Question

A certain substance has a ΔvapH of 38.7 kJ/mol, and at 76.0 °C its vapor pressure is p* = 760 mmHg. What is its vapor pressure (in mmHg) at 29.0 °C? A. 762 mmHg B. 95.5 mmHg C. 688 mmHg D. 758 mmHg

Answer 1

Using the Clausius-Clapeyron equation and given the ΔHvap of 38.7 kJ/mol, the vapor pressure of the substance at 29.0 °C is approximately 95.5 mmHg (B).

Step-by-step explanation:

The question pertains to calculating the vapor pressure of a substance at a different temperature using the Clausius-Clapeyron equation. The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance given its heat of vaporization (ΔHvap).

To find the vapor pressure at 29.0 °C, we need to apply the Clausius-Clapeyron equation which can be written as:
ln(P2/P1) = -ΔHvap / R * (1/T2 - 1/T1)

Where:

• ΔHvap is the heat of vaporization (38.7 kJ/mol)
• R is the ideal gas constant (8.314 J/mol·K)
• P1 is the initial vapor pressure (760 mmHg)
• T1 is the initial temperature (76.0 °C converted to Kelvin)
• P2 is the final vapor pressure
• T2 is the final temperature (29.0 °C converted to Kelvin)

First, we convert the temperatures from °C to Kelvin by adding 273.15. Then we arrange the Clausius-Clapeyron equation to solve for P2 and substitute the known values. After calculating, we find that the vapor pressure at 29.0 °C is approximately 95.5 mmHg, which corresponds to option B.