Question

What is the dimensional formula of \( ab^{-1} \) in the equation \[ \left( P + \frac{a}{V^2} \right) (V - b) = RT, \] where letters have their usual meaning.

• A. \( [M^0 L^3 T^{-2}] \)
• B. \( [M L^2 T^{-2}] \)
• C. \( [M^{-1} L^5 T^3] \)
• D. \( [M^6 L^7 T^4] \)

Answer 1

The Correct Option is B

Given:

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

where:
- \( P \) is pressure,
- \( V \) is volume,
- \( R \) is the universal gas constant,
- \( T \) is temperature.

Step 1: Dimensions of the Given Quantities
- \( [V] = [b] \), so the dimension of \( b \) is:

\[ [b] = [L^3] \quad (\text{volume}) \]

- The dimensional formula for pressure \( P \) is:

\[ [P] = \left[\frac{F}{A}\right] = \left[\frac{MLT^{-2}}{L^2}\right] = [ML^{-1}T^{-2}]. \]

Step 2: Dimension of \( a \)
From the term \( \frac{a}{V^2} \) having the same dimension as pressure \( P \):

\[ \left[\frac{a}{V^2}\right] = [P] = [ML^{-1}T^{-2}]. \]

Thus, the dimensional formula of \( a \) is:

\[ [a] = [P] \times [V^2] = [ML^{-1}T^{-2}] \times [L^6] = [ML^5T^{-2}]. \]

Step 3: Calculating the Dimensional Formula of \( ab^{-1} \)
The dimensional formula of \( b \) is \( [L^3] \). Therefore, the dimensional formula of \( ab^{-1} \) is:

\[ ab^{-1} = \frac{[a]}{[b]} = \frac{[ML^5T^{-2}]}{[L^3]} = [ML^2T^{-2}]. \]

Therefore, the correct dimensional formula of \( ab^{-1} \) is \( [ML^2T^{-2}] \).

Answer 2

Step 1: Given equation.
\[ \left( P + \frac{a}{V^2} \right)(V - b) = RT \]
This is the Van der Waals equation for real gases.

Step 2: Analyze dimensional consistency.
Both sides of the equation have the same dimensions as \( RT \), which represents energy (work done). Hence, the left-hand side must also have the dimensions of energy.

Step 3: Focus on the term \( \frac{a}{V^2} \).
Since \( P + \frac{a}{V^2} \) has the same dimensions as pressure (P),
\[ [\frac{a}{V^2}] = [P] \] \[ [a] = [P][V^2] \]
Pressure has dimensions \( [P] = [M L^{-1} T^{-2}] \), and volume has \( [V] = [L^3] \).

Thus, \[ [a] = [M L^{-1} T^{-2}] [L^6] = [M L^5 T^{-2}] \]

Step 4: Find dimensions of \( b \).
The term \( (V - b) \) indicates that \( b \) has the same dimensions as volume:
\[ [b] = [L^3] \]

Step 5: Find dimensions of \( a b^{-1} \).
\[ [a b^{-1}] = \frac{[a]}{[b]} = \frac{[M L^5 T^{-2}]}{[L^3]} = [M L^2 T^{-2}] \]

Step 6: Final Answer.
The dimensional formula of \( a b^{-1} \) is:
\[ \boxed{[M L^2 T^{-2}]} \]

Final Answer: \( [M L^2 T^{-2}] \)