Final answer:
To find the heat of vaporization of an unknown liquid from its vapor pressure at two temperatures, you use the Clausius-Clapeyron equation, convert Celsius temperatures to Kelvin, and solve for the heat of vaporization (ΔHvap) using the provided vapor pressure data.
Step-by-step explanation:
To determine the heat of vaporization of an unknown liquid given its vapor pressure at two different temperatures, we use the Clausius-Clapeyron equation which describes the relationship between vapor pressure and temperature. This equation can be rearranged to solve for the enthalpy of vaporization (ΔHvap).
Firstly, we need to convert the temperatures from Celsius to Kelvin by adding 273.15. So, we have T1 = 25°C + 273.15 = 298.15 K and T2 = 45°C + 273.15 = 318.15 K. Then, we can use the following form of the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)
Where:
- P1 = 39 mmHg (vapor pressure at T1)
- P2 = 88 mmHg (vapor pressure at T2)
- ΔHvap is the heat of vaporization
- R = 8.314 J/(mol·K), the ideal gas constant
- T1 and T2 are the absolute temperatures in Kelvin
To solve for ΔHvap, we rearrange the equation:
ΔHvap = -R ln(P2/P1) / (1/T2 - 1/T1)
By plugging in the values, we get:
ΔHvap = -8.314 J/(mol·K) * ln(88/39) / (1/318.15 K - 1/298.15 K)
Perform the calculations to find the heat of vaporization, expressed in joules per mole (J/mol).