Practicing Success
Which of the following represents the Clausius-Clapeyron equation: |
\(P = Ae^{−\frac{\Delta H_{vap}}{RT}}\) \(\frac{dlog_{10}P}{dT} = −\frac{\Delta H_{vap}}{2.303RT^2}\) \(\frac{dlog_{e}P}{dT} = −\frac{\Delta H_{vap}}{RT^2}\) All of the above |
All of the above |
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. |