Question 11.2.5: Use the two-point version of the Clausius–Clapeyron equation......

Use the two-point version of the Clausius–Clapeyron equation.

The vapor pressure of liquid aluminum is 400. mm Hg at 2590 K. Assuming that ΔH_{\text{vap}} for Al (296 kJ/mol) does not change significantly with temperature, calculate the vapor pressure of liquid Al at 2560 K.

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You are asked to calculate the vapor pressure of a liquid at a given temperature.
You are given the vapor pressure of the liquid at a different temperature and the enthalpy of vaporization for the liquid.
Use the alternative form of the Clausius–Clapeyron equation (Equation 11.2) along with pressure, temperature, and ΔH_{\text{vap}} data to calculate the vapor pressure of aluminum at 2560 K.

\ln\frac{P_{2}}{P_{1}} =\frac{-\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_{2}}-\frac{1}{T_{1}} \right)

P_{1} = 400. mm Hg            T_{1} = 2590 K
P_{2} = ?                                 T_{2} = 2560 K

ΔH_{\text{vap}} = 296 kJ/mol

\ln\frac{P_{2}}{400.  \text{mm Hg}} =\frac{-\left(296\text{ kJ/mol}\right) }{8.3145  \times  10^{-3}  \text{kJ/K}  \cdot  \text{mol}} \left(\frac{1}{2560\text{ K}}-\frac{1}{2590\text{ K}} \right)

\ln (P_{2}) – \ln (400. mm Hg) = -0.1611

\ln (P_{2})= 5.830
P_{2} = e^{5.830} = 340. mm Hg

Is your answer reasonable? Notice that the vapor pressure has decreased because the temperature decreased (T_{2} < T_{1}).

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