The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5 kJ/mol)

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

where ;

is the vapor pressure at temperature 1

is the vapor pressure  at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

= 1 atm

 = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

= 282 °C  = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values  into the Clausius - Clapeyron equation, we have:

=  474.37 K

To °C ; we have =   (474.37 - 273)°C

=  201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C