Clausius-Clapeyron Equation and Solids: Step-by-Step Chemistry Guidance

Q1. Liquid sodium can be used as a heat transfer fluid. Its vapor pressure is 40.0 torr at 633°C and 400.0 torr at 823°C. What is the heat of vaporization (ΔHvap) in kJ/mole?

Background

Topic: Clausius-Clapeyron Equation

This question tests your ability to use the Clausius-Clapeyron equation to determine the enthalpy (heat) of vaporization from vapor pressure data at two temperatures.

Key Terms and Formulas

• Vapor pressure: The pressure exerted by a vapor in equilibrium with its liquid at a given temperature.

• ΔHvap: Molar heat of vaporization (energy required to vaporize one mole of liquid).

Key formula (Clausius-Clapeyron equation):

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

• $P_1$, $P_2$ = vapor pressures at temperatures $T_1$, $T_2$

• $T_1$, $T_2$ = temperatures in Kelvin

• $R$ = gas constant = $8.314$ J/mol·K

Step-by-Step Guidance

1. Convert both temperatures from Celsius to Kelvin by adding 273.15.

2. Write down the vapor pressures $P_1$ and $P_2$ and their corresponding temperatures $T_1$ and $T_2$ in Kelvin.

3. Plug the values into the Clausius-Clapeyron equation:

   $\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

4. Rearrange the equation to solve for $\Delta H_{\text{vap}}$:

   $\Delta H_{\text{vap}} = -R \cdot \frac{\ln\left(\frac{P_2}{P_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}$

Try solving on your own before revealing the answer!

Final Answer: 54.2 kJ/mol

After converting temperatures and plugging the values into the rearranged Clausius-Clapeyron equation, you find $\Delta H_{\text{vap}}$ is approximately 54.2 kJ/mol.

Q2. Calculate the highest cooking temperature (in K) in a pressure cooker that establishes a total pressure of 2.00 atm during cooking. The normal boiling point of water is 100.0°C. The enthalpy of vaporization (ΔHvap) of water is 40.65 kJ/mole.

Background

Topic: Clausius-Clapeyron Equation and Boiling Point

This question tests your ability to use the Clausius-Clapeyron equation to relate boiling point changes to pressure changes.

Key Terms and Formulas

• Boiling point: Temperature at which vapor pressure equals external pressure.

• Normal boiling point: Boiling point at 1 atm pressure.

• ΔHvap: Molar heat of vaporization.

Key formula:

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Step-by-Step Guidance

1. Identify $P_1$ (1 atm, normal boiling point) and $P_2$ (2 atm, pressure cooker).

2. Convert the normal boiling point (100.0°C) to Kelvin ($T_1$).

3. Set up the Clausius-Clapeyron equation to solve for $T_2$ (the new boiling point in Kelvin).

4. Rearrange the equation to isolate $T_2$:

   $\frac{1}{T_2} = \frac{1}{T_1} - \frac{R}{\Delta H_{\text{vap}}} \ln\left(\frac{P_2}{P_1}\right)$

Try solving on your own before revealing the answer!

Final Answer: 394 K

After substituting the values and solving for $T_2$, the highest cooking temperature is approximately 394 K.

Q3. Determine the vapor pressure (mmHg) of a liquid at 74.2°C if the vapor pressure of the liquid is 83.6 mmHg at 42.5°C. The molar heat of vaporization (ΔHvap) of this liquid is 63.5 kJ/mole.

Background

Topic: Clausius-Clapeyron Equation

This question tests your ability to use the Clausius-Clapeyron equation to predict vapor pressure at a new temperature.

Key Terms and Formulas

• Vapor pressure: Pressure exerted by a vapor in equilibrium with its liquid.

• ΔHvap: Molar heat of vaporization.

Key formula:

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Step-by-Step Guidance

1. Convert both temperatures from Celsius to Kelvin.

2. Write down $P_1$ (83.6 mmHg at $T_1$) and set up the equation to solve for $P_2$ at $T_2$.

3. Plug the values into the Clausius-Clapeyron equation and rearrange to solve for $P_2$:

   $P_2 = P_1 \cdot \exp\left(-\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right)$

4. Substitute the values for $ΔHvap$, $R$, $T_1$, and $T_2$ into the equation.

Try solving on your own before revealing the answer!

Final Answer: 495 mmHg

After substituting and calculating, the vapor pressure at 74.2°C is approximately 495 mmHg.

Q4. Estimate the molar heat of vaporization (ΔHvap, in kJ/mole) of a liquid whose vapor pressure doubles when the temperature is raised from 85.0°C to 95.0°C.

Background

Topic: Clausius-Clapeyron Equation

This question tests your ability to estimate the enthalpy of vaporization from the change in vapor pressure over a temperature interval.

Key Terms and Formulas

• Vapor pressure: Pressure exerted by a vapor in equilibrium with its liquid.

• ΔHvap: Molar heat of vaporization.

Key formula:

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Step-by-Step Guidance

1. Convert both temperatures from Celsius to Kelvin.

2. Since the vapor pressure doubles, set $P_2 = 2P_1$.

3. Plug these values into the Clausius-Clapeyron equation:

   $\ln(2) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

4. Rearrange to solve for $\Delta H_{\text{vap}}$:

   $\Delta H_{\text{vap}} = -R \cdot \frac{\ln(2)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}$

Try solving on your own before revealing the answer!

Final Answer: 44.2 kJ/mol

After substituting the values, the estimated molar heat of vaporization is about 44.2 kJ/mol.

Q5. Vanadium (molar mass = 50.94 g/mole) crystallizes in a body-centered cubic (bcc) lattice, and the length of the edge of a unit cell is 305 pm. What is the density (in g/cm³) of vanadium?

Background

Topic: Crystal Lattices and Density Calculations

This question tests your ability to calculate the density of a metal from its unit cell dimensions and lattice type.

Key Terms and Formulas

• Body-centered cubic (bcc): A type of crystal lattice with atoms at each corner and one atom in the center.

• Unit cell: The smallest repeating unit in a crystal lattice.

• Density: Mass per unit volume.

Key formulas:

Number of atoms per bcc unit cell: 2

Volume of unit cell: $V = a^3$

Density: $\text{density} = \frac{\text{mass of atoms in cell}}{\text{volume of cell}}$

Step-by-Step Guidance

1. Calculate the volume of the unit cell in cm³. Convert edge length from pm to cm using $1 \text{ cm} = 10^{10} \text{ pm}$.

2. Determine the number of atoms per unit cell (2 for bcc).

3. Calculate the mass of atoms in one unit cell using the molar mass and Avogadro's number.

4. Set up the density formula using the mass and volume calculated.

Try solving on your own before revealing the answer!

Final Answer: 6.11 g/cm³

After calculating the mass and volume, the density of vanadium is approximately 6.11 g/cm³.

Q6. Copper (molar mass = 63.55 g/mole) crystallizes in a face-centered cubic (fcc) lattice. If the density of copper is 8.96 g/cm³, what is the length (in pm) of the unit cell edge?

Background

Topic: Crystal Lattices and Unit Cell Dimensions

This question tests your ability to relate density, molar mass, and unit cell dimensions for a face-centered cubic lattice.

Key Terms and Formulas

• Face-centered cubic (fcc): A crystal lattice with atoms at each corner and at the center of each face.

• Unit cell: The smallest repeating unit in a crystal lattice.

Key formulas:

Number of atoms per fcc unit cell: 4

Volume of unit cell: $V = a^3$

Density: $\text{density} = \frac{\text{mass of atoms in cell}}{\text{volume of cell}}$

Step-by-Step Guidance

1. Calculate the mass of atoms in one unit cell using the molar mass and Avogadro's number.

2. Set up the density formula and solve for the volume of the unit cell.

3. Take the cube root of the volume to find the edge length $a$ in cm, then convert to pm using $1 \text{ cm} = 10^{10} \text{ pm}$.

Try solving on your own before revealing the answer!

Final Answer: 361 pm

After solving, the edge length of the copper unit cell is approximately 361 pm.

Q7. A face-centered cubic (fcc) cell contains 8 “X” atoms at the corners and 6 “Y” atoms at the faces. What is the empirical formula of the solid?

Background

Topic: Crystal Lattice Composition and Empirical Formula

This question tests your ability to determine the empirical formula from the arrangement of atoms in a unit cell.

Key Terms and Formulas

• Empirical formula: The simplest whole-number ratio of atoms in a compound.

• fcc lattice: Atoms at corners and faces are shared among adjacent unit cells.

Key concepts:

• Each corner atom is shared by 8 unit cells (contributes 1/8 per cell).

• Each face atom is shared by 2 unit cells (contributes 1/2 per cell).

Step-by-Step Guidance

1. Calculate the number of “X” atoms per unit cell: $8 \times \frac{1}{8}$.

2. Calculate the number of “Y” atoms per unit cell: $6 \times \frac{1}{2}$.

3. Write the ratio of “X” to “Y” atoms and simplify to the smallest whole numbers for the empirical formula.

Try solving on your own before revealing the answer!

Final Answer: XY₃

The empirical formula is XY₃, based on the ratio of atoms contributed by corners and faces.

Q8. Nickel (II) Oxide, NiO (molar mass = 74.69 g/mole) crystallizes in a face-centered cubic lattice. Given the density of NiO is 6.67 g/cm³, calculate the radius (in pm) of its unit cell.

Background

Topic: Crystal Lattice Geometry and Density

This question tests your ability to relate density, unit cell dimensions, and atomic radius in a face-centered cubic lattice.

Key Terms and Formulas

• Face-centered cubic (fcc): 4 formula units per unit cell.

• Unit cell: Smallest repeating unit in a crystal lattice.

• Density: Mass per unit volume.

Key formulas:

Volume of unit cell: $V = a^3$

Density: $\text{density} = \frac{\text{mass of atoms in cell}}{\text{volume of cell}}$

Relationship between edge length and radius in fcc: $a = 8r$

Step-by-Step Guidance

1. Calculate the mass of NiO in one unit cell using the molar mass and Avogadro's number.

2. Set up the density formula and solve for the volume of the unit cell.

3. Take the cube root of the volume to find the edge length $a$ in cm, then convert to pm.

4. Use the relationship $a = 8r$ to solve for the radius $r$ in pm.

Try solving on your own before revealing the answer!

Final Answer: 52.6 pm

After calculating, the radius of the NiO unit cell is approximately 52.6 pm.