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Van Der Waals Equation Calculator

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The Van Der Waals theory account for the volume of the gas molecules and intermolecular interactions between the molecules. Modification of ideal gas law gives the Van Der Waals equation. This calculator determines the pressure as well as the temperature of the gas using the formula given,

$\left ( p + \frac{an^{2}}{V^{2}} \right )\left ( V - nb \right )=nRT$.

Where,
P = pressure,
V = volume,
n = number of moles,
T = temperature,
R = gas constant,
a and b = constants.
 

Steps

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Step 1 : Put down the given values from the problem.

Step 2 : Plug in the given values into the Van Der Waals equation and simply to get the appropriate value.

Problems

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Problems based on Van Der Waals are given below.

Solved Examples

Question 1: At 350K, ammonia of 2.00 mol fills in a 9.00 liter bottle. Calculate the pressure using Van Der Waals equation. Given, a = 4.17 $atmL^{2}/mol^{2}$ and b = 0.0371 L/mol.
Solution:
 
Step 1 : Given, P = ?
V = 9.00 L,
n = 2.00 mol,
T = 350K,
R = 8.3144 $JK^{-1}mol^{-1}$,
a = 4.17 $atmL^{2}/mol^{2}$,
b = 0.0371 L/mol.

Step 2 : Pressure of the gas is,
P = $\frac{nRT}{V-nb}$ - $\frac{n^2a}{V^2}$

P = $\frac{2 mol \times 8.3144 JK^{-1}mol^{-1} \times 350K}{9L - 2 mol \times 0.0371 L/mol}$ - $\frac{(2 mol)^2 \times 4.17 atmL^{2}/mol^{2}}{(9L)^2}$

P = 651.845 pascal.
 

Question 2: Calculate the temperature of $CO_2$ of $2.73 \times 10^{-4} m^3/mol$ at 10 bar using Van der Waals equation. Given, a = 2.343 $atmL^{2}/mol^{2}$ and b = 0.395 L/mol.

Solution:
 
Step 1 : Given, P = 10 bar = 1 000 000 pascals,
V = $2.73 \times 10^{-4} m^3/mol$,
n = 1 mol,
T = ?
R = 8.3144 $JK^{-1}mol^{-1}$,
a = 2.343 $atmL^{2}/mol^{2}$,
b = 0.395 L/mol.

Step 2 : Temperature of gas is,
$\left ( p + \frac{an^{2}}{V^{2}} \right )$ $( V - nb )$ = $nRT$

$T$ = $\left ( p + \frac{an^{2}}{V^{2}} \right )\left (\frac{V - nb}{nR} \right )$

$T$ = $\left ( 1 000 000 \ pascals + \frac{2.343 atmL^{2}/mol^{2} (1)^{2}}{(2.73 \times 10^{-4} m^3/mol)^{2}} \right )\left (\frac{2.73 \times 10^{-4} m^3/mol - 1 \times 0.395 L/mol}{1 \times 8.3144 JK^{-1}mol^{-1}} \right )$

$T$ = 369 K