Which of the following represents the Clausius-Clapeyron equation:

The correct answer is option 4. All of the above.

We know Clausius-Clapeyron equation as,

\(log_eP = −\frac{\Delta H_{vap}}{RT} + log_eA\) ------- (i)

The above equation can also be written as:

 \(P = Ae^{−\frac{\Delta H_{vap}}{RT}}\) -------(ii)

In equation (ii) if we take log on both sides, we get equation (i).

Now, differentiating equation (i), we get

\(\frac{dlog_{e}P}{dT} = −\frac{\Delta H_{vap}}{RT^2}\)-------(iii)

changing \(log_e\) to \(log_{10}\), we get

\(\frac{dlog_{10}P}{dT} = −\frac{\Delta H_{vap}}{2.303RT^2}\) -------(iv)

Equations (i), (ii), (iii), and (iv) are the various forms of Clausius-Clapeyron equation.