The coefficient of viscosity for hot air is

The correct option is Greater than the coefficient of viscosity for cold air.

Let us understand the step-by-step reasoning behind this answer:

1. Viscosity in Gases: Unlike liquids, where viscosity is primarily due to cohesive forces between molecules, viscosity in gases is governed by the transfer of momentum between layers of gas through molecular collisions. According to the kinetic theory of gases, the coefficient of viscosity ($\eta$) of a gas is given by the relation:

$\eta =\frac{1}{3}\rho \lambda \overline{v}$

where:
- $\rho$ is the density of the gas,
- $\lambda$ is the mean free path of the gas molecules, and
- $\overline{v}$ is the average speed (or root-mean-square speed) of the gas molecules.

2. Dependence on Temperature: The mean free path $\lambda$ is inversely proportional to the density $\rho$ (i.e., $\lambda \propto \frac{1}{\rho }$). Therefore, the product $\rho \lambda$ is independent of the density (and pressure) of the gas. This simplifies the dependence of viscosity to the average molecular speed:
$\eta \propto \overline{v}$

3. Relationship with Absolute Temperature: The average speed of gas molecules is directly proportional to the square root of the absolute temperature ($T$):
$\overline{v}\propto \sqrt{T}$

Consequently, the coefficient of viscosity of a gas is directly proportional to the square root of its absolute temperature:
$\eta \propto \sqrt{T}$

4. Conclusion: Since hot air has a higher temperature than cold air, the average kinetic energy and speed of its molecules are higher, leading to more frequent momentum transfer. Thus, the coefficient of viscosity for hot air is greater than that for cold air.