Van Der Waals Equation Calculator

Van Der Waals Equation:\((P+\frac{a\cdot n^{2}}{v^{2}})(V-n\cdot b)=n\cdot R\cdot T\)


Where Gas constant (R) = 8.3144621
Temperature (T):KNumber of moles (n):Volume (V):m3

Measure of the attraction between the particles (a):J*m3/mol2

Volume excluded by a mole of particles (b):m3/mol

Pressure of the gas (P):pascal

 

Van der Waals Equation Calculator is a free online tool that displays the physical properties of the gas. BYJU’S online Van der Waals equation calculator tool performs the calculation faster and it displays the physical properties in a fraction of seconds.

How to Use the Van der Waals Equation Calculator? 

The procedure to use the Van der Waals equation calculator is as follows:

Step 1: Enter the volume, temperature, number of moles, measure of attraction between particles  in the input field

Step 2: Now click the button “Calculate Pressure”  to get the pressure of the ideal gas

Step 3: Finally, the pressure of the ideal gas using Van der Waals equation will be displayed in the output field

What is Meant by Van der Waals Equation?

Van der Waals equation is an equation that provides the relationship between the volume, pressure and the absolute temperature of the real gas. It also uses two other parameters such as the intermolecular attraction and the finite size of the molecule. These two parameters are called the van der Waals parameters. Thus, the van der Waals equation is used to find the physical properties of the gas. The Van der Waals formula is given by

\((P+\frac{an^{2}}{V^{2}})(V-nb)=nRT\)

Here,

P = Pressure

V = Volume

T = Temperature

n = Number of moles of gas

R is the gas constant which is equal to 8.3144598 J/ mol. K

“a” and “b” are the Van Der Waals constants. The constant “a” is called the attraction parameter and “b” is called the repulsion parameter.

Solved Example on Van der Waals Equation

Example:

Determine the pressure of the gas using the Van der Waals equation where 1 mole of ammonia fills 7-litre bottle at a temperature of 350K. (a= 4.17 atm L2/mol2 and b = 0.0371 L/ mol

Solution:

Given that, a = 4.17 atm L2/mol2

b =  0.0371 L/ mol

n=1

V = 7 Litre

T = 350 k

We know that, Van der Waals equation is 

\((P+\frac{an^{2}}{V^{2}})(V-nb)=nRT\)

By rearranging the above formula to find the pressure, the formula becomes

\(P=\frac{nRT}{V-nb} -\frac{n^{2}a}{V^{2}}\)

Now, substitute the values in the formula, we get

\(P=\frac{(1)(0.0821)(350)}{7-(1)(0.0371)} -\frac{1^{2}(4.17)}{7^{2}}\)

P = 4.04 atm

Thus, the pressure of the real gas using Van der Waals equation is 4.04 atm

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