Final answer:
To calculate the vapor pressure of 2,3,4-trimethylpentane at 105.5°C, we use the Clausius-Clapeyron equation with known values for the molar enthalpy of vaporization, normal boiling point, and atmospheric pressure, then perform calculations to solve for P2.
Step-by-step explanation:
To find the vapor pressure of 2,3,4-trimethylpentane at 105.5°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure and temperature with the molar enthalpy of vaporization. The formula to use is:
ln(P2/P1) = (ΔHvap/R) [(1/T1) - (1/T2)]
Where P1 is the vapor pressure at the normal boiling point, P2 is the vapor pressure at the desired temperature, T1 and T2 are the absolute temperatures in Kelvin at the normal boiling point and the desired temperature respectively, ΔHvap is the molar enthalpy of vaporization, and R is the ideal gas constant (8.314 J/mol·K).
First, we convert the temperatures from Celsius to Kelvin by adding 273.15. This gives us T1 = 113.47°C + 273.15 = 386.62 K and T2 = 105.5°C + 273.15 = 378.65 K.
Next, we rearrange the Clausius-Clapeyron equation to solve for P2:
ln(P2) = ln(P1) + (ΔHvap/R) [(1/T1) - (1/T2)]
We need P1, the vapor pressure at the normal boiling point, which is the atmospheric pressure (760 mmHg since the boiling point is at normal conditions). The ΔHvap is given as 37,600 J/mol.
Substituting our values into the equation, we get:
ln(P2) = ln(760 mmHg) + (37600 J/mol) / (8.314 J/mol·K) [(1/386.62 K) - (1/378.65 K)]
After calculation, the approximate value for ln(P2) is obtained, and then we take the exponential (e) to get P2 in mmHg.
Doing this computation gives us an answer which should be close to one of the options provided, A through D.