Answer :
Final Answer:
The vapor pressure of ethanol at 62.1 °C is about 382.75 mmHg, calculated using the Clausius-Clapeyron equation with given data and constants.
Explanation:
To calculate the vapor pressure of ethanol at 62.1 °C using the Clausius-Clapeyron equation, we can use the following formula:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where:
P1 = Vapor pressure at temperature T1 (63.5 °C)
P2 = Vapor pressure at the desired temperature (62.1 °C)
ΔHvap = Enthalpy of vaporization
R = Gas constant (8.314 J/K•mol)
T1 = Temperature in Kelvin corresponding to T1
T2 = Temperature in Kelvin corresponding to T2
First, let's convert the temperatures to Kelvin:
T1 = 63.5 + 273.15 = 336.65 K
T2 = 62.1 + 273.15 = 335.25 K
Now, plug in the values and solve for ln(P2/P1):
ln(P2/400.0) = -(39300 J/mol) / (8.314 J/K•mol) * (1/335.25 K - 1/336.65 K)
ln(P2/400.0) = -4727.48 * (0.002985 - 0.002975)
ln(P2/400.0) = -4727.48 * 0.000010
ln(P2/400.0) = -0.0472748
Now, solve for P2:
P2/400.0 = [tex]e^(-0.0[/tex]472748)
P2 = 400.0 * [tex]e^(-0.0[/tex]472748)
P2 ≈ 382.75 mmHg
So, the vapor pressure of ethanol at 62.1 °C is approximately 382.75 mmHg.
Learn more about Vapor Pressure
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