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The vapor pressure of 1-propanol is 10.0 torr at 14.7 °C and 100.2 torr at 52.8 °C. Calculate the heat of vaporization of 1-propanol.

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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