Question

Any ideal gas as compared to a real gas at very high pressure occupies:

Answer 1

Hint: The equation which gives the relation between pressure, volume and temperature for ideal gas is, $PV\text{ = }nRT$. However real gases do not obey the above relation as there are extra terms for pressure and volume also called pressure and volume correction. The relation is given by van der waals equation for real gases.


Complete step-by-step answer:
- For ideal gases, the compressibility factor ,Z = 1 so the equation becomes:
 PV = nRT
However, for real gases we follow the Van der Waals equation i.e.
  $[P+\dfrac{a{{n}^{2}}}{{{V}^{2}}}][V-nb]=nRT$
Where,
P stands for pressure,
V stands for volume,
n stands for number of moles,
a is pressure correction constant
b is volume correction constant
T stands for temperature
R stands for universal gas constant
Ideal gases molecules do not have any force of repulsion or attraction at molecular level. However, in case of real gases, the repulsive and attractive forces are significant and to be considered.
When there is high pressure, the distance between the adjacent gas molecules is relatively less. As the gas molecules come closer to each other, the force of repulsions between electrons of adjacent atoms becomes significant.
This leads to an increase in the size of molecules. As a result, it occupies higher volume as compared to an ideal gas subjected to high pressure due to absence of repulsion force.

Therefore, the correct answer is option (B).

Note: Compressibility factor(Z) is a correction factor which describes the deviation of a real gas from its behavior of an ideal gas. It is simply the ratio of molar volume of the gas to the molar volume of an ideal gas subjected to the same identical temperature and pressure.