Answer :
Final answer:
To calculate the vapor pressure of ethanol at 59.5 °C knowing its vapor pressure at 63.5 °C and its ΔHvap, the Clausius-Clapeyron equation is used, resulting in an approximate vapor pressure of 385 mmHg.
Explanation:
To find the vapor pressure of ethanol at 59.5 °C given its vapor pressure at 63.5 °C, along with the ΔHvap, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R(1/T2 - 1/T1)
First, convert both temperatures to Kelvin:
T1 = 63.5 + 273.15
= 336.65 K
T2 = 59.5 + 273.15
= 332.65 K
Next, use the Clausius-Clapeyron equation to solve for P2, which is the vapor pressure at 59.5 °C.
Given:
ΔHvap = 39.3 kJ/mol = 39300 J/mol (conversion for units)
R = 8.314 J/K • mol
P1 = 400.0 mmHg
ln(P2/400) = -(39300)/(8.314)(1/332.65 - 1/336.65)
After solving, we find that P2, the vapor pressure at 59.5 °C, is approximately 385 mmHg.