It is given as:
ln $\frac{p_{1}}{p_{2}}$ = $\frac{\Delta H_{vap}}{R}$($\frac{1}{T_{2}}$ - $\frac{1}{T_{1}}$).
Where p1 and p2 are the initial and final pressures,
T1 and T2 are the initial and final temperatures,
$\Delta$ Hvap is enthalpy of vaporization
Clausius Clapeyron Equation Calculator calculates the molar enthalpy of vaporization $\Delta$ Hvap if initial pressure p1, final pressure p2, initial temperature T1 and final temperature T2 are known.
Read the problem and observe the given quantities.
Step 2 :
Use the formula:
$\Delta$ Hvap = $\frac{R ln \frac{p_{2}}{p_{1}}}{\frac{1}{T_{1}} - \frac{1}{T_{2}}}$
Where p1 and p2 are the initial and final pressures,
T1 and T2 are the initial and final temperatures,
$\Delta$ Hvap is enthalpy of vaporization.
Substitute the given quantities to get the desired quantity.
Find the Enthalpy of vaporization if the Vapor pressures of ice at 270 and 275K is 1.56 pa and 3.780 pa respectively?
Step 1 :Given quantities are:
Initial pressure p1 = 1.56 pa
Final pressure p2 = 3.78 pa
Initial temperature T1 = 270 K,
Final temperature T2 = 275 K.
To find Molar enthalpy of vaporization $\Delta$ Hvap.
Step 2 :To Molar enthalpy of vaporization is given as:
$\Delta$ Hvap = $\frac{R ln \frac{p_{2}}{p_{1}}}{\frac{1}{T_{1}} - \frac{1}{T_{2}}}$
= R ln $\frac{p_{1}}{p_{2}}$ $\frac{T_{1} T_{2}}{T_{2} - T_{1}}$
= 8.314 $\times$ ln $\frac{1.56}{3.78}$ $\times$ $\frac{270 \times 275}{275 - 270}$
= $\frac{-7.358}{-0.0000673}$
= 109334.43 J/mol.
Answer :The Enthalpy of Vaporization $\Delta$ Hvap = 109334.43 J/mol.
The vapor pressure of water is 10 pa at 280 K and is having pressure of 15 pa at 290 K. Find the enthalpy of vaporization?
Step 1 :Given quantities are:
Initial pressure p1 = 10 pa,
Vapor pressure p2 = 15 pa,
Initial temperature T1 = 280 K,
Final temperature T2 = 290 K.
To find the Enthalpy of vaporization $\Delta$ Hvap.
Step 2 :To Molar enthalpy of vaporization is given as:
$\Delta$ Hvap = R $\frac{ ln \frac{p_{2}}{p_{1}}}{\frac{1}{T_{1}} - \frac{1}{T_{2}}}$
= R ln $\frac{p_{1}}{p_{2}}$ $\frac{T_{1} T_{2}}{T_{2} - T_{1}}$
= 8.314 $\times$ ln $\frac{15}{10}$ $\times$ $\frac{280 \times 290}{290 - 280}$
= 8.314 $\times$ 0.405 $\times$ 8120
= 27374.374 J/mol.
Answer :The Enthalpy of vaporization 27374.374 KJ/mol.