The vapor pressure of ethanol is [tex]1.00 \times 10^2 \, \text{mmHg}[/tex] at [tex]34.9^\circ\text{C}[/tex]. What is its vapor pressure at [tex]53.7^\circ\text{C}[/tex]? (The enthalpy of vaporization for ethanol, [tex]H_{\text{vap}}[/tex], is [tex]39.3 \, \text{kJ/mol}[/tex].)

To find the vapor pressure of ethanol at a different temperature, we can use the Clausius-Clapeyron equation. By converting the temperatures to Kelvin and substituting the values into the equation, we can calculate the vapor pressure at 53.7°C.

Explanation:

To find the vapor pressure of ethanol at 53.7°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R)(1/T2 - 1/T1)
where P1 is the vapor pressure at the initial temperature (34.9°C), P2 is the vapor pressure at the final temperature (53.7°C), ΔHvap is the heat of vaporization for ethanol (39.3 kJ/mol), R is the gas constant (8.314 J/(mol·K)), T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin:
T1 = 34.9 + 273.15 = 308.05 K
T2 = 53.7 + 273.15 = 326.85 K

Next, we can substitute the values into the equation:
ln(P2/1.00 x 10^2) = (-39.3 x 10^3 J/mol)/(8.314 J/(mol·K))(1/326.85 K - 1/308.05 K)

Solving for P2, we find:
P2 = (1.00 x 10^2)e^((-39.3 x 10^3 J/mol)/(8.314 J/(mol·K))(1/326.85 K - 1/308.05 K))

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