Estimate the boiling point of water in Leadville, Colorado, elevation 3170 m. To do this, use the barometric formula relating pressure and altitude: \(P=P_{0} \times 10^{-\mathrm{Mgh} / 2.303 \mathrm{RT}} \quad(\text { where } P=\text { pressure } \mathrm{in}\) atm; \(P_{0}=1 \mathrm{atm} ; g=\) acceleration due to gravity; molar mass of air, \(M=0.02896 \mathrm{kg} \mathrm{mol}^{-1} ; \quad R=\) \(8.3145 \mathrm{Jmol}^{-1} \mathrm{K}^{-1} ;\) and \(T\) is the Kelvin temperature). Assume the air temperature is \(10.0^{\circ} \mathrm{C}\) and that \(\Delta H_{\text {vap }}=41 \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
After solving these equations, the approximate boiling point of water in Leadville, Colorado, which is at an elevation of 3170 meters, can be found. The exact value depends on the exact results from each of the calculations.

Step by step solution

01

Calculate the Pressure at Altitude

Plug the given values into the barometric formula to find the pressure at an altitude of 3170 m. The formula is \(P=P_{0} \times 10^{-\mathrm{Mgh} / 2.303 \mathrm{RT}}\). Given that \(P_0=1 \mathrm{atm}\), \(M = 0.02896 \mathrm{kg/mol}\), \(g = 9.81 \mathrm{m/s}^2\), \(h = 3170 \mathrm{m}\), \(R = 8.3145 \mathrm{J/(mol∙K)}\), and \(T = 283.15 \mathrm{K}\) (this is the given air temperature of 10.0°C converted into Kelvin), the pressure at this altitude can be calculated.
02

Apply the Clausius–Clapeyron Relation

Once the pressure at the given altitude is known, it can be used along with the given value for the enthalpy of vaporization of water to estimate the boiling point of water at this altitude. The boiling point of water at 1 atm (standard pressure) is 100°C, which is 373.15 K in Kelvin. Assume this as the initial temperature (T1). The Clausius–Clapeyron equation states that \(ln(P_2/P_1) = -\Delta H_{vap}/R (1/T2 - 1/T1)\). Given that \(\Delta H_{vap} = 41 \mathrm{kJ/mol} = 41000 \mathrm{J/mol}\), \(P_1 = 1 \mathrm{atm}\), \(T_1 = 373.15 \mathrm{K}\), \(P_2\) equals the pressure calculated in step 1, and \(R = 8.3145 \mathrm{J/(mol∙K)}\), solve this equation for \(T_2\), which is the estimated boiling point in Kelvin. Convert this to Celsius for final answer.

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Key Concepts

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

Barometric Formula
The barometric formula is a key concept that relates the atmospheric pressure at different altitudes. Understanding this concept is significant when it comes to predicting how certain physical phenomena, such as the boiling point of water, will differ with elevation.

At sea level, the atmospheric pressure is generally considered to be 1 atm, but this decreases with increasing altitude. This relationship is precisely described by the barometric formula, which is mathematically represented as \(P=P_{0} \times 10^{-\frac{Mgh}{2.303RT}}\). In this equation, \(P\) is the pressure at altitude, \(P_{0}\) is the reference pressure at sea level (usually 1 atm), \(M\) is the molar mass of air, \(g\) is the acceleration due to gravity, \(h\) is the altitude in meters, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

The use of this formula allows us to calculate how pressure changes with altitude, which is essential for estimating boiling points and understanding meteorological patterns among other applications.
Clausius–Clapeyron Relation
Moving onto another crucial concept, the Clausius–Clapeyron relation describes how the pressure of a substance's vapor changes with temperature, particularly at the boiling point. It forms the base for understanding how the boiling point of water changes with atmospheric pressure, which directly relates to altitude changes.

The relation is usually presented as \( ln(\frac{P_2}{P_1}) = -\frac{\Delta H_{vap}}{R} (\frac{1}{T_2} - \frac{1}{T_1}) \), where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively, \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(R\) is the universal gas constant.

For a given enthalpy of vaporization and initial boiling point, this relation allows us to estimate the new boiling point of a liquid (like water) at a different pressure, such as that found at the altitude of Leadville, Colorado.
Enthalpy of Vaporization
Enthalpy of vaporization (\(\Delta H_{vap}\)) is a term that might seem daunting, but it simply refers to the amount of energy required to convert one mole of a liquid into vapor at a constant temperature and pressure. It is a key ingredient in the Clausius–Clapeyron relation and is paramount in calculating the changes in boiling points due to pressure variations.

The value of \(\Delta H_{vap}\) is unique for every substance; for water, it is about 41 kJ/mol. This amount of energy must be absorbed by water to transition from its liquid state to gas at its boiling point under normal atmospheric pressure (1 atm).

Understanding the concept of enthalpy of vaporization is essential, as it is an intrinsic property that affects how we can boil water at different altitudes, for example. It also has wider applications in areas such as chemical engineering, meteorology, and the culinary arts.
Atmospheric Pressure
Last but not least, atmospheric pressure is the force exerted by the weight of the air in the atmosphere of Earth. It is an everyday concept that affects many aspects of daily life, from weather forecasts to cooking. At sea level, atmospheric pressure is standardized at approximately 1 atm, but this decreases with altitude because there is less air above to exert pressure.

Atmospheric pressure is influenced by altitude, temperature, and local weather conditions. It is essential to consider when estimating the boiling point of liquids, as boiling occurs when the vapor pressure equals the external pressure. This means that at higher altitudes, where atmospheric pressure is lower, water will boil at a lower temperature.

Recognizing the impact of atmospheric pressure is vital for scientists and engineers who work with gases and liquids, and for outdoor enthusiasts, who may need to adjust their cooking methods when on mountains or high altitudes like those in Leadville, Colorado.

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Most popular questions from this chapter

In each of the following pairs, which would you expect to have the higher boiling point? (a) \(\mathrm{C}_{7} \mathrm{H}_{16}\) or \(\mathrm{C}_{10} \mathrm{H}_{22} ;\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O} ;\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

The normal boiling point of acetone, an important laboratory and industrial solvent, is \(56.2^{\circ} \mathrm{C}\) and its \(\Delta H_{\text {vap }}\) is \(25.5 \mathrm{kJmol}^{-1} .\) At what temperature does acetone have a vapor pressure of \(375 \mathrm{mmHg} ?\)

One of the following substances is a liquid at room temperature and the others are gaseous: \(\mathrm{CH}_{3} \mathrm{OH}\) \(\mathrm{C}_{3} \mathrm{H}_{8} ; \mathrm{N}_{2} ; \mathrm{N}_{2} \mathrm{O} .\) Which do you think is the liquid? Explain.

Germanium has a cubic unit cell with a side edge of \(565 \mathrm{pm} .\) The density of germanium is \(5.36 \mathrm{g} / \mathrm{cm}^{3}\) What is the crystal system adopted by germanium?

All solids contain defects or imperfections of structure or composition. Defects are important because they influence properties, such as mechanical strength. Two common types of defects are a missing ion in an otherwise perfect lattice, and the slipping of an ion from its normal site to a hole in the lattice. The holes discussed in this chapter are often called interstitial sites, since the holes are in fact interstices in the array of spheres. The two types of defects described here are called point de kcts because they occur within specific sites. In the 1930 s, two solidstate physicists, W. Schottky and J. Fraenkel, studied the two types of point defects: A Schottky defect corresponds to a missing ion in a lattice, while a Fraenkel defect corresponds to an ion that is displaced into an interstitial site. (a) An example of a Schottky defect is the absence of a \(\mathrm{Na}^{+}\) ion in the NaCl structure. The absence of a \(\mathrm{Na}^{+}\) ion means that a \(\mathrm{Cl}^{-}\) ion must also be absent to preserve electrical neutrality. If one NaCl unit is missing per unit cell, does the overall stoichiometry change, and what is the change in density? (b) An example of a Fraenkel defect is the movement of \(a \mathrm{Ag}^{+}\) ion to a tetrahedral interstitial site from its normal octahedral site in \(\mathrm{AgCl}\), which has a structure like \(\mathrm{NaCl}\). Does the overall stoichiometry of the compound change, and do you expect the density to change? (c) Titanium monoxide (TiO) has a sodium chloridelike structure. X-ray diffraction data show that the edge length of the unit cell is \(418 \mathrm{pm}\). The density of the crystal is \(4.92 \mathrm{g} / \mathrm{cm}^{3}\) Do the data indicate the presence of vacancies? If so, what type of vacancies?

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