Given the normal boiling temperature and enthalpy of vaporization of a substance

Question

Given the normal boiling temperature and enthalpy of vaporization of a substance, we can use the Clausius-Clapeyron equation to predict the vapor pressure at a range of different temperatures. For example, the normal boiling temperature of benzene is 353.25 K, with a standard enthalpy of vaporization DEltaH_vap = 30.8 kJ mol^-1. If we assume that the enthalpy of vaporization varies little with temperature and pressure, we may use the integrated form of the Clausius-Clapeyron equation to determine the vapor pressure of benzene at another temperature, such as 298.15 K. What is the vapor pressure of benzene at 298.15 K?

Explanation / Answer

Let the vapor pressure of benzene at 298.15 K = P

Thus, ln(1/P) = (30800/0.0821)*[(1/298.15) - (1/353.25)]

or, ln(1/P) = 196.265

or, P = 5.8*10-86 atm

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