Question

The unit of Van der Waal's constant "a" can be (1) litrc atm \(\mathrm{mol}^{-1}\) (2) atm litre \(^{2} \mathrm{~mol}^{-2}\) (3) atm litre \(\mathrm{mol}^{-2}\) (4) litre \(\mathrm{mol}^{-1}\)

Short answer

Option (2) atm litre \(^{2} \mathrm{~mol}^{-2})} instructor of ai to 16000.

Step-by-step solution

Understand Van der Waal's Equation

Van der Waal's equation for real gases is given by: \[ \left( P + \frac{a}{V^2} \right) \left( V - b \right) = RT \] Here, 'a' and 'b' are Van der Waal's constants.

Isolate the Constant 'a'

From the equation, isolate the term containing the constant 'a': \[ P + \frac{a}{V^2} = \frac{RT}{V - b} \] Focus on the term \( \frac{a}{V^2} \).

Dimensional Analysis

Perform dimensional analysis for \( a \). The term \( \frac{a}{V^2} \) must have the same units as pressure (P). Assume V to have the units of volume (liters) and a has units that we need to determine. Pressure (P) has units of \( \text{atm} \). Hence, \[ \frac{a}{(liter)^2} = atm \]Therefore, \[ a = atm \times liter^2 \]

Match with Options

Now, find the option that matches the unit \( atm \times liter^2 \). Option (2) matches the derived unit.

Key concepts

Dimensional Analysis

Dimensional analysis is a powerful tool in physics and chemistry to understand the relationships between different physical quantities. It involves checking the consistency of units in equations and can be especially helpful when deriving or verifying formulas.
In this exercise, we used dimensional analysis to determine the units of Van der Waal's constant 'a'.
By isolating 'a' in Van der Waal's equation and comparing its units with the units of pressure, we ensured that the resultant unit was consistent and logical. This method protects against errors in calculations and provides a deeper understanding of the equation's structure.
For example, we found that:
\( \frac{a}{V^2} = \text{Pressure} \)
Because V is volume in liters and Pressure (P) is in atmospheres (atm), the dimensional analysis gave us:
\( \frac{a}{liter^2} = \text{atm} \)
Therefore, \( a = \text{atm} \times \text{liter}^2 \).

Real Gases

Gases often don't behave exactly as predicted by ideal gas laws, especially under high pressure or low temperature. Real gases have interactions between particles and finite volumes, which must be accounted for.
The Van der Waals equation modifies the ideal gas law to account for these real-world deviations:
\( \left( P + \frac{a}{V^2} \right) ( V - b ) = RT \)
Where:

• \