Question

What is the compressibility factor (Z) for 0.02 mole of a van der Waals's gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible.
Given : $ {\text{RT = 20L atm mo}}{{\text{l}}^{{\text{( - 1)}}}} $ $ {\text{a = 1000 atm }}{{\text{L}}^{{\text{(2)}}}}{\text{ mo}}{{\text{l}}^{{\text{( - 2)}}}} $
(A) 2
(B) 1
(C) 2
(D) 5

Answer 1

Hint :The general gas equation, commonly known as the ideal gas law, is the state equation of a hypothetical ideal gas. Although it has numerous drawbacks, it is a decent approximation of the behaviour of various gases under many situations. Benoît Paul Émile Clapeyron originally articulated it in 1834 as a mixture of empirical Boyle's law, Charles' law, Avogadro's law, and Gay-law. Lussac's.

Complete Step By Step Answer:
The van der Waals equation modifies the Ideal Gas Law and predicts the features of actual gases by defining particles of non-zero volume regulated by pairwise attractive forces. It was developed by Johannes Diderik van der Waals in 1873.
 $ (P + \frac{{a{n^2}}}{{{V^2}}})(V - nb) = nRT $
 P is the pressure, V is the volume, R is the universal gas constant, and T is the absolute temperature in this equation of state. The van der Waals model may be used to create isotherms (pressure vs volume graphs at constant temperature).
Positive values are assigned to the constants a and b, which are unique to each gas. Intermolecular attraction is corrected by the term involving the constant a. Attractive interactions between molecules lower the pressure of a real gas, causing molecules to slow down and reduce collisions with the walls. The greater the attraction between molecules and the easier the gas compresses, the higher the value of a. The excluded volume of the gas, or the volume occupied by the gas particles, is represented by the b term. As these two correction factors approach zero, the van der Waals equation becomes the Ideal Gas Law.
Now we substitute all the given numericals
 $ \left( {0.1 + \frac{{1000 \times {{(0.02)}^2}}}{{{V^2}}}} \right)V = 20 \times 0.02 $
 $ = 0.1{V^2} - 0.4V + 0.4 = 0 $
 $ = {V^2} - 4V + 4 = 0 $
 $ \Rightarrow \quad V = 2L $
 $ Z = \frac{{PV}}{{nRT}} = \frac{{0.1 \times 2}}{{20 \times 0.02}} = 0.5 $
 $ \Rightarrow Z = 0.5 $
Hence option D is correct.

Note :
The compressibility factor (Z), also known as the compression factor or the gas deviation factor in thermodynamics, is a correction factor that indicates how a real gas differs from ideal gas behaviour. The ratio of a gas's molar volume to the molar volume of an ideal gas at the same temperature and pressure is easily defined. It's a helpful thermodynamic characteristic for accounting for real-world behaviour in the ideal gas law.