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Question

The Vander Waal’s equation of state for real gases is given as $\left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = nRT$ which of the following terms has dimensions different from that of energy

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Dimensional Analysis of Physical Quantities.
	**Related Topics**: Van der Waals Equation, Pressure, Volume, Energy.
	**Prerequisites**: Understanding of dimensions of pressure, volume, energy and the Van der Waals equation.
## SOLUTION PATHWAY
**Given Information**:
	Van der Waals equation: $\left( {P + \frac{a}{{{V^2}}}} ight)\left( {V - b} ight) = nRT$
	We need to determine which term does not have the dimensions of energy.
**Method & Steps**:
	1. Analyze the terms in Van der Waals equation. From the equation, we know that 'P' has units of Pressure, 'V' has units of volume, 'n' has units of moles, 'R' has units of energy per mole per Kelvin and 'T' has units of Kelvin.
		Reasoning: We need to find the dimensions of a, b, and the given options.
		Formula/Rule:  $[Pressure] = [Force]/[Area] = M L^{-1}T^{-2}$, $[Volume] = L^3$, $[Energy] = M L^2 T^{-2}$
	2. Analyze the term 'a'.  Since 'a/V^2' is added to pressure, it must have the same dimensions as pressure. Therefore, $ [\frac{a}{{{V^2}}}] = [P]$
		Working:  $[a] = [P][{V^2}] = M L^{-1}T^{-2} (L^3)^2 = M L^5 T^{-2}$
		Key Point:  Dimensions of a are $M L^5 T^{-2}$.
	3. Analyze the term 'b'. Since 'b' is subtracted from 'V', it must have the same dimensions as volume. Therefore $[b] = [V] = L^3$.
		Working: Dimensions of b are $L^3$
		Key Point: Dimensions of b are $L^3$
	4. Now, we find the dimensions of the given options:
		[A] $PV = M L^{-1}T^{-2} * L^3 = M L^2 T^{-2}$. This is the dimension of energy.
		[B] $\frac{a}{V^2} = M L^5 T^{-2} / L^6 = M L^{-1} T^{-2}$. This is the dimension of pressure, not energy.
		[C] $\frac{ab}{V^2} = M L^5 T^{-2} * L^3 / L^6 = M L^2 T^{-2}$. This is the dimension of energy.
		[D] $bP = L^3 * M L^{-1} T^{-2} = M L^2 T^{-2}$. This is the dimension of energy.
	5. Compare the dimensions of each term with the dimension of energy.
		Result: Option B has dimensions of pressure, while others have dimensions of energy.
		Verification: All terms except B have dimensions $M L^2 T^{-2}$ which is the dimension of energy.
 ## OPTION ANALYSIS
[A] :x: PV has dimensions of energy.
[B] ✓: $\frac{a}{V^2}$ has dimensions of pressure, which is not energy.
[C] :x: $\frac{ab}{V^2}$ has dimensions of energy.
[D] :x: $bP$ has dimensions of energy.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to equate terms added/subtracted to have the same dimensions.
			How to avoid: Always remember that terms that are added or subtracted must have the same physical dimensions.
		2.[Error 2]: Incorrectly calculating dimensions of combined terms.
			How to avoid: Carefully apply the rules of exponents when combining dimensions.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember that PV has the dimensions of energy.
		2.[Memory aid]: Use the equation PV=nRT to remember that PV has units of energy.

Mr. ANIL KUMAR.G · Answer

PV

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\frac{a}{{{V^2}}}$"

Mr. ANIL KUMAR.G · Answer

$\frac{{ab}}{{{V^2}}}$

Mr. ANIL KUMAR.G · Answer

bP

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