variables. The value of \(k_{g}\) is needed to design a catalytic reactor. since this coefficient is difficult to determine directly, values of the other variables are measured or estimated and \(k_{g}\) is calculated from the given correlation. The variable values are as follows: $$\begin{aligned}d_{p} &=5.00 \mathrm{mm} \\\y &=0.100 \quad(\text { dimensionless }) \\\D &=0.100 \mathrm{cm}^{2} / \mathrm{s} \\\\\mu &=1.00 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2} \\\\\rho &=1.00 \times 10^{-3} \mathrm{g} / \mathrm{cm}^{3} \\\u &=10.0 \mathrm{m} / \mathrm{s}\end{aligned}$$ (a) What is the estimated value of \(k_{g} ?\) (Give its value and units.) (b) Why might the true value of \(k_{g}\) in the reactor be significantly different from the value estimated in Part (a)? (Give several possible reasons.) (c) Create a spreadsheet in which up to five sets of values of the given variables ( \(d_{p}\) through \(u\) ) are entered in columns and the corresponding values of \(k_{g}\) are calculated. Test your program using the following variable sets: (i) the values given above; (ii) as above, only double the particle diameter \(d_{p}\) (making it \(10.00 \mathrm{mm}\) ); (iii) as above, only double the diffusivity \(D ;\) (iv) as above, only double the viscosity \(\mu ;(\mathrm{v})\) as above, only double the velocity \(u\). Report all five calculated values of \(k_{g}\).

A seed crystal of diameter \(D\) (mm) is placed in a solution of dissolved salt, and new crystals are observed to nucleate (form) at a constant rate \(r\) (crystals/min). Experiments with seed crystals of different sizes show that the rate of nucleation varies with the seed crystal diameter as \(r(\text { crystals/min })=200 D-10 D^{2} \quad(D \text { in } \mathrm{mm})\) (a) What are the units of the constants 200 and \(10 ?\) (Assume the given equation is valid and therefore dimensionally homogeneous.) (b) Calculate the crystal nucleation rate in crystals/s corresponding to a crystal diameter of 0.050 inch. (c) Derive a formula for \(r\) (crystals/s) in terms of \(D\) (inches). (See Example \(2.6-1 .\) ) Check the formula using the result of Part (b). (d) The given equation is empirical; that is, instead of being developed from first principles, it was obtained simply by fitting an equation to experimental data. In the experiment, seed crystals of known size were immersed in a well-mixed supersaturated solution. After a fixed run time, agitation was ceased and the crystals formed during the experiment were allowed to settle to the bottom of the apparatus, where they could be counted. Explain what it is about the equation that gives away its empirical nature. (Hint: Consider what the equation predicts as \(D\) continues to increase.)

A concentration \(C(\mathrm{mol} / \mathrm{L})\) varies with time (min) according to the equation \(C=3.00 \exp (-2.00 t)\) (a) What are the implicit units of 3.00 and 2.00? (b) Suppose the concentration is measured at \(t=0\) and \(t=1\) min. Use two- point linear interpolation or extrapolation to estimate \(C(t=0.6 \mathrm{min})\) and \(t(C=0.10 \mathrm{mol} / \mathrm{L})\) from the measured values, and compare these results with the true values of these quantities. (c) Sketch a curve of \(C\) versus \(t,\) and show graphically the points you determined in Part (b).

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