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 problem involves determining the dimensions of the quantity \(\frac{a}{b^2}\) in the equation of state for a real gas given by:

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

First, let's analyze the dimensions of each term:

1. The term \(P\) (pressure) has dimensions: \([M^1 L^{-1} T^{-2}]\).
2. The volume \(V\) has dimensions: \([L^3]\).
3. The universal gas constant \(R\) has dimensions of energy per mole per Kelvin, which is generally expressed in terms of pressure, volume, and temperature: \([M^1 L^2 T^{-2} \Theta^{-1}]\), where \(\Theta\) represents the temperature dimension.

The term \(\frac{a}{V^2}\) should have the same dimensions as \(P\), which implies:

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

Thus, the dimensions of \(a\) can be derived as follows:

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

Now, the term \(b\) is a volume correction and is typically related to the volume: \([b] = [L^3]\)

So for the expression \(\frac{a}{b^2}\), we calculate:

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

It turns out that the dimensions of \(\frac{a}{b^2}\) match those of \(P\) (pressure).

Therefore, the correct answer is:

Option:

P

 

Answer 2

In the given equation of state for a real gas:

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

the term \( \frac{a}{V^2} \) must have the same dimensions as pressure \( P \) since it is being added to \( P \).

The dimensional formula of pressure \( P \) is:

\([P] = [F][A^{-1}] = [M][L^{-1}][T^{-2}],\)

where \( F \) is force and \( A \) is area. Therefore, the dimensions of \( \frac{a}{V^2} \) must also be the same as \( P \).

Since \( \frac{a}{V^2} \) has the same dimensions as pressure

The correct option is (D) : P