Problem 39
Question
Cyclohexanol has a vapor pressure of \(10.0 \mathrm{mmHg}\) at \(56.0^{\circ} \mathrm{C}\) and \(100.0 \mathrm{mmHg}\) at \(103.7^{\circ} \mathrm{C} .\) Calculate its enthalpy of vaporization, \(\Delta H_{\mathrm{vap}}\)
Step-by-Step Solution
Verified Answer
\(\Delta H_{\mathrm{vap}} \approx 41\,454 \, J/mol\).
1Step 1: Converting Temperatures to Kelvin
Firstly, temperatures need to be converted from degrees Celsius to Kelvin. This can be done by adding 273.15 to the Celsius temperature. Therefore, \(T1 = 56.0 + 273.15 = 329.15K\) and \(T2 = 103.7 + 273.15 = 376.85K\).
2Step 2: Rearranging the Clausius-Clapeyron Equation
We rearrange the Clausius-Clapeyron equation to isolate \(\Delta H_{\mathrm{vap}}\): \(\Delta H_{\mathrm{vap}} = -R\left(\frac{1}{T2} - \frac{1}{T1}\right) / \ln\left(\frac{P2}{P1}\right)\). We express the pressures in the same units as the constant R (which is in atmospheres), so we have \(P1 = 10.0\,mmHg/760 = 0.01316 \, atmospheres\) and \(P2 = 100.0\,mmHg/760 = 0.1316 \, atmospheres\).
3Step 3: Substitution and Calculation
Substitute the values into the rearranged equation: \(\Delta H_{\mathrm{vap}} = - 8.314 \, J/(K \cdot mol) \cdot [(1/376.85 - 1/329.15) / \ln (0.1316/0.01316)]\). After performing the operations, you will find \(\Delta H_{\mathrm{vap}} \approx 41\,454 \, J/mol\).
Key Concepts
Clausius-Clapeyron equationVapor pressureThermodynamics
Clausius-Clapeyron equation
The Clausius-Clapeyron equation is a vital formula in the study of phase transitions, such as boiling or condensation. It relates vapor pressure and temperature to the enthalpy of vaporization, which is the energy required to transform a liquid into a vapor. The equation is expressed as:
To effectively use this equation, ensure temperature is in Kelvin and pressure is converted appropriately if needed, as seen with the cyclohexanol example.
- \( \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{\mathrm{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \)
To effectively use this equation, ensure temperature is in Kelvin and pressure is converted appropriately if needed, as seen with the cyclohexanol example.
Vapor pressure
Vapor pressure is a crucial concept in understanding how substances vaporize. It is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. When a liquid reaches its vapor pressure, it begins to vaporize.
Factors affecting vapor pressure include:
Factors affecting vapor pressure include:
- Temperature: As temperature increases, so does vapor pressure. Molecules gain more energy, escape into the vapor phase, and thus exert more pressure.
- Type of liquid: Different substances have different vapor pressures at the same temperature. A volatile liquid has a higher vapor pressure compared to a less volatile one.
Thermodynamics
Thermodynamics is the branch of physical science concerned with heat and its relation to other forms of energy and work. It is foundational in exploring how energy transfers occur and how they affect matter.
In the context of enthalpy of vaporization:
In the context of enthalpy of vaporization:
- Enthalpy: It measures the total heat content within a system. When considering vaporization, it's the heat required to change a liquid into a vapor at constant pressure.
- Laws of thermodynamics: These laws govern energy transfers. The first law, conservation of energy, is pivotal in calculations involving vaporization, ensuring all energy is accounted for.
- Heat and work: In thermodynamic processes like vaporization, energy often shifts between heat and work, influencing system behavior.
Other exercises in this chapter
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