To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.
This duet found the relationship between temperature and pressure expressed as,
[tex]ln P = constant - \frac{\Delta H}{RT}[/tex]
For the two states that we have then we could define the pressure and temperature in each of them as
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where,
[tex]P_{1,2}[/tex]= Pressure at state 1 and 2
[tex]T_{1,2}[/tex]= Temperature at state 1 and 2
[tex]\Delta H[/tex]= Enthalpy of Vaporization of a substance
R = Gas constant (8.134J/mol.K)
Our values are given by,
[tex]P_1 = 1atm \\\Delta H = 38.56*10^{-3} J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K[/tex]
Therefore replacing we have that,
[tex]ln(\frac{P_2}{P_1}) = \frac{-\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
[tex]ln(\frac{P_2}{1atm}) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]ln(P_2) - Ln(1atm) = \frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})[/tex]
[tex]P_2 = e^{\frac{-38.56*10^3}{8.314}(\frac{1}{311.95}-\frac{1}{351.55})}[/tex]
[tex]P_2 = 0.187355atm[/tex]
Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm
