/*! 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 16 Fit the following data for argon... [FREE SOLUTION] | 91Ó°ÊÓ

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Fit the following data for argon to a fifth-order polynomial in \(T\). Use your result to determine the critical temperature. \begin{tabular}{lccc} \(T / \mathrm{K}\) & \(\Delta_{\text {vap }} \bar{H} / \mathrm{J} \cdot \mathrm{mol}^{-1}\) & \(T / \mathrm{K}\) & \(\Delta_{\text {vap }} \bar{H} / \mathrm{J} \cdot \mathrm{mol}^{-1}\) \\ \hline \(83.80\) & \(6573.8\) & \(122.0\) & \(4928.7\) \\ \(86.0\) & \(6508.4\) & \(126.0\) & \(4665.0\) \\ \(90.0\) & \(6381.8\) & \(130.0\) & \(4367.7\) \\ \(94.0\) & \(6245.2\) & \(134.0\) & \(4024.7\) \\ \(98.0\) & \(6097.7\) & \(138.0\) & \(3618.8\) \\ \(102.0\) & \(5938.8\) & \(142.0\) & \(3118.2\) \\ \(106.0\) & \(5767.6\) & \(146.0\) & \(2436.3\) \\ \(110.0\) & \(5583.0\) & \(148.0\) & \(1944.5\) \\ \(114.0\) & \(5383.5\) & \(149.0\) & \(1610.2\) \\ \(118.0\) & \(5166.5\) & \(150.0\) & \(1131.5\) \end{tabular}

Short Answer

Expert verified
Fit a fifth-order polynomial to the data and solve the equation to find the critical temperature.

Step by step solution

01

Organize the Data

First, organize the temperature \(T\) and enthalpy \(\Delta_{\text{vap}} \bar{H}\) data into two separate lists for use in polynomial fitting. This can typically involve inputting the given data points into a suitable data structure like a spreadsheet or programming tool.
02

Fit the Polynomial

Use a software or a calculator that can perform polynomial regression, such as Python's NumPy or any advanced graphing calculator. Apply a fifth-order polynomial regression on the temperature (T) as the independent variable and \(\Delta_{\text{vap}} \bar{H}\) as the dependent variable to find the polynomial expression. This will provide coefficients for the polynomial equation \ y = a_5T^5 + a_4T^4 + a_3T^3 + a_2T^2 + a_1T + a_0 \.
03

Analyze the Polynomial

Once the polynomial is fitted, analyze the coefficients obtained. The polynomial will have the form: \( P(T) = a_5T^5 + a_4T^4 + a_3T^3 + a_2T^2 + a_1T + a_0 \). You need to identify the temperature value \(T\) at which the polynomial value of \(\Delta_{\text{vap}} \bar{H}\) becomes zero or has a specific critical behavior, indicating the critical temperature.
04

Determine the Critical Temperature

The critical temperature can be found by solving the polynomial equation for when it reaches zero or another critical point, which implies finding the roots of the polynomial obtained in the previous step. You may use computational algebra systems to numerically find the roots of the polynomial. The relevant temperature from the list of roots is identified as the critical temperature.

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Key Concepts

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

Critical Temperature
In the field of thermodynamics, the critical temperature of a substance is a key concept. It is defined as the highest temperature at which a substance can exist as a liquid, regardless of pressure. Above this temperature, the substance cannot transition into a liquid state and will remain gaseous. When analyzing datasets such as those for argon, determining the critical temperature involves recognizing when an important thermodynamic quantity, like the enthalpy of vaporization, approaches zero. This usually signifies that molecular interactions no longer support a liquid form. In this exercise, we fit a polynomial to temperature and enthalpy data to identify the critical temperature. Finding the critical temperature involves exploring the behavior of the polynomial we fitted to our data. By identifying the roots where the polynomial equals zero, we can find the critical temperature. This offers insight into the thermodynamic properties of argon at varying temperatures.
Enthalpy of Vaporization
The enthalpy of vaporization is an essential concept in understanding the phase transitions of substances like argon. This term describes the amount of energy required to convert one mole of a liquid into a gas at constant pressure. In thermodynamic studies, this is crucial when analyzing the heat energy exchange during the phase change process, such as when a substance evaporates. In our dataset, the enthalpy of vaporization ( Δ_{ ext{vap}} ar{H} ) has been measured at various temperatures for argon. A pattern emerges as the enthalpy values decrease with increasing temperature, reflecting the lessening need for energy input to facilitate phase change. For our regression model using polynomial fitting, ( Δ_{ ext{vap}} ar{H} ) becomes the dependent variable, helping to illustrate how much energy is retained or dissipated as temperature progresses toward the critical point.
Fifth-order Polynomial
Polynomial regression is a statistical technique used to fit a polynomial curve through a series of data points. In this exercise, we are concerned with fitting a fifth-order polynomial to data obtained from argon thermodynamic studies. A fifth-order polynomial takes the form \( P(T) = a_5T^5 + a_4T^4 + a_3T^3 + a_2T^2 + a_1T + a_0 \).This polynomial is useful for modeling the relationship between temperature and enthalpy of vaporization, providing a detailed representation that captures the nuanced changes in enthalpy as the temperature changes. The choice of a fifth-order polynomial means that we are allowing the relationship to be flexible enough to pick up any subtle curves or trends in the data. Polynomial regression effectively summarizes complex thermodynamic behaviors and allows us to derive essential insights, such as predicting the critical temperature. With polynomial coefficients in hand, computation tools can extrapolate important values like the critical temperature from the fitted curve, showcasing the practical application of higher-order polynomial models.
Argon Thermodynamics
When exploring argon in thermodynamic contexts, we delve into its behavior under various conditions, such as temperature changes. Argon is a noble gas, and its simple atomic structure makes it an excellent model for studying basic thermodynamic principles. Thermodynamic properties of argon, including its enthalpy of vaporization and critical temperature, provide insight into its phase behavior and energy requirements. In experiments, measurements and calculations like those conducted in this exercise allow us to understand how argon transitions between its gas and liquid phases. Using a polynomial regression to analyze temperature and enthalpy data, we can more precisely study argon's behavior. This exercise highlights how mathematical models explain real-world phenomena, such as the gradual decline of enthalpy as we approach the critical temperature, ultimately revealing the physical limits and properties of argon as a substance.

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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 \(\Delta_{\mathrm{vap}} \bar{H} ?\) $$\begin{array}{lcc}T / \mathrm{K} & \bar{H} / \mathrm{kJ} \cdot \mathrm{mol}^{-1} & \bar{S} / \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \\\\\hline 240 & 4.7183 & 112.259 \\\280 & 7.7071 & 123.870 \\\300 & 9.3082 & 129.375 \\\320 & 10.9933 & 134.756\\\330 & 11.8671 & 137.412 \\\337.668 & 12.5509 & 139.437 \\\337.668 & 47.8100 & 243.856 \\\350 & 48.5113 & 245.937 \\\360 &49.0631 & 247.492 \\\380 & 50.1458 & 250.419 \\\400 & 51.2257 & 253.189\end{array}$$

Consider the phase change C(graphite) \(\rightleftharpoons\) C(diamond)Given that \(\Delta_{\mathrm{f}} G^{\circ} / \mathrm{J} \cdot \mathrm{mol}^{-1}=1895+3.363 T\), calculate \(\Delta_{\mathrm{r}} H^{\circ}\) and \(\Delta_{\mathrm{r}} S^{\circ} .\) Calculate the pressure at which diamond and graphite are in equilibrium with each other at \(25^{\circ} \mathrm{C}\). Take the density of diamond and graphite to be \(3.51 \mathrm{~g} \cdot \mathrm{cm}^{-3}\) and \(2.25 \mathrm{~g} \cdot \mathrm{cm}^{-3}\), respectively. Assume that both diamond and graphite are incompressible.

Plot the following data for the densities of liquid and gaseous ethane in equilibrium with each other as a function of temperature, and determine the critical temperature of ethane. $$\begin{array}{cccccc} T / \mathrm{K} & \rho^{1} / \mathrm{mol} \cdot \mathrm{dm}^{-3} & \rho^{8} / \mathrm{mol} \cdot \mathrm{dm}^{-3} & T / \mathrm{K} & \rho^{1} / \mathrm{mol} \cdot \mathrm{dm}^{-3} & \rho^{8} / \mathrm{mol} \cdot \mathrm{dm}^{-3} \\ \hline 100.00 & 21.341 & 1.336 \times 10^{-3} & 283.15 & 12.458 & 2.067 \\ 140.00 & 19.857 & 0.03303 & 293.15 & 11.297 & 2.880 \\ 180.00 & 18.279 & 0.05413 & 298.15 & 10.499 & 3.502 \\ 220.00 & 16.499 & 0.2999 & 302.15 & 9.544 & 4.307 \\ 240.00 & 15.464 & 0.5799 & 304.15 & 8.737 & 5.030 \\ 260.00 & 14.261 & 1.051 & 304.65 & 8.387 & 5.328 \\ 270.00 & 13.549 & 1.401 & 305.15 & 7.830 & 5.866 \end{array}$$

The isothermal compressibility, \(\kappa_{T},\) is defined by \(\kappa_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\) Because \((\partial P / \partial V)_{T}=0\) at the critical point, \(\kappa_{T}\) diverges there. A question that has generated a great deal of experimental and theoretical research is the question of the manner in which \(\kappa_{T}\) diverges as \(T\) approaches \(T_{\mathrm{c}} .\) Does it diverge as \(\ln \left(T-T_{\mathrm{c}}\right)\) or perhaps \(\operatorname{as}\left(T-T_{\mathrm{c}}\right)^{-\gamma}\) where \(\gamma\) is some critical exponent? An early theory of the behavior of thermodynamic functions such as \(\kappa_{T}\) very near the critical point was proposed by van der Waals, who predicted that \(\kappa_{T}\) diverges as \(\left(T-T_{\mathrm{c}}\right)^{-1} .\) To see how van der Waals arrived at this prediction, we consider the (double) Taylor expansion of the pressure \(P(\bar{V}, T)\) about \(T_{\mathrm{c}}\) and \(V_{\mathrm{c}}\): \(P(\bar{V}, T)=P\left(\bar{V}_{\mathrm{c}}, T_{\mathrm{c}}\right)+\left(T-T_{\mathrm{c}}\right)\left(\frac{\partial P}{\partial T}\right)_{\mathrm{c}}+\frac{1}{2}\left(T-T_{\mathrm{c}}\right)^{2}\left(\frac{\partial^{2} P}{\partial T^{2}}\right)_{\mathrm{c}}\) \(+\left(T-T_{\mathrm{c}}\right)\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)\left(\frac{\partial^{2} P}{\partial V \partial T}\right)_{\mathrm{c}}+\frac{1}{6}\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{3}\left(\frac{\partial^{3} P}{\partial \bar{V}^{3}}\right)_{\mathrm{c}}+\cdots\) Why are there no terms in \(\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)\) or \(\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2} ?\) Write this Taylor series as \(P=P_{\mathrm{c}}+a\left(T-T_{\mathrm{c}}\right)+b\left(T-T_{\mathrm{c}}\right)^{2}+c\left(T-T_{\mathrm{c}}\right)\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)+d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{3}+\cdots\) Now show that \(\left(\frac{\partial P}{\partial \bar{V}}\right)_{T}=c\left(T-T_{\mathrm{c}}\right)+3 d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2}+\cdots \quad\left(\begin{array}{c}T \rightarrow T_{\mathrm{c}} \\ \bar{V} \rightarrow \bar{V}_{\mathrm{c}}\end{array}\right)\) and that \(\kappa_{T}=\frac{-1 / \bar{V}}{c\left(T-T_{\mathrm{c}}\right)+3 d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2}+\cdots}\) Now let \(\bar{V}=\bar{V}_{\mathrm{c}}\) to obtain \(\kappa_{T} \propto \frac{1}{T-T_{\mathrm{c}}} \quad T \rightarrow\left(T_{\mathrm{c}}\right)\) Accurate experimental measurements of \(\kappa_{T}\) as \(T \rightarrow T_{\mathrm{c}}\) suggest that \(\kappa_{T}\) diverges a little more strongly than \(\left(T-T_{\mathrm{c}}\right)^{-1}\). In particular, it is found that \(\kappa_{T} \rightarrow\left(T-T_{\mathrm{c}}\right)^{-\gamma}\) where \(\gamma=1.24\) Thus, the theory of van der Waals, although qualitatively correct, is not quantitatively correct.

Sketch the phase diagram for oxygen using the following data: triple point, \(54.3 \mathrm{~K}\) and \(1.14\) torr; critical point, \(154.6 \mathrm{~K}\) and 37828 torr; normal melting point, \(-218.4^{\circ} \mathrm{C} ;\) and normal boiling point, \(-182.9^{\circ} \mathrm{C}\). Does oxygen melt under an applied pressure as water does?

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