Question

Question: Explain why the plot of \({\rm{PV}}\) for \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) differs from that of an ideal gas.

Short answer

When the pressure of a real gas, such as \({\rm{C}}{{\rm{O}}_{\rm{2}}}\), rises, attractive forces reduce the force of collision between molecules and container walls, resulting in a lower pressure than an ideal gas. As a result, \({\rm{PV}}\) is less than it would be in a perfect gas.

The volume of the gas reduces as the pressure rises, and the gas molecules clash with each other and the container wall more often, increasing the pressure. At high pressures, \({\rm{PV}}\) outperforms a perfect gas.

Step-by-step solution

Step 1: Define Gas

One of the four fundamental states of matter, a gas is made up of particles with no definite volume or structure.

Explanation for difference 

When the pressure of a real gas, such as \({\rm{C}}{{\rm{O}}_{\rm{2}}}\), rises, the attractive forces reduce the force of collision between the molecules and the container walls, lowering the pressure exerted in comparison to an ideal gas. As a result, \({\rm{PV}}\) is lower than that of an ideal gas.

The volume of the gas reduces with greater pressures, and the gas molecules smash with each other and the container wall more often, increasing the pressure. At high pressure, \({\rm{PV}}\) exceeds that of an ideal gas.

Therefore, When the pressure of a real gas, such as \({\rm{C}}{{\rm{O}}_{\rm{2}}}\), rises, the attractive forces minimise the force of collision between the molecules and the container walls, resulting in a lower pressure than in an ideal gas. As a result, \({\rm{PV}}\) is smaller than in a perfect gas.

With higher pressures, the volume of the gas decreases, and the gas molecules collide with each other and the container wall more often, raising the pressure. \({\rm{PV}}\) outperforms an ideal gas at high pressures.