Question

For non-zero value of force of attraction between gas molecules for large volume, gas equation will be:
a.) PV = nRT - $\dfrac{{{n^2}a}}{V}$
b.) PV = nRT + nbP
c.) P = $\dfrac{{nRT}}{{V - b}}$
d.) PV = nRT

Answer 1

Hint : In a real gas equation, Van der Waals constant ‘a’ accounts for the interactive forces between the particles of the gas and the Van der Waals constant ‘b’ accounts for the volume correction.

Complete step by step answer:
The Van der Waals equation for real gases is:
$(P + \dfrac{{a{n^2}}}{{{V^2}}})(V - nb) = nRT$
In this equation, the Van der Waals constant ‘a’ signifies the intermolecular attractive forces between the particles of the gas. It is also known as the pressure correction term. The Van der Waals constant ‘b’ signifies the effective size of the molecules. It is also known as the volume correction term.
The formula for finding the Van der Waals constant ‘b’ is:
$b = 4 \times {N_A} \times \dfrac{4}{3}\pi {r^3}$

Now, if the gas has non zero value of force of attraction, ‘a’ is significant. And if the volume of the gas is large, the volume correction term can be neglected, i.e. the Van der Waal’s constant ‘b’ can be neglected as the value of ‘b’ is comparatively small.
Therefore, the real gas equation simplifies to:
$(P + \dfrac{{a{n^2}}}{{{V^2}}})V = nRT$
$\Rightarrow PV + \dfrac{{a{n^2}}}{V} = nRT$
$\Rightarrow PV = nRT - \dfrac{{a{n^2}}}{V}$
So, this is the desired equation.

Hence, the correct answer is (A) PV = nRT - $\dfrac{{{n^2}a}}{V}$

Note: A student must know that a small term can only be neglected if it is in addition or multiplication with the comparatively large term. Larger the value of a, larger will be the intermolecular forces between the molecules. The unit of a is given by $atm{{L}^{2}}mo{{l}^{-2}}_{{}}$ and the unit of b is given by $Lmo{{l}^{-1}}_{{}}$.