Question

For real gases van der Waals equation is written as $\bigg( p + \frac{an^2}{V^2} \bigg) (V - nb) = nRT$ Where 'a' and 'b' are van der Waals constants . Two sets of gases are : (I)${ O2, CO2, H2}$ and $He$ (II) ${CH4, O2}$ and ${H2}$ The gases given in set-I in increasing order of 'b' and gases given in set-II in decreasing order of 'a', are arranged below. Select the correct order from the following :

• A. (I) ${ O2 < He < H2 < CO2 }$ (II) ${H2 > O2 > CH4 }$
• B. (I) ${ H2 < He < O2 < CO2 }$ (II) ${CH4 > O2 > H2 }$
• C. (I) ${ H2 < O2 < He < CO2 }$ (II) ${O2 > CH4 > H2 }$
• D. (I) ${ He < H2 < CO2 < O2 }$ (II) ${CH4 > H2 > O2 }$

Answer 1

The Correct Option is B

The van der Waals equation for real gases is given by:

\left( p + \frac{an^2}{V^2} \right) (V - nb) = nRT

where:

• p is the pressure of the gas,
• V is the volume of the gas,
• T is the temperature of the gas,
• n is the number of moles,
• R is the ideal gas constant,
• a is the van der Waals constant that represents the attraction between particles,
• b is the van der Waals constant that represents the volume occupied by the gas particles.

**For Set I (Increasing Order of 'b'):**

The 'b' value is proportional to the size of the molecules. Lighter gases such as H_2 and He usually have smaller 'b' values compared to heavier gases like CO_2. Therefore, the correct increasing order of 'b' for set I is:

{ H_2 < He < O_2 < CO_2 }

**For Set II (Decreasing Order of 'a'):**

The 'a' value is higher for gases with stronger intermolecular forces. CH_4 generally has significant intermolecular forces due to its structure, followed by O_2, and then H_2 which has very low intermolecular forces. Therefore, the correct decreasing order of 'a' for set II is:

{ CH_4 > O_2 > H_2 }

Accordingly, the correct answer is:

(I) { H_2 < He < O_2 < CO_2 } (II) { CH_4 > O_2 > H_2 }