/*! 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 149 One method of preparing elementa... [FREE SOLUTION] | 91影视

91影视

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at 600.^ C followed by condensation of the mercury vapor. Given the heat of vaporization of mercury (296 J/g) and the vapor pressure of mercury at \(25.0^{\circ} \mathrm{C}\left(2.56 \times 10^{-3} \text {torr }\right),\) what is the vapor pressure of the condensed mercury at \(300 .^{\circ} \mathrm{C} ?\) How many atoms of mercury are present in the mercury vapor at \(300 .^{\circ} \mathrm{C}\) if the reaction is conducted in a closed 15.0 -L container?

Short Answer

Expert verified
The vapor pressure of the condensed mercury at 300掳C is 171.27 torr. There are 3.27 x 10虏虏 atoms of mercury in the vapor phase at 300掳C in a 15.0-L container.

Step by step solution

01

Convert temperatures to Kelvin

We need to work with temperatures in Kelvin for these calculations. To do this, add 273.15 to the Celsius temperature: Temperature 1 (T鈧): 25.0掳C + 273.15 = 298.15 K Temperature 2 (T鈧): 300.0掳C + 273.15 = 573.15 K
02

Use the Clausius-Clapeyron equation

The Clausius-Clapeyron equation relates the vapor pressure, temperature, and heat of vaporization of a substance: \( \ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_\text{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \) where \(P_1\) is the initial vapor pressure (2.56 x 10鈦宦 torr), \(P_2\) is the final vapor pressure, \( \Delta H_\text{vap}\) is the heat of vaporization (296 J/g), R is the ideal gas constant (8.314 J/mol K), T鈧 is the initial temperature (298.15 K), and T鈧 is the final temperature (573.15 K). We will solve for \(P_2\).
03

Calculate the vapor pressure at 300掳C

Rearrange the Clausius-Clapeyron equation to solve for \(P_2\): \( P_2 = P_1 \cdot \exp\left( \frac{-\Delta H_\text{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \right) \) Plug in the given values and calculate the vapor pressure at 300掳C: \( P_2 = (2.56 \times 10^{-3}\, \text{torr}) \exp\left( \frac{-(296\,\text{J/g})}{(8.314\, \text{J/mol K})} \left( \frac{1}{573.15\, \text{K}} - \frac{1}{298.15\, \text{K}} \right) \right) \) \( P_2 = 171.27\, \text{torr} \) The vapor pressure of the condensed mercury at 300掳C is 171.27 torr.
04

Use the ideal gas law

Now, we will use the ideal gas law to find the number of atoms in the vapor phase in a 15.0-L container: \(P * V = n * R * T\) where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. First, convert the pressure to atm and R = (0.0821 L atm/mol K): \(P_{\text{atm}} = \frac{171.27\, \text{torr}}{760\, \text{torr/atm}} = 0.2253\, \text{atm}\) Next, plug in the values, and solve for the number of moles (n): \(n = \frac{P * V}{R * T} = \frac{0.2253\, \text{atm} * 15.0\, \text{L}}{0.0821\, \text{L atm/mol K} * 573.15\, \text{K}}\) \(n = 0.0544\, \text{moles} \)
05

Determine the number of atoms

To find the number of atoms, multiply the number of moles by Avogadro's number (6.022 x 10虏鲁 atoms per mole): Number of Mercury atoms = 0.0544 moles * 6.022 x 10虏鲁 atoms/mol = 3.27 x 10虏虏 atoms There are 3.27 x 10虏虏 atoms of mercury in the vapor phase at 300掳C in a 15.0-L container.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is a vital formula in thermodynamics that describes the relationship between the vapor pressure and temperature of a substance. It illustrates how vapor pressure increases with temperature for a given substance. Specifically, it connects the change in vapor pressure due to a change in temperature to the substance's heat of vaporization.

Mathematically, the equation is expressed as:
\[ \ln\left(\frac{P_2}{P_1}\right) = \frac{-\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
where \(P_1\) and \(P_2\) are the initial and final vapor pressures, \(\Delta H_{\text{vap}}\) is the heat of vaporization, R is the ideal gas constant, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin. This formula can predict the vapor pressure at a new temperature if the vapor pressure is known at some initial condition, making it invaluable in many fields including meteorology, engineering, and material science.
Heat of Vaporization
The heat of vaporization, represented by \(\Delta H_{\text{vap}}\), is the amount of energy required to convert one gram of a liquid into vapor without a change in temperature. It represents the strength of intermolecular forces within the liquid; a higher heat of vaporization indicates stronger forces that must be overcome.

For instance, mercury has a heat of vaporization of 296 joules per gram, which is a measure of the energy needed to overcome the forces holding mercury atoms together in the liquid phase. This property is crucial in calculations involving phase changes, such as converting liquid mercury to gaseous mercury, and ties directly into the Clausius-Clapeyron equation for determining vapor pressure changes.
Ideal Gas Law
The ideal gas law is a cornerstone in the study of gases, connecting pressure, volume, temperature, and the amount of gas present. Given by the equation:
\[ P * V = n * R * T \]
it allows us to calculate any one of these variables if the others are known. Here, P is the pressure of the gas, V is the volume it occupies, n is the number of moles of gas, R is the universal gas constant (with a value of 8.314 J/mol路K in SI units or 0.0821 L路atm/mol路K for calculations involving liters and atmospheres), and T is the temperature in Kelvin.

In the context of mercury vapor pressure, once the vapor pressure at a certain temperature is determined using the Clausius-Clapeyron equation, the ideal gas law can then be used to find out how many moles of gaseous mercury are present in a certain volume at that temperature.
Avogadro's Number
Avogadro's number, approximately \(6.022 \times 10^{23}\), represents the quantity of particles, typically atoms or molecules, in one mole of a substance. It's a fundamental constant that provides a link between the microscopic world of atoms and the macroscopic world of grams and liters that we interact with daily.

Understanding Avogadro's number is essential when transitioning from the amount of a substance in moles to the actual number of atoms or molecules present. This conversion is particularly crucial when quantifying the number of mercury atoms present in the vapor phase after evaporation, as demonstrated in the mercury vapor pressure calculation.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. It is a specific form of pressure unique to each substance and varies with temperature. A higher temperature usually results in a higher vapor pressure, as more molecules have enough kinetic energy to escape from the liquid or solid phase into the gas phase.

By understanding the vapor pressure, we gain insights into a substance's volatility and how readily it will evaporate. For mercury, knowledge of its vapor pressure at various temperatures allows for safe handling and use in applications, such as in thermometers or fluorescent lamps, where mercury's vapor state is utilized.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the kinetic molecular theory to explain why a liquid gets cooler as it evaporates from an insulated container.

A certain form of lead has a cubic closest packed structure with an edge length of \(492 \mathrm{pm} .\) Calculate the value of the atomic radius and the density of lead.

Consider the following enthalpy changes: $$\begin{aligned} \mathrm{F}^{-}+\mathrm{HF} \longrightarrow \mathrm{FHF}^{-} & \Delta H=-155 \mathrm{kJ} / \mathrm{mol} \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{HF} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}--\mathrm{HF} & \Delta H=-46 \mathrm{kJ} / \mathrm{mol} \\\ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HOH}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}--\mathrm{HOH}(\text { in ice }) & \Delta H=-21 \mathrm{kJ} / \mathrm{mol} \end{aligned}$$ How do the strengths of hydrogen bonds vary with the electronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?

A plot of \(\ln \left(P_{\text {vap }}\right)\) versus \(1 / T(\mathrm{K})\) is linear with a negative slope. Why is this the case?

A common response to hearing that the temperature in New Mexico is \(105^{\circ} \mathrm{F}\) is, "It's not that bad; it's a dry heat," whereas at the same time the summers in Atlanta, Georgia, are characterized as "dreadful," even though the air temperature is typically lower. What role does humidity play in how our bodies regulate temperature?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.