Question

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at

a.$\mathbf{0}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$

b.$\mathbf{100}\mathbf{°}\mathit{C}$

What is the translational kinetic energy per mole of an ideal gas at

c.$\mathbf{0}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$

d.$\mathbf{100}\mathbf{°}\mathit{C}$

Short answer

1. The average kinetic energy at is .$5.65\times {10}^{-21}\text{J}$
2. The average kinetic energy at is$7.72\times {10}^{-21}\text{J}$
3. The translational kinetic energy per mole of an ideal gasis$3.40\times {10}^3\text{J}$.
4. The translational kinetic energy per mole of an ideal gas is .$4.65\times {10}^3\text{J}$

Step-by-step solution

Given data

Temperature,

$\begin{array}{c}T=0°\text{C}\\ =273\text{\hspace{0.17em}K}\end{array}$

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

Understanding the concept

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 $1.38\times {10}^{-23}J/molecule\cdot K$.

(a) Determine the average value of the translational kinetic energy of the molecules of an ideal gas at 0.00°C .

The expression for the average kinetic energy is given by,

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

Substitute$1.38\times {10}^{-23}J/molecule\cdot K$for K and $273K$for T into the above equation,

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

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

(b) Determine the average value of the translational kinetic energy of the molecules of an ideal gas at .100.00°C

The expression for the average kinetic energy is given by,

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

Substitute$1.38\times {10}^{-23}J/molecule\cdot K$for K and $273K$for T into the above equation,

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

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

(c) Determine is the translational kinetic energy per mole of an ideal gas at .0.00°C

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

So,

$K=N\times K_{avg}$at $0°C$

Thus,

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

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

(d) Determine is the translational kinetic energy per mole of an ideal gas at 100.00°C

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

So,

$K=N\times K_{avg}$at$100°C$

Thus,

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

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