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Ch.11 - Liquids and Intermolecular Forces
Brown - Chemistry: The Central Science 15th Edition
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Tutti i libri di testoBrown 15th EditionCh.11 - Liquids and Intermolecular ForcesProblema 87
Capitolo 11, Problema 87

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, 𝑃1 and 𝑃2, and the absolute temperatures at which they were measured, 𝑇1 and 𝑇2:
ln𝑃1𝑃2=−Δ𝐻vap𝑅(1𝑇1−1𝑇2)
b. Gasoline is a mixture of hydrocarbons, a component of which is octane (CH3CH2CH2CH2CH2CH2CH2CH3). Octane has a vapor pressure of 13.95 torr at 25°C and a vapor pressure of 144.78 torr at 75°C. Use these data and the equation in part (a) to calculate the heat of vaporization of octane.
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 11.83.
d. Calculate the vapor pressure of octane at −30°C.

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Identify the given values: vapor pressures P1 and P2 at temperatures T1 and T2 respectively. Convert temperatures from Celsius to Kelvin by adding 273.15 to each Celsius temperature.
Use the Clausius-Clapeyron equation in the form ln(P1/P2) = -ΔH_vap/R (1/T1 - 1/T2) to set up the equation with the known values of P1, P2, T1, and T2.
Rearrange the equation to solve for ΔH_vap, the heat of vaporization. ΔH_vap = -R * ln(P1/P2) / (1/T1 - 1/T2), where R is the gas constant (8.314 J/mol·K).
To find the normal boiling point of octane, set the vapor pressure equal to the standard atmospheric pressure (760 torr) and solve for the temperature using the Clausius-Clapeyron equation rearranged to solve for T2.
For calculating the vapor pressure at a new temperature (e.g., -30°C), convert this temperature to Kelvin and use the Clausius-Clapeyron equation with the known ΔH_vap to solve for the new vapor pressure.

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Clausius–Clapeyron Equation

The Clausius–Clapeyron equation describes the relationship between vapor pressure and temperature for a substance. It is expressed as ln(P1/P2) = -ΔHvap/R(1/T1 - 1/T2), where P1 and P2 are vapor pressures at temperatures T1 and T2, ΔHvap is the heat of vaporization, and R is the ideal gas constant. This equation is crucial for understanding how changes in temperature affect the vapor pressure of a liquid.
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Clausius-Clapeyron Equation

Heat of Vaporization (ΔHvap)

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor at constant temperature and pressure. It is a critical property of substances, influencing their vapor pressures and boiling points. In the context of the Clausius–Clapeyron equation, ΔHvap is essential for calculating how much energy is needed for a substance to transition from liquid to gas, impacting its vapor pressure at different temperatures.
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Heat Capacity

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It reflects the tendency of a substance to evaporate; higher vapor pressures indicate a greater propensity for evaporation. Understanding vapor pressure is vital for predicting the behavior of substances under varying temperature conditions, as seen in the calculations for octane in the given problem.
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Raoult's Law and Vapor Pressure
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