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Determine the vapor pressure of isooctane at 38 degrees Celsius, given its boiling point is 98.2 degrees Celsius and its enthalpy of vaporization is 35.8 kJ/mol.

Answer :

Final answer:

To find the vapor pressure of isooctane at 38 degrees Celsius, the Clausius-Clapeyron equation is used, incorporating the given enthalpy of vaporization and the temperature conversion to Kelvin.

Explanation:

To determine the vapor pressure of isooctane at 38 degrees Celsius, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization to the pressure and temperature:

ln(P2/P1) = ΔHvap/R (1/T1 - 1/T2)

Where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant (8.314 J/(mol·K)).

In this problem, we consider the normal boiling point temperature (T1) to be 98.2 degrees Celsius, which we convert to Kelvin (T1 = 371.35 K), and the corresponding pressure (P1) is atmospheric pressure, or 101.3 kPa. We want to calculate the pressure at 38 degrees Celsius (T2 = 311.15 K). The enthalpy of vaporization (ΔHvap) is given as 35.8 kJ/mol, which is 35,800 J/mol.

Rearranging the equation for P2 and substituting the known values, we can calculate the vapor pressure of isooctane at 38 degrees Celsius.

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