The equation for real gas is given by $$\left(P + \frac{a}{V^2}\right)(V - b) = RT$$, 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:

Solution :

Given equation for real gas :

$$(P+\frac{a}{V^2})(V-b)=RT$$

Since quantities added inside bracket must have same dimensions,

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

has dimensions of pressure.

Therefore,

$$[a] = [P][V^2]$$

Pressure :

$$[P] = ML^{-1}T^{-2}$$

Volume :

$$[V] = L^3$$

Hence,

$$[a] = (ML^{-1}T^{-2})(L^6)$$

$$= ML^5T^{-2}$$

Also, since \(V-b\) is present,

$$[b] = [V]$$

$$= L^3$$

Now,

$$[ab^{-2}] = [a][b]^{-2}$$

$$= (ML^5T^{-2})(L^3)^{-2}$$

$$= ML^5T^{-2} \times L^{-6}$$

$$= ML^{-1}T^{-2}$$

This is the dimension of pressure.

Also,

Energy density

$$= \frac{\text{Energy}}{\text{Volume}}$$

Dimensions of energy :

$$[E] = ML^2T^{-2}$$

Therefore,

$$[\text{Energy density}] = \frac{ML^2T^{-2}}{L^3}$$

$$= ML^{-1}T^{-2}$$

Hence,

$$[ab^{-2}]$$

has dimensions of energy density.

Final Answer :

Energy density.