
Derivation of Van der Waals equation.
Answer
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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.
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