Question

Statement: The excluded volume (b) = 4 (volume of one gas molecule)
A. True
B. False

Answer 1

Hint: The equation of state for 1 mole of a gas after pressure correction ‘a’ and volume correction ‘b’ is modified as$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$. Where, P is pressure, V is volume, R gas constant, and T temperature.

Complete step by step answer:
The van der waal equation of state is $\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$, where a and b are van der waal constants. ‘a’ is the measure of magnitude of the attractive forces present in a gas, while ‘b’ is the volume occupied by the molecule of the gas, also called as excluded or co volume.
We have to find that the excluded volume is 4 times the volume of one mole of a gas or not. For this excluded volume correction will be done in the ideal gas equation, $P=\dfrac{RT}{V}$.
We know that volume occupied by molecules is V, real volume of the gas is more than the ideal volume, so corrected volume will be V-b, putting this as volume correction in ideal gas equation, we have,
$P=\dfrac{RT}{V-b}$
Assuming that the molecules are surrounded by sphere of radius = 2r, so, the excluded volume of 2 particles will be,$b\dfrac{1}{2}=4\pi \dfrac{{{d}^{3}}}{3}=8\dfrac{4\pi {{r}^{3}}}{3}$
This value divided by two will give the excluded volume as, $\dfrac{8\dfrac{4\pi {{r}^{3}}}{3}}{2}$or ,
$\dfrac{8V}{2}$ = 4V, so, the excluded volume is 4 times the original volume.
Hence, the statement: excluded volume (b) = 4 (volume of one gas molecule) is true.

Note: The formula for volume of any substance is $\dfrac{4}{3}\pi {{r}^{3}}$. For ‘n’ number of molecules of a gas, the equation of states with pressure and volume correction will become, $\left( P+\dfrac{a{{n}^{2}}}{{{V}^{2}}} \right)\left( V-nb \right)=nRT$.