Question

Van der Waals equation for gas is stated as, $p={\displaystyle \frac{nRT}{V-nb}}-a{\left({\displaystyle \frac{n}{b}}\right)}^2$. This equation reduces to the perfect gas equation, $p={\displaystyle \frac{nRT}{V}}$, when ______________.
(A) temperature is sufficiently high and pressure is low
(B) temperature is sufficiently low and pressure is high
(C) both temperature and pressure are very high
(D) both temperature and pressure are very low

Answer 1

Hint: Ideal gas has negligible intermolecular forces of attraction between the gaseous particles. However, Real gas does not follow the ideal gas equation. This is because they have noticeable intramolecular forces of attraction or repulsion between the gas particles. Increase in the kinetic energy and reducing the volume of gas may allow the gas to act as an ideal gas.

Complete Solution :
We have been provided that Van der Waals equation for gas is stated as,$p={\displaystyle \frac{nRT}{V-nb}}-a{\left({\displaystyle \frac{n}{b}}\right)}^2$,
We need to find the condition to reduce it in the perfect gas equation,$p={\displaystyle \frac{nRT}{V}}$,

So, for that:
 We know that: Van Der Waals equation is written as follows,
 $(p+{\displaystyle \frac{a{n}^2}{v^2}})(V-nb)=nRT$
The equation can be rewritten as
$P={\displaystyle \frac{nRT}{V-nb}}-a{\left({\displaystyle \frac{n}{v}}\right)}^2$,
- Where v is volume, P is pressure, T is temperature, n is a mole of gas and R is the universal gas constant and a and b are van der Waals constant.
- Ideal gas particles have negligible attractive forces but real gases have small forces of attraction or repulsion between the gas particles. Ideal gas does not have a volume but real gases have. Thus, there is a possibility that real gas can behave like an ideal gas.

- With the increase in temperature, the kinetic energy of particles increases and decreases the potential energy of gaseous particles. Because of this increase in kinetic energy the intermolecular forces go on decreasing, and at a certain temperature and pressure it becomes negligible.at low pressure, the volume occupied by the gas is high such that the individual particle becomes negligible concerning volume. this real gas can behave as an ideal gas,
- When the temperature is high and pressure is low.
On increasing temperature, the equation, and decrease in the volume the equation is reduced to $\text{ PV = nRT }$ .
So, this is usually true when the pressure is low and the temperature is high.
So, the correct answer is “Option A”.

Note: the real gas equation is also written as $\text{ PV = zRT }$ , where z is known as the compressibility factor. For an ideal gas, the z is 1 .but for real gas the compressibility factor is less or greater than the unity. For example, Nitrogen gas deviates from the ideal behaviour at the low temperature of $\text{ 100 K }$. But on increases in the temperature it approaches the ideal behaviour.