Question

The equation of state of a real gas is given by \[ \left( P + \frac{a}{V^2} \right)(V - b) = RT, \] where \( P, V \) and \( T \) are pressure, volume and temperature respectively and \( R \) is the universal gas constant. The dimensions of \[ \frac{a}{b^2} \] is similar to that of :

• A. PV
• B. R
• C. RT
• D. P

Answer 1

The Correct Option is D

The objective is to determine the dimensional analysis of the quantity \(\frac{a}{b^2}\) within the provided real gas equation of state:

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

We will first analyze the dimensions of each component term:

1. Pressure ( \(P\) ) has dimensions: \([M^1 L^{-1} T^{-2}]\).
2. Volume ( \(V\) ) has dimensions: \([L^3]\).
3. The universal gas constant ( \(R\) ), representing energy per mole per Kelvin, is dimensionally expressed in terms of pressure, volume, and temperature as: \([M^1 L^2 T^{-2} \Theta^{-1}]\), where \(\Theta\) denotes temperature.

For dimensional consistency, the term \(\frac{a}{V^2}\) must have the same dimensions as \(P\) :

\([\frac{a}{V^2}] = [M^1 L^{-1} T^{-2}]\)

From this, the dimensions of \(a\) are derived:

\(a \times [L^{-6}] = [M^1 L^{-1} T^{-2}] \implies [a] = [M^1 L^5 T^{-2}]\)

The term \(b\) represents a volume correction and has dimensions of volume: \([b] = [L^3]\)

Now, we compute the dimensions for the expression \(\frac{a}{b^2}\):

\([\frac{a}{b^2}] = \frac{[M^1 L^5 T^{-2}]}{[L^6]} = [M^1 L^{-1} T^{-2}]\)

The dimensions of \(\frac{a}{b^2}\) are identical to those of \(P\) (pressure).

Therefore, the correct identification is:

Option:

P