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At standard temperature and pressure, the molar volumes of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are 22.06 and \(22.40 \mathrm{~L}\), respectively. (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) On cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling the gases to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are 2.02 and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

Short Answer

Expert verified
The molar volumes of both Cl₂ and NH₃ gases are nearly the same at standard temperature and pressure because, according to the ideal gas law, all gases have the same molar volume under the same conditions. Upon cooling to 160 K, the molar volumes decrease as the kinetic energy of gas molecules decreases. In the solid state, Cl₂ and NH₃ have molar volumes of 35.10 cm³/mol and 20.27 cm³/mol, which are not as similar as in the gaseous state, as the differences in their molecular weights, dipole moments, and molecular shapes play a more significant role in the solid state. We expect that the molar volumes in the liquid state would be closer to those in the solid state due to stronger interactions between molecules in the liquid state compared to the gaseous state.

Step by step solution

01

a) The reason for nearly the same molar volumes

Both \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) have nearly the same molar volumes despite their different molecular weights, dipole moments, and molecular shapes. The reason behind this similarity is because they are both gases at standard temperature and pressure, and according to the ideal gas law, all gases have the same molar volume under the same conditions. The ideal gas law is given by: \[PV=nRT\] 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. At standard temperature and pressure (STP), any gas will occupy a volume of 22.4 L per mole.
02

b) Molar volumes upon cooling to 160 K

Upon cooling the gases to 160 K, we expect that the molar volumes would decrease. The reason is that the kinetic energy of the gas molecules would decrease as we reduce the temperature, which makes the molecules move slower and occupy a relatively smaller volume.
03

c) Calculation of molar volumes in the solid state

We can calculate the molar volumes for crystalline Cl2 and NH3 at 160 K using the densities given: Density = \(\frac{Mass}{Volume}\) Molar Volume = \(\frac{Molar Mass}{Density}\) For Cl2: Molar mass of Cl2 = 2 x Molar mass of Cl = 2 x 35.45g/mol = 70.90 g/mol Molar volume of Cl2 = \(\frac{70.90 g/mol}{2.02 g/cm^3} = 35.10 cm^3/mol\) For NH3: Molar mass of NH3 = 14.01g/mol (N) + 3 x 1.008g/mol (H) = 17.030 g/mol Molar volume of NH3 = \(\frac{17.030 g/mol}{0.84 g/cm^3} = 20.27 cm^3/mol\)
04

d) Comparison of molar volumes in solid and gaseous states

The molar volumes of Cl2 and NH3 in the solid state are 35.10 cm^3/mol and 20.27 cm^3/mol, respectively, while in the gaseous state, they are 22.06 L/mol and 22.40 L/mol, respectively. We can see that the molar volumes in the solid state are not as similar as they are in the gaseous state. The differences in their molecular weights, dipole moments, and molecular shapes play a more significant role in the solid state than in the gaseous state, as the molecules are closer together in the solid state and interact more with each other.
05

e) Molar volumes in the liquid state

We expect that the molar volumes in the liquid state would be closer to those in the solid state than in the gaseous state. This is because the molecules are closer together in the liquid state compared to the gaseous state and have stronger interactions, similar to the solid state. The difference in molecular weights, dipole moments, and molecular shapes would play a more significant role in determining the molar volumes in the liquid state.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental concept used to describe the behavior of gases under various conditions. It states that the product of pressure (P) and volume (V) is proportional to the product of the number of moles (n) and the temperature (T) of a gas. This relationship is expressed by the equation: \[ PV = nRT \]where R is the ideal gas constant.This law is crucial for understanding why gases like \( \mathrm{Cl}_{2} \) and \( \mathrm{NH}_{3} \) have similar molar volumes at standard temperature and pressure (STP), despite having different molecular characteristics. At STP, one mole of any gas occupies approximately 22.4 liters, illustrating that differences in molecular weight or structure do not influence the behavior of gases under these conditions.
This uniformity is why we consider gases 'ideal' under STP, helping us predict and calculate gas behavior in various scientific and industrial applications.
Molecular Weight
Molecular weight, also known as molecular mass, is the sum of the atomic masses of all the atoms in a molecule. It is measured in atomic mass units (amu) or grams per mole (g/mol). For example, the molecular weight of \( \mathrm{Cl}_{2} \) is calculated by adding the weights of two chlorine atoms, each approximately 35.45 g/mol, resulting in 70.90 g/mol. Similarly, \( \mathrm{NH}_{3} \) is composed of one nitrogen (14.01 g/mol) and three hydrogen atoms (1.008 g/mol each), giving a total molecular weight of 17.03 g/mol.Understanding molecular weight is essential for calculating molar volumes and densities, particularly in solid and liquid states. It influences how densely packed molecules are in a given volume, affecting how substances behave when transitioning from gas to liquid or solid form. Moreover, differences in molecular weight and structure can lead to varying interactions between molecules, particularly in condensed phases.
Density Calculation
Density is a measure of how much mass is contained in a given volume. It is crucial in determining how substances will behave in various states. The formula for density is:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]To calculate the molar volume of a substance from its density, you can rearrange this to:\[ \text{Molar Volume} = \frac{\text{Molar Mass}}{\text{Density}} \]This calculation is demonstrated for crystalline \( \mathrm{Cl}_{2} \) and \( \mathrm{NH}_{3} \) at 160 K. We find the molar volume by dividing the molecular weight by the given density.
  • For \( \mathrm{Cl}_{2} \): \( Molar\ Volume = \frac{70.90 \ g/mol}{2.02 \ g/cm^3} = 35.10 \ cm^3/mol \)
  • For \( \mathrm{NH}_{3} \): \( Molar\ Volume = \frac{17.03 \ g/mol}{0.84 \ g/cm^3} = 20.27 \ cm^3/mol \)
These calculations highlight how higher density leads to a smaller molar volume, which is consistent with more closely packed molecules in solids than in gases.
Crystalline Solids
Crystalline solids are materials whose constituents, such as atoms, ions, or molecules, are arranged in an ordered pattern. This rigid structure contributes to the distinct properties of solids compared to liquids and gases. When gases like \( \mathrm{Cl}_{2} \) and \( \mathrm{NH}_{3} \) cool to form crystalline solids at 160 K, their molecules come closer together, reflecting a significant decrease in molar volume. The well-organized lattice structure in crystalline solids means molecules have less space between them, increasing density compared to the gaseous state. Consequently, despite \( \mathrm{Cl}_{2} \) and \( \mathrm{NH}_{3} \) having similar molar volumes as gases, their solid-state volumes differ more due to these molecular interactions and arrangements.Understanding crystalline structures aids in predicting the physical properties of materials, including hardness, melting point, and solubility. It also helps in technological applications such as semiconductors and materials science.

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Most popular questions from this chapter

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is pressurized into liquid and stored in cylinders to be used as a fuel. The normal boiling point of propane is listed as \(-42^{\circ} \mathrm{C}\). (a) When converting propane into liquid at room temperature of \(25^{\circ} \mathrm{C},\) would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid propane is in it? (b) Suppose the fuel tank leaks and a few liters of propane escape rapidly. What do you expect would happen to the temperature of the remaining liquid propane in the tank? Explain. (c) How much heat must be added to vaporize \(20 \mathrm{~g}\) of propane if its heat of vaporization is \(18.8 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this amount of propane occupy at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} ?\)

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, \(P_{1}\) and \(P_{2}\), and the absolute temperatures at which they were measured, \(T_{1}\) and \(T_{2}\) : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)\). Octane has a vapor pressure of \(1.86 \mathrm{kPa}\) at \(25^{\circ} \mathrm{C}\) and a vapor pressure of \(19.3 \mathrm{kPa}\) at \(75^{\circ} \mathrm{C}\). Use these data and the equation in part (a) to calculate the heat of vaporization of octane. \((\mathbf{c})\) By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at \(-30^{\circ} \mathrm{C}\).

Appendix \(\mathrm{B}\) lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B,vapor pressure versus temperature \(\left({ }^{\circ} \mathrm{C}\right) .\) From your plot, estimate the vapor pressure of water at body temperature, \(37^{\circ} \mathrm{C}\). (b) Explain the significance of the data point at \(101.3 \mathrm{kPa}, 100^{\circ} \mathrm{C} .(\mathbf{c}) \mathrm{A}\) city at an altitude of \(1525 \mathrm{~m}\) above sea level has a barometric pressure of \(84.3 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city? (d) A city at an altitude of \(150 \mathrm{~m}\) below sea level would have a barometric pressure of \(103.14 \mathrm{kPa}\). To what temperature would you have to heat water to boil it in this city?

List the three states of matter in order of (a) increasing molecular disorder and \((\mathbf{b})\) increasing intermolecular attraction. (c) Which state of matter is most easily compressed?

For a given substance, the liquid crystalline phase tends to be more viscous than the liquid phase. Why?

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