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Chapter 23: Problem 34

The vapor pressures of solid and liquid hydrogen iodide can be expressed empirically as \(\ln \left(P^{\mathrm{s}} / \mathrm{torr}\right)=-\frac{2906.2 \mathrm{K}}{T}+19.020\) and \(\ln \left(P^{\prime} / \text { torr }\right)=-\frac{2595.7 \mathrm{K}}{T}+17.572\) Calculate the ratio of the slopes of the solid-gas curve and the liquid-gas curve at the triple point.

Short Answer

Expert verified
The ratio of the slopes is approximately 1.12.

Step by step solution

01

Identify the Equation for Slope

The slopes of the graphs of these equations can be identified as the coefficients of the term \(-\frac{B}{T}\). For solid, the slope \(m_s = -2906.2\), and for liquid, the slope \(m_l = -2595.7\).
02

Calculate the Ratio of Slopes

To find the ratio of the slopes of the solid-gas and liquid-gas curves, divide \(m_s\) by \(m_l\). That gives us:\[\text{Ratio} = \frac{-2906.2}{-2595.7}\]
03

Evaluate the Expression

Simplify the expression to get the ratio:\[\text{Ratio} = \frac{2906.2}{2595.7} \approx 1.1197\]

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Key Concepts

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

Vapor Pressure
Vapor pressure refers to the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a particular temperature. It is a unique property of each substance and varies based on temperature.
Understanding vapor pressure is crucial in predicting phase changes. When the vapor pressure of a liquid equals the external pressure, boiling occurs, leading to a phase transition from liquid to gas. Similarly, solid to gas transitions (sublimation) and gas to solid (deposition) depend on vapor pressure.
In the context of hydrogen iodide, the vapor pressure expressions provided for both its solid and liquid forms can be used to predict at which temperature each phase will transition to a gaseous state. This means by observing the vapor pressure, one can predict how and when phase transitions occur in hydrogen iodide.
Triple Point
The triple point of a substance is a unique set of conditions where all three phases鈥攕olid, liquid, and gas鈥攃an exist in equilibrium. This is a distinctive temperature and pressure characteristic for each substance.
At the triple point, specific conditions allow all phases to interchange, and this serves as a fundamental reference in thermodynamics. For hydrogen iodide, understanding its triple point is key to examining the behavior of the substance as it approaches multiple phase transitions at once. By determining the equilibrium between phases, scientists can predict how the substance might behave in various scenarios.
Phase Transition
A phase transition occurs when a substance changes from one state of matter to another, such as from solid to liquid (melting), liquid to gas (vaporization), or gas to solid (deposition). The transition between phases is governed by changes in temperature and pressure.
Hydrogen iodide, like other substances, undergoes phase transitions as described by its specific vapor pressure equations. By calculating the ratio of slopes in the phase diagrams of solid-gas and liquid-gas, one can determine the comparative ease or resistance in achieving these transitions at the triple point. The more rapid the change in slope, the easier the substance undergoes a phase transition under those conditions.
Hydrogen Iodide
Hydrogen iodide (HI) is a diatomic molecule consisting of hydrogen and iodine. It has interesting thermodynamic properties due to its ability to exist in multiple phases under varying conditions.
HI's behavior under different thermal conditions can be studied using its phase diagrams. For example, the vapor pressure equations help explain the convective behavior at specific temperatures, showing how HI transitions from solid or liquid to gas. Understanding this behavior is vital for applications involving HI, as it impacts storage, handling, and usage in chemical reactions.
The provided vapor pressure equations for HI's solid and liquid states enable scientists and engineers to model and predict how HI will behave as environmental conditions change, ensuring safe and efficient use in practical settings.

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Most popular questions from this chapter

Fit the following vapor pressure data of ice to an equation of the form \(\ln P=-\frac{a}{T}+b \ln T+c T\) where \(T\) is temperature in kelvins. Use your result to determine the molar enthalpy of sublimation of ice at \(0^{\circ} \mathrm{C}\). $$\begin{array}{cccc}t /^{\circ} \mathrm{C} & P / \text { torr } & t /^{\circ} \mathrm{C} & P / \text { torr } \\\\\hline-10.0 & 1.950 & -4.8 &3.065 \\\\-9.6 & 2.021 & -4.4 & 3.171 \\\\-9.2 & 2.093 & -4.0 & 3.280 \\\\-8.8 & 2.168 & -3.6 & 3.393 \\\\-8.4 & 2.246 & -3.2 & 3.509 \\\\-8.0 &2.326 & -2.8 & 3.360 \\\\-7.6 & 2.408 & -2.4 & 3.753 \\\\-7.2 & 2.493 & -2.0 & 3.880 \\\\-6.8 & 2.581 & -1.6 & 4.012 \\\\-6.4 & 2.672 & -1.2 &4.147 \\\\-6.0 & 2.765 & -0.8 & 4.287 \\\\-5.6 & 2.862 & -0.4 & 4.431 \\\\-5.2 & 2.962 & 0.0 & 4.579\end{array}$$

The densities of the coexisting liquid and vapor phases of methanol from the triple point to the critical point are accurately given by the empirical expressions \(\frac{\rho^{1}}{\rho_{\mathrm{c}}}-1=2.51709(1-x)^{0.350}+2.466694(1-x)\) \(-3.066818\left(1-x^{2}\right)+1.325077\left(1-x^{3}\right)\) and \(\ln \frac{\rho^{\mathrm{g}}}{\rho_{\mathrm{c}}}=-10.619689 \frac{1-x}{x}-2.556682(1-x)^{0.350}\) \(+3.881454(1-x)+4.795568(1-x)^{2}\) where \(\rho_{\mathrm{c}}=8.40 \mathrm{mol} \cdot \mathrm{L}^{-1}\) and \(x=T / T_{\mathrm{c}},\) where \(T_{\mathrm{c}}=512.60 \mathrm{K} .\) Use these expressions to plot \(\rho^{\prime}\) and \(\rho^{8}\) against temperature, as in Figure \(23.7 .\) Now plot \(\left(\rho^{1}+\rho^{8}\right) / 2\) against \(T\). Show that this line intersects the \(\rho^{1}\) and \(\rho^{\mathrm{g}}\) curves at \(T=T_{\mathrm{c}}\).

When we refer to the equilibrium vapor pressure of a liquid, we tacitly assume that some of the liquid has evaporated into a vacuum and that equilibrium is then achieved. Suppose, however, that we are able by some means to exert an additional pressure on the surface of the liquid. One way to do this is to introduce an insoluble, inert gas into the space above the liquid. In this problem, we will investigate how the equilibrium vapor pressure of a liquid depends upon the total pressure exerted on it. Consider a liquid and a vapor in equilibrium with each other, so that \(\mu^{\prime}=\mu^{8}\). Show that \(\bar{V}^{1} d P^{1}=\bar{V}^{g} d P^{8}\) because the two phases are at the same temperature. Assuming that the vapor may be treated as an ideal gas and that \(\bar{V}^{1}\) does not vary appreciably with pressure, show that \(\ln \frac{P^{\S}\left(\text { at } P^{1}=P\right)}{P^{8}\left(\text { at } P^{1}=0\right)}=\frac{\bar{V}^{1} P^{1}}{R T}\) Use this equation to calculate the vapor pressure of water at a total pressure of 10.0 atm at \(25^{\circ} \mathrm{C} .\) Take \(P^{\mathrm{g}}\left(\text { at } P^{1}=0\right)=0.0313 \mathrm{atm}\).

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.

Determine the value of \(d T / d P\) for water at its normal boiling point of \(373.15 \mathrm{K}\) given that the molar enthalpy of vaporization is \(40.65 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), and the densities of the liquid and vapor are \(0.9584 \mathrm{g} \cdot \mathrm{mL}^{-1}\) and \(0.6010 \mathrm{g} \cdot \mathrm{L}^{-1}\), respectively. Estimate the boiling point of water at 2 atm.
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