Question

For the non-zero values of force of attraction between gas molecules, gas equation will be:
A. $PV=nRT-\dfrac{{{n}^{2}}a}{V}$
B. $PV=nRT-nbP$
C. \[PV=nRT\]
D. $P=\dfrac{nRT}{V-b}$

Answer 1

Hint: The ideal gas equation is only applicable for the ideal gasses, and the real gasses does not obey the ideal gas equation in all temperature and pressure, in other words, the real gasses shows ideal behaviour in certain conditions of temperature and pressure.

The Vander Waals equation for real gases takes into account the attractive forces which are present between the particles of gases and so we add some pressure and volume correction terms in the original ideal gas equation.

Complete step by step answer:
The Ideal gas equation is a type of equation which is applicable only for ideal gases hypothetically. It has a number of limitations, as most of the gases deviate from their ideal behaviour when exposed to different conditions of temperature and pressure. It can be expressed as,
\[PV=nRT\]
Where $P$ is the pressure in which the gas is exposed, $V$ is the volume of the gas, $n$ represents the number of moles of the gas under consideration, $R$ is the gas constant, and $T$ represents the temperature at which the ideal gas is exposed. As we know that the particles of gases experience Vander Waals forces of attraction, but the ideal gas law doesn’t take into account those attractions. Moreover, if we look at all the options, we could eliminate option C as it is incorrect for the same reason.
Now we will write the Vander Waals equation which is applicable for real gasses,
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$
Where $P$ is the pressure in which the gas is exposed, $V$ is the volume of the gas, $n$ represents the number of moles of the gas under consideration, $R$ is the gas constant, and $T$ represents the temperature at which the ideal gas is exposed, $a$ represents the intermolecular attractive forces which are present between the molecules or particles of the gas, and $b$ represents the effective size of the constituent gas particles. Here the $\dfrac{a{{n}^{2}}}{{{V}^{2}}}$ is called the pressure correction term and the $nb$ is called the volume correction term.
The Vander Waals constant $b$ can be calculated by using the formula,
$b=4\times {{N}_{A}}\times \dfrac{4}{3}\pi {{r}^{3}}$
Where ${{N}_{A}}$ Avogadro’s constant, and the formula for volume of sphere is $\dfrac{4}{3}\pi {{r}^{3}}$ where $r$ the radius of the sphere is.
Now if we consider the question, it says that the particles of gases have non-zero value of attractive forces which means that the value of $a$ cannot be neglected in the Vander Waals equation which is expressed above. And if the gas has a higher value of volume, the volume correction term can be neglected, meaning, we will neglect the term $b$ because it is a much smaller term than the volume of the gas. So now, we would get,
$PV+\dfrac{a{{n}^{2}}}{V}=nRT$
Now if we remove the bracket on the left side of the equation, we get,
$(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})V=nRT$
Now we will keep the pressure and volume on the left side and take rest of the terms on the right side of the equation, we get,
$PV=nRT-\dfrac{a{{n}^{2}}}{V}$
Now if we consider the options given in the question

So, the correct answer is Option A .

Note: The Vander Waals equation for real gases considers the attractive forces present between the constituent particles of the gases, so pressure and volume correction terms that is $\dfrac{a{{n}^{2}}}{{{V}^{2}}}$ and $nb$ is added to the original ideal gas equation and the modified equation becomes $(P+\dfrac{a{{n}^{2}}}{{{V}^{2}}})(V-nb)=nRT$. In other words, this equation has non-zero value of attractive forces.