91Ó°ÊÓ

Temperature,

$\begin{array}{c}T=0°\text{C}\\ =273\text{‿é}\end{array}$

$\begin{array}{c}T=100°\text{C}\\ =373\text{K}\end{array}$

The expression for the average kinetic energy is given by,

$K_{avg}=\frac{3}{2}KT$

Here k is the Boltzmann constant, T is the temperature, $K_{avg}$ is the average kinetic energy.

The value of K is equal to the $1.38×{10}^{−23}\text{ }J/molecule\text{ }â‹\dots K$.

The expression for the average kinetic energy is given by,

$K_{avg}=\frac{3}{2}KT$

Substitute$1.38×{10}^{−23}\text{ }J/molecule\text{ }â‹\dots K$for K and $273\text{ }K$for T into the above equation,

$\begin{array}{c}{K}_{avg}=\frac{3}{2}×1.38×{10}^{−23}×273\\ K_{avg}=5.65×{10}^{−21}\text{J}\end{array}$

Therefore the average kinetic energy at$0.00°C$ is $5.65×{10}^{−21}\text{J}$.

The expression for the average kinetic energy is given by,

$K_{avg}=\frac{3}{2}KT$

Substitute$1.38×{10}^{−23}\text{ }J/molecule\text{ }â‹\dots K$for K and $273\text{ }K$for T into the above equation,

$\begin{array}{c}{K}_{avg}=\frac{3}{2}×1.38×{10}^{−23}×373\\ K_{avg}=7.72×{10}^{−21}\text{J}\end{array}$

Therefore the average kinetic energy at $0.00°C$is .$7.72×{10}^{−21}\text{J}$

To find translational kinetic energy, we have to use the number of moles, which is given as$\text{N}=6.02×{10}^{23}\text{molecules}$

So,

$K=N×K_{avg}$at $0°C$

Thus,

$\begin{array}{c}K=6.03×{10}^{23}×5.65×{10}^{−21}\\ K=3.40×{10}^3\text{J}\end{array}$

Therefore the translational kinetic energy per mole of an ideal gas is $3.40×{10}^3\text{J}$.

To find translational kinetic energy, we have to use the number of moles, which is given as$\text{N}=6.02×{10}^{23}\text{molecules}$

So,

$K=N×K_{avg}$at$100°C$

Thus,

$\begin{array}{c}K=6.03×{10}^{23}×7.72×{10}^{−21}\\ K=4.65×{10}^3\text{J}\end{array}$

Therefore the translational kinetic energy per mole of an ideal gas is .$4.65×{10}^3\text{J}$