Chapter 12: Problem 99
Estimate the molar heat of vaporization of a liquid whose vapor pressure doubles when the temperature is raised from \(85^{\circ} \mathrm{C}\) to \(95^{\circ} \mathrm{C}\).
Short Answer
Expert verified
The molar heat of vaporization is approximately 74.4 kJ/mol.
Step by step solution
01
Understand the Given Information
We know that the vapor pressure of the liquid doubles as the temperature increases from \(85^{\circ}\)C to \(95^{\circ}\)C. We need to find the molar heat of vaporization \(\Delta H_{vap}\). This can be calculated using the Clausius-Clapeyron equation.
02
Set Up Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is given as \[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \(P_1\) and \(P_2\) are the initial and final vapor pressures, \(T_1 = 85^{\circ}\)C, \(T_2 = 95^{\circ}\)C, and \(R\) is the gas constant \(8.314\, \mathrm{J/(mol\, K)}\). Since the vapor pressure doubles, \(\frac{P_2}{P_1} = 2.\)
03
Convert Temperatures to Kelvin
Convert the temperatures from Celsius to Kelvin by adding 273.15. So, \(T_1 = 85 + 273.15 = 358.15\,\text{K}\) and \(T_2 = 95 + 273.15 = 368.15\,\text{K}\).
04
Solve for \(\Delta H_{vap}\)
Substitute the values into the Clausius-Clapeyron equation: \[ \ln(2) = -\frac{\Delta H_{vap}}{8.314} \left( \frac{1}{368.15} - \frac{1}{358.15} \right) \] Calculate \(\ln(2)\), which is approximately 0.693. Then solve for \(\Delta H_{vap}\): \(\Delta H_{vap} = - \left(\frac{0.693 \times 8.314}{\frac{1}{368.15} - \frac{1}{358.15}}\right)\).
05
Calculate \(\Delta H_{vap}\)
Calculate the expression: \[ \Delta H_{vap} = - \left(\frac{0.693 \times 8.314}{\frac{1}{368.15} - \frac{1}{358.15}}\right) \] \[ = - \left(\frac{5.76}{-7.76 \times 10^{-5}}\right) \approx 74.4\,\text{kJ/mol}\] Therefore, the molar heat of vaporization is approximately \(74.4\,\text{kJ/mol}\).
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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 powerful tool used in thermodynamics to relate the vapor pressure of a substance to its temperature. This equation is particularly useful in determining the molar heat of vaporization, \( \abla H_{vap} \\), of a liquid. The formula is expressed as: \[ \\ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \\] where \(P_1\)\ and \(P_2\)\ represent the initial and final vapor pressures, respectively, and \(T_1\)\ and \(T_2\)\ are the corresponding absolute temperatures.
The variable \( R \) is the universal gas constant, valued at \( 8.314 \, \mathrm{J/(mol\, K)}\).
The variable \( R \) is the universal gas constant, valued at \( 8.314 \, \mathrm{J/(mol\, K)}\).
- The equation assumes a linear relationship between the natural logarithm of vapor pressure and inverse temperature, allowing us to predict how pressure changes with temperature.
- This linear approximation is valid over a limited temperature range and pressure changes.
- The equation helps in calculating the energy required to convert a given amount of liquid into vapor.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form in a closed system. It is a critical concept that helps explain why certain liquids evaporate faster than others.
In essence, it measures a substance's tendency to evaporate and its volatility.
In essence, it measures a substance's tendency to evaporate and its volatility.
- As the temperature of a liquid increases, the vapor pressure also increases because more molecules have enough energy to escape into the vapor phase.
- High vapor pressure indicates a liquid evaporates quickly, often seen in volatile substances like alcohol.
- The point at which the vapor pressure of a liquid equals the external pressure is called the boiling point.
Temperature Conversion
Temperature conversion is an essential skill in scientific calculations, especially when dealing with equations like the Clausius-Clapeyron. In scientific contexts, temperature is often expressed in the Kelvin scale, where 0 K is absolute zero. Celsius, on the other hand, is widely used in day-to-day life and has its freezing and boiling points of water at 0° C and 100° C, respectively.
Converting between these scales requires a simple formula:
Converting between these scales requires a simple formula:
- To convert Celsius to Kelvin, use: \( K = °C + 273.15 \)
- Kelvin does not use the degree symbol, unlike Celsius and Fahrenheit.
- The Kelvin scale starts from absolute zero, which makes it preferable when working with thermodynamic equations.