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Chapter 23: Problem 33

The sublimation pressure of \(\mathrm{CO}_{2}\) at \(138.85 \mathrm{K}\) and \(158.75 \mathrm{K}\) is \(1.33 \times 10^{-3}\) bar and \(2.66 \times 10^{-2}\) bar, respectively. Estimate the molar enthalpy of sublimation of \(\mathrm{CO}_{2}\).

Short Answer

Expert verified
The molar enthalpy of sublimation of CO2 is approximately 27.62 kJ/mol.

Step by step solution

01

Understanding the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the change in pressure with the change in temperature for a phase transition. It is represented as \( \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{sub}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \), where \(\Delta H_{sub}\) is the molar enthalpy of sublimation, \(R\) is the universal gas constant \(8.314 \, \text{J/mol路K}\), and \(P_1, P_2\) and \(T_1, T_2\) are initial and final pressures and temperatures, respectively.
02

Substituting Known Values

We are given \(P_1 = 1.33 \times 10^{-3} \, \text{bar}\), \(P_2 = 2.66 \times 10^{-2} \, \text{bar}\), \(T_1 = 138.85 \, \text{K}\), and \(T_2 = 158.75 \, \text{K}\). Substitute these into the Clausius-Clapeyron equation to solve for \(\Delta H_{sub}\).
03

Calculating the Natural Logarithm

Calculate \( \ln \left( \frac{P_2}{P_1} \right) \) using \(P_1\) and \(P_2\). This becomes \( \ln \left( \frac{2.66 \times 10^{-2}}{1.33 \times 10^{-3}} \right) = \ln (20) \approx 2.9957\).
04

Calculating the Inverse Temperature Difference

Calculate \( \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \) using the given temperatures. First, calculate each part: \( \frac{1}{T_1} = \frac{1}{138.85} \approx 0.00720\), and \( \frac{1}{T_2} = \frac{1}{158.75} \approx 0.00630\). Then, find the difference: \(0.00720 - 0.00630 = 0.00090\; \frac{1}{\text{K}}\).
05

Solving for \(\Delta H_{sub}\)

Rearrange the Clausius-Clapeyron equation to solve for \(\Delta H_{sub}\): \(\Delta H_{sub} = \ln (20) \times \frac{8.314}{0.00090}\). Plug in the calculated values: \(\Delta H_{sub} = 2.9957 \times \frac{8.314}{0.00090} \approx 27620.39 \, \text{J/mol}\) or \(27.62 \, \text{kJ/mol}\).

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Key Concepts

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

Sublimation Pressure
Sublimation pressure is a crucial concept when studying phase changes, particularly when a substance transitions directly from a solid to a gas phase. It specifically refers to the pressure exerted when the sublimation phase transition occurs under equilibrium. This pressure is determined under specific temperature conditions.
In the context of the exercise, we observe sublimation pressure values at two distinct temperatures for carbon dioxide: 1.33 \(\times 10^{-3}\) bar at 138.85 K and 2.66 \(\times 10^{-2}\) bar at 158.75 K. These pressures represent the equilibrium point where the solid and vapor phases of carbon dioxide coexist.
Understanding sublimation pressure is important because it provides insight into the conditions under which a material will sublimate. This is particularly relevant in applications involving low temperatures and pressures, where substances can transition directly to a gas without passing through the liquid state.
Enthalpy of Sublimation
The enthalpy of sublimation, denoted by \(\Delta H_{sub}\), is the energy required to convert one mole of a substance from a solid to a gaseous state at constant pressure. This value is a measure of how much energy the molecules need to overcome intermolecular forces binding them in the solid phase.
To calculate the molar enthalpy of sublimation, the Clausius-Clapeyron equation is employed. This equation links pressure and temperature changes during the sublimation process, allowing us to estimate \(\Delta H_{sub}\) using given sublimation pressures and temperatures.
From the exercise, substituting the given values into the equation affords us a closer look at energy absorption in real-world substances like carbon dioxide. This calculated enthalpy value, 27.62 kJ/mol, indicates the amount of heat necessary for \(\mathrm{CO}_2\) to transform from solid directly to gas under specified conditions.
Carbon Dioxide
Carbon dioxide (\(\mathrm{CO}_2\)) is a colorless gas commonly known for its role in respiration, photosynthesis, and as a greenhouse gas in Earth's atmosphere. However, it also possesses interesting phase properties, which include sublimation.
At standard atmospheric conditions, solid carbon dioxide, or "dry ice," sublimates at much lower temperatures than water ice. This property is exploited in various applications, including refrigeration, fog effects in entertainment, and even fire extinguisher systems.
In the given problem, understanding the specific temperatures, pressures, and enthalpy values associated with carbon dioxide's sublimation sheds light on its chemical behavior. It allows us to harness its unique properties effectively across different industries.

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Most popular questions from this chapter

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 \(85.46 \mathrm{~K}\).

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.

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 \(r \rightarrow \infty ?\)

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.

Sketch the phase diagram for \(\mathrm{I}_{2}\) given the following data: triple point, \(113^{\circ} \mathrm{C}\) and 0.12atm; critical point, \(512^{\circ} \mathrm{C}\) and 116 atm; normal melting point, \(114^{\circ} \mathrm{C} ;\) normal boiling point, \(184^{\circ} \mathrm{C} ;\) and density of liquid \(>\) density of solid.
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