The normal boiling point of ethanol is 78.4 °C and 101.3 KPa. The heat of vaporization for ethanol is 42.32 kJ/mol. Determine the vapor pressure of ethanol at 95.0 °C. Use the Clasius Clapeyron Equation: Ln [P2/P1] = - [∆Hvap /R] * [1/T2 - 1/T1]

6.1×10^4Pa or 61KPa

Explanation:

The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.