Question

The Clausius-Clapeyron equation may be given as:

• A. ${\log }_{10}\left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
• B. $\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$
• C. $\frac{dP}{dT}=\frac{q}{T\Delta V}$
• D. $\frac{dP}{dT}=\frac{\Delta V}{\Delta S}$

Answer 1

Answer: (A), (B), (C)

${\log }_{10}\left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$
$\frac{dP}{dT}=\frac{q}{T\Delta V}$

The Clausius-Clapeyron equation allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at same temperature, and if the enthalphy of vaporization is known, ${\log }_{10}\left(\frac{P_2}{P_1}\right)=\frac{△H_{vap}}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

Other forms of Clausius-Clapeyron equation are,

$\frac{dP}{dT}=\frac{△S}{△V}$ where $△S=$ Change in entropy.

$\frac{dP}{dT}=\frac{q}{T△V}$ where $△V=$ Change in volume.