a substance has a vapor pressure of 35.0 mmHg at 52.0 degrees Celsius and 70.0 mm Hg at 77.0 degrees Celsius. What is the heat of vaporization for this substance?
To find the heat of vaporization for a substance, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
- P2 and P1 are the vapor pressures at temperatures T2 and T1, respectively.
- ΔHvap is the heat of vaporization.
- R is the ideal gas constant (8.314 J/(mol·K)).
- T2 and T1 are the temperatures in Kelvin.
First, we need to convert the given temperatures from Celsius to Kelvin:
- T1 = 52.0°C + 273.15 = 325.15 K
- T2 = 77.0°C + 273.15 = 350.15 K
Next, we plug in the values into the Clausius-Clapeyron equation:
ln(70.0 mmHg / 35.0 mmHg) = (-ΔHvap / 8.314 J/(mol·K)) * (1/350.15 K - 1/325.15 K)
Simplifying the equation further:
ln(2) = (-ΔHvap / 8.314 J/(mol·K)) * (0.002876 - 0.003069)
Subtracting the fractions:
ln(2) = (-ΔHvap / 8.314 J/(mol·K)) * (-0.000193)
Cancelling out the negative signs:
ln(2) = (ΔHvap / 8.314 J/(mol·K)) * 0.000193
Now, we can solve for ΔHvap by isolating the variable:
ΔHvap = -(ln(2) * 8.314 J/(mol·K)) / 0.000193
Calculating this expression will give us the heat of vaporization for the substance.