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Physics Question 26 – JEE-MAIN 2025

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The equation for real gas is given by (P+aV2)(V−b)=RT, where P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2 is equivalent to that of :

Terms added or subtracted in an equation must have the same physical dimensions.

🥷
Ninja Strategy•Dimensional Matching

Quickly determine the dimensions of a and b from the Van der Waals equation, then calculate the dimension of ab−2 and match it with the distinct dimensions of the given options.

● Video Walkthrough
Step 1: Determine Dimensions of 'a' and 'b'✦ Active

According to the principle of homogeneity of dimensions, terms added or subtracted must have the same dimensions. From the Van der Waals equation (P+aV2)(V−b)=RT:

[aV2]=[P]⟹[a]=[P][V2]=[ML−1T−2][L3]2=[ML5T−2] [b]=[V]=[L3]
Step 2: Calculate Dimension of ab−2○ Expand

Now, substitute the dimensions of a and b to find the dimension of ab−2:

[ab−2]=[a][b−2]=[ML5T−2][L3]−2=[ML5T−2][L−6]=[ML−1T−2]
Step 3: Compare with Options○ Expand

Let's find the dimensions of the given options: 1. Compressibility: [P−1]=[M−1LT2] 2. Energy density: [EnergyVolume]=[ML2T−2L3]=[ML−1T−2] 3. Planck's constant: [Energy×Time]=[ML2T−2×T]=[ML2T−1] 4. Strain: Dimensionless [M0L0T0] Comparing the dimension of ab−2 ([ML−1T−2]) with the options, it matches the dimension of Energy density.

💡 Teacher's Secret Hint

Recall the standard dimensions of common physical quantities.

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