The vapor pressure and enthalpy of vaporization of an unknown substance are measured to be 425.7 torr and 36.8 kJ/mol, respectively at room temperature (298.15 K). Estimate the normal boiling point of the substance at an atmospheric pressure of 760 torr.

The Famous Clausius Clapeyron Equation:

Assume Pressures P1 and P2 at Temperatures T1 and T2 respectively. Thus there exists a relationship between these 4 parameters at known value of enthalpy of Vaporization. This relation was put forward by Clausius and Clapeyron which can be mathematically expressed as-

$$\begin{gathered} \ln \left(\frac{P2}{P1}\right) = -\frac{∆H_{vap}}{R}(\frac{1}{T2}-\frac{1}{T1}) \\ where- \\ P1 = Vapour pressure of the unknown subs\tan ce at temperature T1 \\ P2=Vapour pressure of the unknown subs\tan ce at temperature T2 \\ ∆H_{vap} = Enthalpy of Vaporization \\ R = Gas Cons\tan t \end{gathered}$$

Given-

P2 = 760 torr i.e. atmospheric pressure

P1=425.7 torr 

T2 = ? = This is the normal boiling point temperature we need to calculate

T1 = 298.15 K

R = 8.314 J/Kmol

$∆$H_vap = 36.8 kJ/mol 

since, 1 kJ = 1000 J

thus,  $∆$H_vap = 36.8 kJ/mol x 1000 = 36800 J/mol

Substitute the values in Clausius-Clapeyron Equation-