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For ethanol, [tex]C_2H_5OH[/tex], the normal boiling point is 78.37 °C and the enthalpy of vaporization is 38.6 kJ/mol. What is the vapor pressure at 40.0 °C?

Answer :

Final answer:

The vapor pressure of ethanol at 40.0 °C can be determined using the Clausius-Clapeyron equation. After converting the temperatures to Kelvin and the enthalpy of vaporization to Joules, the calculation results in a vapor pressure of about 0.444 atmospheres.

Explanation:

To determine the vapor pressure at 40.0 °C, we can utilize the Clausius-Clapeyron equation, which relates the vapor pressure at one temperature to the vapor pressure at another. First, we need to convert the temperature to Kelvin. Thus, the boiling point is 78.37 °C + 273.15 = 351.52 K and the given temperature is 40.0 °C + 273.15 = 313.15 K.

The Clausius-Clapeyron equation is: ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1), where Presure (P) is in atmospheres, enthalpy of vaporization (ΔHvap) is in J/mol, gas constant (R) is 8.314 J/(mol*K), and temperature (T) is in Kelvin.

In this case, P1 is the pressure at the boiling point (which is 1 atmosphere), ΔHvap is 38.6 kJ/mol (38600 J/mol), T1 is 351.52 K, and T2 is 313.15 K.

Upon substituting these values into the equation to solve for P2 (vapor pressure at 40.0 °C), we find that the vapor pressure is approximately 0.444 atmospheres.

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