We’re being asked to determine the vapor pressure of water at 40°C. We’re given the heat of vaporization of water ΔH_{vap} = 40.7 kJ/mol.

We can use the ** Clausius-Clapeyron Equation** to solve for the vapor pressure of water.

$\overline{){\mathbf{ln}}{\mathbf{}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{P}}_{\mathbf{1}}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{\u2206}{\mathbf{H}}_{\mathbf{vap}}}{\mathbf{R}}\mathbf{[}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{]}}$

where:

P_{1} = vapor pressure at T_{1}

P_{2} = vapor pressure at T_{2}

ΔH_{vap} = heat of vaporization (in J/mol)

R = gas constant (8.314 J/mol•K)

T_{1} and T_{2} = temperature (in K).

The molar enthalpy of vaporization (∆H) of water is 40.7 kJ/mol. What is the vapor pressure of water at 40.°C (R=8.314J/K mol)?

A) 11.5 torr

B) 311 torr

C) 3.58 torr

D) 38.6 torr

E) 61.4 torr

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