![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
Derivation of Van der Waals equation.
Answer
458.7k+ views
Hint: The correct answer for this question is based on the derivation which makes use of correction in pressure and also correction in volume by deriving for ideal pressure and volume and substituting in the ideal gas equation.
Complete step by step answer:
We have studied about the ideal gas derivation and the equation which is given by,$PV=nRT$ ……..(1)
Now, let us derive Van der Waals equation based on pressure and volume correction.
(i) Pressure correction:
If we consider two gaseous molecules A and B in a container having ‘n’ number of gas molecules and consider one gas molecule is near the surface of the container and the other is at the interior.
The gas molecule at the interior has no net force being exerted that is the value is zero because it is attracted by surrounding gas molecules and therefore net effect is inward pull and hence pressure is inversely proportional to volume and is given by,$P\propto \dfrac{1}{V}$ …..(2)
The other gas molecule B, is about to strike the container and also there are other molecules which are striking it and there is inward pull of gas B by striking molecules and also this pressure is inversely proportional to volume$P\propto \dfrac{1}{V}$…….(3)
From (2) and (3) we can write as, Actual pressure$P$ = Ideal pressure${{P}_{i}}$ - pressure correction${{P}_{c}}$
\[\Rightarrow {{P}_{i}}=P+{{P}_{c}}\] …….(4)
Now, from (2) and (3) pressure correction is,
\[{{P}_{c}}\propto \dfrac{1}{V}\times \dfrac{1}{V}\propto \dfrac{1}{{{V}^{2}}}\]
Thus, ${{P}_{c}}=\dfrac{a}{{{V}^{2}}}$ where a is constant…..(5)
Put (5) in (4)$\Rightarrow {{P}_{i}}=P+\dfrac{a}{{{V}^{2}}}$ …..(6)
(ii) Volume correction:
At higher pressure molecules occupy minimum area or volume. Thus by subtracting volume correction b from observed volume we have,
\[{{V}_{i}}=V-b\] ……(7)
Since from equation (1), the ideal volume and pressure can be written for one mole of gas as, ${{P}_{i}}{{V}_{i}}=RT$
Now, substituting equation (6) and (7) in the above one we have,
$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$
Note: Note that the above equation derived is applicable for only one mole of a gas and for n moles of gas we can write the Van der Waals equation as,$\left( P+\dfrac{{{n}^{2}}a}{{{V}^{2}}} \right)\left( V-nb \right)=nRT$and this helps you to solve problems based on this formula.
Complete step by step answer:
We have studied about the ideal gas derivation and the equation which is given by,$PV=nRT$ ……..(1)
Now, let us derive Van der Waals equation based on pressure and volume correction.
(i) Pressure correction:
If we consider two gaseous molecules A and B in a container having ‘n’ number of gas molecules and consider one gas molecule is near the surface of the container and the other is at the interior.
The gas molecule at the interior has no net force being exerted that is the value is zero because it is attracted by surrounding gas molecules and therefore net effect is inward pull and hence pressure is inversely proportional to volume and is given by,$P\propto \dfrac{1}{V}$ …..(2)
The other gas molecule B, is about to strike the container and also there are other molecules which are striking it and there is inward pull of gas B by striking molecules and also this pressure is inversely proportional to volume$P\propto \dfrac{1}{V}$…….(3)
From (2) and (3) we can write as, Actual pressure$P$ = Ideal pressure${{P}_{i}}$ - pressure correction${{P}_{c}}$
\[\Rightarrow {{P}_{i}}=P+{{P}_{c}}\] …….(4)
Now, from (2) and (3) pressure correction is,
\[{{P}_{c}}\propto \dfrac{1}{V}\times \dfrac{1}{V}\propto \dfrac{1}{{{V}^{2}}}\]
Thus, ${{P}_{c}}=\dfrac{a}{{{V}^{2}}}$ where a is constant…..(5)
Put (5) in (4)$\Rightarrow {{P}_{i}}=P+\dfrac{a}{{{V}^{2}}}$ …..(6)
(ii) Volume correction:
At higher pressure molecules occupy minimum area or volume. Thus by subtracting volume correction b from observed volume we have,
\[{{V}_{i}}=V-b\] ……(7)
Since from equation (1), the ideal volume and pressure can be written for one mole of gas as, ${{P}_{i}}{{V}_{i}}=RT$
Now, substituting equation (6) and (7) in the above one we have,
$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$
Note: Note that the above equation derived is applicable for only one mole of a gas and for n moles of gas we can write the Van der Waals equation as,$\left( P+\dfrac{{{n}^{2}}a}{{{V}^{2}}} \right)\left( V-nb \right)=nRT$and this helps you to solve problems based on this formula.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)