Chapter 12: Problem 99

Estimate the molar heat of vaporization of a liquid whose vapor pressure doubles when the temperature is raised from \(85^{\circ} \mathrm{C}\) to \(95^{\circ} \mathrm{C}\).

Short Answer

Expert verified

The molar heat of vaporization of the liquid is approximately \(41.4 \mathrm{KJ/mol}\)

Step by step solution

Convert temperature to Kelvin

The temperatures given are in Celsius. Convert them into Kelvin. The formula to convert Celsius to Kelvin is \(T(K) = T(^{\circ}C) + 273.15\). So, \(85^{\circ} \mathrm{C} = 358.15 \mathrm{K}\) and \(95^{\circ} \mathrm{C} = 368.15 \mathrm{K}\).

Apply the Clausius-Clapeyron equation

The Clausius-Clapeyron equation in its logarithmic form is \(\ln(\frac{P_2}{P_1}) = -\frac{\Delta H_{\text{vap}}}{R}(\frac{1}{T_2} - \frac{1}{T_1})\). Here, \(P_2\) and \(P_1\) are the final and initial pressures respectively. \(T_2\) and \(T_1\) are the final and initial temperatures in Kelvin. It is given that the pressure doubles when the temperature is increased, so, \(P_2 = 2P_1\). Substituting these values and rearranging for \(\Delta H_{\text{vap}}\), we get \(\Delta H_{\text{vap}}= -R \times \ln(\frac{P_2}{P_1}) \times (\frac{1}{T_2} - \frac{1}{T_1})^{-1}\).

Substitute the values

Substituting, \(\Delta H_{\text{vap}} = -8.314 \times \ln(2) \times (\frac{1}{368.15} - \frac{1}{358.15})^{-1}\). After calculating, you should get \(\Delta H_{\text{vap}}\) approximately equal to \(41.4 \mathrm{KJ/mol}\).

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Estimate the molar heat of vaporization of a liquid whose vapor pressure doubles when the temperature is raised from \(85^{\circ} \mathrm{C}\) to \(95^{\circ} \mathrm{C}\).

Short Answer

Step by step solution

Convert temperature to Kelvin

Apply the Clausius-Clapeyron equation

Substitute the values

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