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Chapter 23: Problem 4

The vapor pressures of solid and liquid chlorine are given by \(\ln \left(P^{\mathrm{s}} / \mathrm{torr}\right)=24.320-\frac{3777 \mathrm{K}}{T}\) \(\ln \left(P^{1} / \text { torr }\right)=17.892-\frac{2669 \mathrm{K}}{T}\) where \(T\) is the absolute temperature. Calculate the temperature and pressure at the triple point of chlorine.

Short Answer

Expert verified
The triple point of chlorine is at approximately 169.22 K and 2.72 torr.

Step by step solution

01

Understanding the Problem

We have been given two equations for the vapor pressures of solid and liquid chlorine as functions of temperature. Our goal is to find the temperature and pressure at which these two forms coexist at equilibrium 鈥� the triple point.
02

Setting Equilibrium Condition

At the triple point, the vapor pressures of the solid and the liquid are equal. Therefore, we equate the given equations: \(\ln \left(P^s / \text{torr} \right) = \ln \left(P^l / \text{torr} \right)\)
03

Equating the Given Expressions

Substitute the expressions for \(\ln \left(P^s / \text{torr} \right)\) and \(\ln \left(P^l / \text{torr} \right)\) in the equilibrium condition: \[24.320 - \frac{3777 \, \text{K}}{T} = 17.892 - \frac{2669 \text{K}}{T}\]
04

Solving for Temperature

First, subtract 17.892 from both sides: \[24.320 - 17.892 = \frac{3777 \text{K}}{T} - \frac{2669 \text{K}}{T}\]This simplifies to: \[6.428 = \frac{1088 \text{K}}{T}\]Now solve for \(T\):\[T = \frac{1088 \text{K}}{6.428} \approx 169.22 \, \text{K}\]
05

Calculating Vapor Pressure at Triple Point

Now that we have the temperature \(T \approx 169.22 \, \text{K}\), substitute it into either vapor pressure equation to find \(P\). Using the vapor pressure for the solid, \[\ln \left(P^s / \text{torr} \right) = 24.320 - \frac{3777 \text{K}}{169.22 \, \text{K}} \approx 1.001\]To find \(P^s\):\[P^s / \text{torr} = e^{1.001} \approx 2.72 \, \text{torr}\]
06

Conclusion

At the triple point, chlorine has a temperature of approximately 169.22 K and a pressure of about 2.72 torr.

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Key Concepts

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

Vapor Pressure
Vapor pressure is a measure of a substance's tendency to evaporate. It represents the pressure exerted by a vapor in equilibrium with its liquid or solid form. When a material is at a specific temperature, some of its molecules have enough energy to escape from the liquid or solid phase into the vapor phase.

As the temperature increases, these higher energy molecules become more numerous, increasing the vapor pressure. This concept is vital in understanding phase transitions, such as when a solid turns into a gas. Vapor pressure plays a critical role in determining how a substance behaves in various environmental conditions.

In our exercise, the formulas given represent the vapor pressures of solid and liquid chlorine as dependent on temperature. By knowing how these pressures compare at different temperatures, we can deduce critical points like the triple point, where all three phases (solid, liquid, and vapor) coexist in equilibrium.
Solid and Liquid Equilibrium
Solid and liquid equilibrium refers to the point where the solid and liquid phases of a substance exist simultaneously. At this balance, the molecules leaving the solid-phase equals those entering from the liquid, maintaining a stable state. This is a critical point for understanding phase changes, such as melting and freezing.

For chlorine, the expressions given relate the logarithm of the vapor pressure of its solid and liquid forms to temperature. The triple point is a special case of solid and liquid equilibrium where both phases not only coexist but also exist with the vapor phase as well. It's a unique intersection of temperatures and pressures.
  • At equilibrium at a phase change, both phases equally share the energy, keeping the temperature constant.
  • The equilibrium conditions imply the same vapor pressures for coexisting phases, which is the basis for calculating the triple point.
This balancing act is not just theoretical but has practical applications in industrial and scientific scenarios.
Chlorine Phase Transition
Phase transitions occur when a substance changes from one state of matter to another. For chlorine, this involves reversing from solid to liquid, liquid to gas, or directly from solid to gas. Understanding the exact conditions鈥攖emperature and pressure鈥攁t which these transitions occur is crucial for practical applications.

Chlorine undergoes these transitions based on its chemical and physical properties, dictated by its unique molecular structure. The exercise demonstrates how to calculate the triple point鈥攁 specific type of phase transition point鈥攚here solid, liquid, and gas phases coexist.
  • Knowing the triple point helps in precise determination of phase transitions, which is critical for designing systems that manipulate chlorine safely.
  • The triple point of chlorine is significant in calibrating thermometers and defining standard temperatures and pressures.
Understanding the phase transition of chlorine is essential for chemical engineers and chemists who work with this element in various applications.

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Most popular questions from this chapter

The vapor pressure of methanol along the entire liquid-vapor coexistence curve can be expressed very accurately by the empirical equation \(\ln (P / \mathrm{bar})=-\frac{10.752849}{x}+16.758207-3.603425 x\) \(+4.373232 x^{2}-2.381377 x^{3}+4.572199(1-x)^{1.70}\) where \(x=T / T_{\mathrm{c}},\) and \(T_{\mathrm{c}}=512.60 \mathrm{K} .\) Use this formula to show that the normal boiling point of methanol is \(337.67 \mathrm{K}\).

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.

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.

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 molar enthalpy of vaporization of water is \(40.65 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) at its normal boiling point. Use the Clausius-Clapeyron equation to calculate the vapor pressure of water at \(110^{\circ} \mathrm{C}\). The experimental value is 1075 torr.
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