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Clausius-Clapeyron Equation Calculator
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Clausius Clapeyron Equation is used to determine the vapor pressure of given liquid or solids.
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.
 

Steps for Clausius Clapeyron Equation Calculator

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Step 1 :  

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.



Problems on Clausius Clapeyron Equation Calculator

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  1. 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.



  2. 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.



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