What is the number of degrees of freedom of a molecule of a diatomic gas at room temperature?
Answer
Generally a molecule of a diatomic gas possesses 5 degrees of freedom at room temperature. 3 due to translational motion and 2 due to rotational motion.
Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 12.1 and Eq. (12.41) above.
Answer
A given mass of water in vapour state has $1.67 \times 10^3$ times the volume of the same mass of water in liquid state (Ex. 12.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by $10^3$ times the radius increases by $V^{1 / 3}$ or 10 times, i.e., $10 \times 2 \mathring A=20 \mathring A$. So the average distance is $2 \times 20=40 \mathring A$.
The density of water is 1000 $kg m ^{-3}$. The density of water vapour at $100^{\circ} C$ and $1 atm$ pressure is $0.6 kg m ^{-3}$. The volume of a molecule multiplied by the total number gives , what is called, molecular volume. The ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure is $6 \times 10^{-4}$. The volume of a water molecule is $3 \times 10^{-29} m ^3$. What is the average distance between atoms (interatomic distance) in water?
Answer
A given mass of water in vapour state has $1.67 \times 10^3$ times the volume of the same mass of water in liquid state (Ex. 12.1). This is also the increase in the amount of volume available for each molecule of water. When volume increases by $10^3$ times the radius increases by $V^{1 / 3}$ or 10 times, i.e., $10 \times 2 \mathring A=20 \mathring A$. So the average distance is $2 \times 20=40 \mathring A$.
The density of water is 1000 $kg m ^{-3}$. The density of water vapour at $100^{\circ} C$ and $1 atm$ pressure is $0.6 kg m ^{-3}$. The volume of a molecule multiplied by the total number gives , what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.
Answer
For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is $1000 / 0.6=1 /\left(6 \times 10^{-4}\right)$ times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. $6 \times 10^{-4}$.