Answer :
Final answer:
The heat of vaporization of 1-propanol is approximately 40,000 J/mol.
Explanation:
To calculate the heat of vaporization of 1-propanol, we can use the Clausius-Clapeyron equation. Let's assign the given values:
- P1 = 10.0 torr
- T1 = 14.7 °C
- P2 = 100.2 torr
- T2 = 52.8 °C
First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15:
- T1 = 14.7 °C + 273.15 = 287.85 K
- T2 = 52.8 °C + 273.15 = 325.95 K
Next, we can substitute the values into the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
ln(100.2/10.0) = -(ΔHvap/R) * (1/325.95 - 1/287.85)
Simplifying the equation:
ln(10.02) = -(ΔHvap/R) * (0.003067 - 0.003472)
ln(10.02) = -(ΔHvap/R) * (-0.000405)
Now, we can solve for ΔHvap:
ΔHvap = -ln(10.02) / (-0.000405 * R)
Using the value of the ideal gas constant R = 8.314 J/(mol·K), we can calculate the heat of vaporization:
ΔHvap = -ln(10.02) / (-0.000405 * 8.314)
ΔHvap ≈ 40,000 J/mol
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