Question

Using Van der Waals equation,$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$ at high pressure, the Van der Waals equation gets reduced to:
(A) $\left( P+\dfrac{a}{{{V}^{2}}} \right)V=RT$
(B) $P(V-b)=RT$
(C) $PV=RT$
(D) $\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$

Answer 1

Hint: Write the Van der Waals equation at high pressure. Write the ideal gas equation. Think about the relationship between pressure and volume. Solve the Van der Waals equation, keeping in mind pressure is high.

Complete Solution :
We have come across the concept of the ideal gas equation in physical chemistry.
The equation is given by, PV = nRT
For 1 mole of a gas, it reduces to PV = RT …..(1)
- All gases are real gases. No gases behave ideally. So, J. D. Vander Waals derived an equation which approximately gives correction for all real gases to the ideal gas.

- Real gas can occupy different volume and therefore apply slight different pressures and thus pressure and volume corrections are needed and this is given by,
${{P}_{i}}=P+{{P}_{a}}$
where,${{P}_{i}}$ is the ideal pressure,$P$ is observed pressure and${{P}_{a}}$ is the actual pressure.
Since, ${{P}_{a}}\propto \dfrac{1}{V}$ that is proportional to number of moles per unit volume that are attracted and also ${{P}_{a}}\alpha \dfrac{1}{V}$ that is proportional to number of molecules per unit volume which are attracting, we have${{P}_{a}}\propto {{\left( \dfrac{1}{V} \right)}^{2}}$
Thus,${{P}_{a}}=\dfrac{a}{{{V}^{2}}}$ where ‘a’ is the characteristic constant.
-Thus, Van der Waals equation at pressure and volume correction is given by,
$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)= RT$
Now, rearranging equation (1) we get,
\[\Rightarrow P=\dfrac{RT}{V}\]
This implies, pressure is inversely proportional to volume.

- Now getting back to the Van der Waals equation, at high pressure, volume will be extremely low. So, we can neglect the term,$\dfrac{a}{{{V}^{2}}}$ but we cannot exclude the term, (V-b).
- Therefore, the reduced equation is,$P(V-b)=RT$
- Therefore, at high pressure, Van der Waals equation, $\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$ gets reduced to$P(V-b)=RT$.
- Therefore, the correct answer is option $P(V-b)=RT$
So, the correct answer is “Option B”.

Note: Remember pressure is always inversely proportional to volume. At high pressure, volume will be negligible. Be very careful while neglecting specific terms. A small mistake made to neglect (V-b) term in this case, will not give the answer. You can’t neglect that term because it is in multiplication with pressure. It is highly significant.