JEE PYQ: Units & Measurements - Question ID 99ad00083450 (JEE Main 2023)

Core Formula & Concept:

The given equation resembles the van der Waals equation of state for real gases, which corrects the ideal gas law for molecular size and intermolecular forces. The ideal gas law is: $PV = nRT$ For real gases, the van der Waals equation modifies this to: $\left(P + \frac{a n^2}{V^2}\right)(V - n b) = nRT$ where:

• $P$ is pressure,
• $V$ is volume,
• $T$ is temperature,
• $R$ is the universal gas constant,
• $a$ accounts for intermolecular attractive forces,
• $b$ accounts for the finite size of molecules.

In the given equation: $\left[X + \frac{a}{Y^2}\right][Y - b] = RT$ we identify $X$ as pressure ($P$), $Y$ as volume ($V$), and $R$, $T$ as the gas constant and temperature, respectively. The terms $\frac{a}{Y^2}$ and $b$ correspond to the van der Waals corrections.

To find the physical quantity equivalent to $\frac{a}{b}$, we must analyze the dimensions of $a$ and $b$ using dimensional analysis.

Step-by-Step Derivation:

Step 1: Dimensional Analysis of the Given Equation

The equation is: $\left[X + \frac{a}{Y^2}\right][Y - b] = RT$ Since $X$ is pressure ($P$), and $Y$ is volume ($V$), we rewrite it as: $\left[P + \frac{a}{V^2}\right][V - b] = RT$

Step 2: Expand the Left Side

Expanding the left-hand side (LHS): $P(V - b) + \frac{a}{V^2}(V - b) = PV - Pb + \frac{a}{V} - \frac{a b}{V^2}$ So the equation becomes: $PV - Pb + \frac{a}{V} - \frac{a b}{V^2} = RT$

Step 3: Dimensional Consistency

For the equation to be dimensionally consistent, each term must have the same dimensions as $RT$, which is energy (since $R$ has dimensions of energy per temperature, and $T$ is temperature).

Let’s analyze each term:

• $PV$: Pressure × Volume = Force/Area × Volume = Force × Length = Energy. ✅
• $Pb$: Pressure × $b$ must have dimensions of energy. So, $b$ must have dimensions of volume. ✅
• $\frac{a}{V}$: $\frac{a}{V}$ must have dimensions of energy. Since $V$ is volume, $a$ must have dimensions of energy × volume. ✅
• $\frac{a b}{V^2}$: This term must also have dimensions of energy. Since $b$ is volume and $V^2$ is volume squared, $\frac{a b}{V^2}$ simplifies to $\frac{a}{V}$, which we already confirmed has dimensions of energy. This is consistent. ✅

Step 4: Determine Dimensions of $a$ and $b$

From the above:

• $b$ has dimensions of volume: $[b] = L^3$.
• $a$ has dimensions of energy × volume: $[a] = (M L^2 T^{-2}) \times L^3 = M L^5 T^{-2}$.

Step 5: Compute Dimensions of $\frac{a}{b}$

Now, compute $\frac{a}{b}$: $\left[\frac{a}{b}\right] = \frac{M L^5 T^{-2}}{L^3} = M L^2 T^{-2}$ This is the dimension of energy.

Step 6: Match with Given Options

The options are:

• A: Impulse ($M L T^{-1}$)
• B: Energy ($M L^2 T^{-2}$)
• C: Pressure gradient ($M L^{-2} T^{-2}$)
• D: Coefficient of viscosity ($M L^{-1} T^{-1}$)

The dimensions of $\frac{a}{b}$ match energy, so the correct answer is B.

Common Traps & Exam Tip:

• Misidentifying the Equation: Students often fail to recognize the van der Waals equation structure. They may treat $\frac{a}{Y^2}$ and $b$ as arbitrary corrections without dimensional significance.
• Incorrect Dimensional Analysis: Some students mistakenly assume $a$ has dimensions of pressure × volume squared, leading to incorrect conclusions. Always verify dimensions term-by-term.
• Overcomplicating the Problem: The question only requires dimensional consistency, not a full derivation of the van der Waals equation. Focus on the dimensions of $a$ and $b$ individually.
• Confusing $a$ and $b$: Students may swap the roles of $a$ and $b$, leading to incorrect dimensional results. Remember: $a$ corrects for intermolecular forces, $b$ for molecular volume.

Exam Tip: When faced with such problems, always:

1. Identify the physical meaning of each variable.
2. Ensure dimensional consistency across all terms.
3. Compare the derived dimensions with standard physical quantities.

This systematic approach minimizes errors and ensures full marks.