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Vapor Pressure Calculator

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In a closed system, the pressure exerted by a vapor at a given temperature in the thermodynamic equilibrium is called vapor pressure. The vapor pressure calculator will help to determine the vapor pressure of a solution. The formula to calculate the vapor pressure is given by,

$P_{2}$ = $P_{1} \ exp$$\frac{\Delta H}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$where,
$T_{1}$ = initial temperature,
$T_{2}$ = final temperature,
$P_{1}$ = vapor pressure at $T_{1}$,
$P_{2}$ = vapor pressure at $T_{2}$,
$\Delta H$ = enthalpy of vaporization,
R = universal gas constant = 8.3144 J/mol K.
 

Steps

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Steps to determine the vapor pressure :

Step 1 : Note all the values from the given question.

Step 2 : Plugin all the values into the formula given and get the result accordingly.
$P_{2}$ = $P_{1}$ $exp \frac{\Delta H}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$.

Problems

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Below are some of the problems given based on vapor pressure.

Solved Examples

Question 1: At 298K, the vapor pressure of acetaldehyde is 0.0313 atm and enthalpy of vaporization is 44000 J/mol. What is the vapor pressure at 310K temperature.
Solution:
 
Given :
$T_{1}$ = 298K,
$T_{2}$ = 310K
$P_{1}$ = 0.0313 atm,
$P_{2}$ = ?
$\Delta H$ = 44000 J/mol,
R = 8.3144 J/mol K

The final pressure of acetaldehyde is,
$P_{2}$ = $P_{1} \ exp$ $\frac{\Delta H}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$

$P_{2}$ = $0.0313 \ exp$ $\frac{44000}{8.3144}\left ( \frac{1}{298}-\frac{1}{310} \right )$

$P_{2}$ = $0.0313 \ exp \ 5292.02$ $\left ( \frac{1}{298}-\frac{1}{310} \right )$

$P_{2} = 0.0622 \ atm$.
 

Question 2: At 273K, the vapor pressure of solution is 9.15 mmHg and enthalpy of vaporization is 44000 J/mol. What is the vapor pressure at 293K temperature.
Solution:
 
Given :
$T_{1}$ = 273K,
$T_{2}$ = 293K
$P_{1}$ = 9.15 mmHg = 0.0120 atm,
$P_{2}$ = ?
$\Delta H$ = 44000 J/mol,
R = 8.3144 J/mol K

The final pressure of solution is,
$P_{2}$ = $P_{1} \ exp$ $\frac{\Delta H}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )$

$P_{2}$ = $0.0120 \ exp$ $\frac{44000}{8.3144}\left ( \frac{1}{273}-\frac{1}{293} \right )$

$P_{2}$ = $0.0120 \ exp \ 5292.02$ $\left ( \frac{1}{273}-\frac{1}{293} \right )$

$P_{2} = 0.0031 \ atm$.