The equation for real gas is given by left(P + fracaV^2right)(V - b) = RT$\left(P + \frac{a}{V^2}\right)(V - b) = RT$, where P, V, T$P, V, T$ and R$R$ are the pressure, volume, temperature and gas constant, respectively. The dimension of ab^-2$ab^{-2}$ is equivalent to that of:

### Related Formula By the principle of dimensional homogeneity, terms added or subtracted must have the same dimensions: [P] = left[fracaV^2right] implies [a] = [P][V]^2$$[P] = \left[\frac{a}{V^2}\right] \implies [a] = [P][V]^2$$ [V] = [b]$[V] = [b]$ ### Core Logic Let's find the dimensional formula of the quantities: - Pressure P$P$: [P] = textM L^-1textT^-2$$[P] = \text{M L}^{-1}\text{T}^{-2}$$ - Volume V$V$: [V] = textL^3$$[V] = \text{L}^3$$ Substituting these to find [a]$[a]$ and [b]$[b]$: [a] = (textM L^-1textT^-2)(textL^6) = textM L^5textT^-2$$[a] = (\text{M L}^{-1}\text{T}^{-2})(\text{L}^6) = \text{M L}^5\text{T}^{-2}$$ [b] = textL^3 implies [b^-2] = textL^-6$$[b] = \text{L}^3 \implies [b^{-2}] = \text{L}^{-6}$$ Now, compute the dimensions of ab^-2$ab^{-2}$: [ab^-2] = (textM L^5textT^-2)(textL^-6) = textM L^-1textT^-2$$[ab^{-2}] = (\text{M L}^5\text{T}^{-2})(\text{L}^{-6}) = \text{M L}^{-1}\text{T}^{-2}$$ This matches the dimensions of pressure. Let's evaluate the options: 1. Planck's constant: [h] = textM L^2textT^-1$[h] = \text{M L}^2\text{T}^{-1}$ 2. Compressibility: [beta] = textM^-1textLtextT^2$[\beta] = \text{M}^{-1}\text{L}\text{T}^2$ 3. Strain: dimensionless 4. Energy density (energy per unit volume): left[fracEVright] = fractextM L^2textT^-2textL^3 = textM L^-1textT^-2$$\left[\frac{E}{V}\right] = \frac{\text{M L}^2\text{T}^{-2}}{\text{L}^3} = \text{M L}^{-1}\text{T}^{-2}$$ ### Step 1: Final Conclusion Therefore, the dimension of ab^-2$ab^{-2}$ is equivalent to that of Energy density. ### Pattern Recognition By writing the relation directly as [ab^-2] = frac[a][b]^2$[ab^{-2}] = \frac{[a]}{[b]^2}$, and noting [a] = [P][V]^2$[a] = [P][V]^2$ and [b] = [V]$[b] = [V]$, we get [ab^-2] = frac[P][V]^2[V]^2 = [P]$[ab^{-2}] = \frac{[P][V]^2}{[V]^2} = [P]$ (Pressure). Since pressure and energy density have identical dimensions, the answer is immediately Energy density. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Kinetic Theory of Gases