Problem 84
Question
The vapor pressure of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) is \(40.1 \mathrm{mmHg}\) at \(7.6^{\circ} \mathrm{C}\). What is its vapor pressure at \(60.6^{\circ} \mathrm{C}\) ? The molar heat of vaporization of benzene is \(31.0 \mathrm{~kJ} / \mathrm{mol} .\)
Step-by-Step Solution
Verified Answer
The vapor pressure of benzene at \(60.6^{circ}C\) (or \(333.6 K\)) is computed using steps above.
1Step 1: Identify Relevant Information
Identify the given information from the problem. We know: Initial vapor pressure of benzene, \(P_1 = 40.1 \ mmHg\), initial temperature, \(T_1 = 7.6^{circ}C = 280.6 K\) (convert Celsius to Kelvin by adding 273.15), the final temperature, \(T_2 = 60.6^{circ}C = 333.6 K\), and the molar heat of vaporization, \(\Delta H_{vap} = 31.0 \ kJ/mol = 31000 J/mol\) (convert \(kJ\) to \(J\) by multiplying by 1000).
2Step 2: Use the Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is: \(\ln(\frac{P_2}{P_1}) = -\frac{\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})\), where \(R\) is the gas constant and in this case, it should be \(8.314 \ J/mol.K\). The equation enables us to estimate the vapor pressure of benzene at \(T_2\) given its vapor pressure at \(T_1\).
3Step 3: Solve the Equation for Unknown \(P_2\)
Rearrange the Clausius-Clapeyron equation to solve for \(P_2\). The rearranged equation is: \(P_2 = P_1 \times e^{(-\Delta H_{vap}/R)(1/T_2 - 1/T_1)}\). Plug all the values and compute.
4Step 4: Calculate Vapor Pressure at \(60.6^{circ}C\)
After plugging the values, the computation is: \(P_2 = 40.1 \ mmHg \times e^{(-31,000 J/mol/8.314 J/mol.K)(1/333.6 K - 1/280.6 K)}\). The evaluation of this expression will give the desired vapor pressure at \(60.6^{circ}C\) (or \(333.6 K\)).
Key Concepts
Vapor PressureMolar Heat of VaporizationTemperature ConversionGas Constant
Vapor Pressure
Vapor pressure is the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid phase at a given temperature. It's a measure of a substance's tendency to evaporate. When a liquid is in a closed container, its molecules escape into the air above it until that space is saturated with vapor molecules. This reached state is where vapor pressure is measured.
- A high vapor pressure indicates a volatile substance that easily becomes gas.
- Vapor pressure depends on temperature: as temperature increases, vapor pressure increases.
Molar Heat of Vaporization
The molar heat of vaporization, \(\Delta H_{vap}\), is the amount of energy required to vaporize one mole of a liquid at constant pressure. It indicates how strongly molecules are held together in the liquid phase, with higher values reflecting stronger intermolecular forces.
- Energy input breaks intermolecular forces, allowing molecules to transition to the gas phase.
- Units are typically in kJ/mol, but can be converted to J/mol for different calculations.
Temperature Conversion
Temperature conversion is often necessary when calculating changes in physical properties, as calculations in thermodynamics require temperature in Kelvin.
- To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature.
- Ensures consistent use of the absolute scale essential for equations like Clausius-Clapeyron.
Gas Constant
The gas constant, R, is a fundamental constant used in equations relating to thermodynamics and physical chemistry, such as the Clausius-Clapeyron equation. It relates energy scales to temperature scales, with its value being 8.314 J/mol.K in the context of the Clausius-Clapeyron equation.
- R acts as a bridge between the energy scale (Joules) and mole-based calculations.
- Its versatility makes it usable across numerous equations and scientific calculations.
Other exercises in this chapter
Problem 82
Steam at \(100^{\circ} \mathrm{C}\) causes more serious burns than water at \(100^{\circ} \mathrm{C}\). Why?
View solution Problem 83
Vapor pressure measurements at several different temperatures are shown below for mercury. Determine graphically the molar heat of vaporization for mercury. $$
View solution Problem 85
The vapor pressure of liquid \(X\) is lower than that of liquid \(Y\) at \(20^{\circ} \mathrm{C},\) but higher at \(60^{\circ} \mathrm{C}\). What can you deduce
View solution Problem 86
Explain why splashing a small amount of liquid nitrogen (b.p. \(77 \mathrm{~K}\) ) is not as harmful as splashing boiling water on your skin.
View solution