Answer :
If the vapor pressure of ethanol is 100 mmHg at 34. 9°C then the vapor pressure at 66. 5°C will be 100[tex]e^{39.3 kJ/mol/8.314}[/tex]
The vapor pressure of a liquid is the pressure exerted by the vapor when the liquid and vapor are in equilibrium. The vapor pressure of ethanol at 34.9°C is 100 mmHg.
The vapor pressure at 66.5°C can be determined by using the Clausius-Clapeyron equation. The Clausius-Clapeyron equation states that the vapor pressure of a liquid is related to its enthalpy of vaporization (ΔHvap) and temperature (T) by the following equation:
ln(P2/P1) = (ΔHvap/R) (1/T2 - 1/T1)
Where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, and R is the universal gas constant (8.314 J/molK). When we plug in the values for ethanol, we can solve for P2, the vapor pressure at 66.5°C.
P2 = P1e^((ΔHvap/R)(1/T2 - 1/T1))
P2 = 100[tex]e^{(39.3 kJ/mol/8.314)}[/tex] J/molK
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