Top

Clausius-Clapeyron Equation Calculator
Top
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

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

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.

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.