Given, Avogadro's number `N = 6.02 xx 10^23` and Boltzmann's constant `k = 1.38 xx 10^-23 J//K`.
(a) Calculate the average kinetic energy of translation of the molecules of an ideal gas at `0^@ C and at 100^@ C`.
(b) Also calculate the corresponding energies per mole of the gas.

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Understand the formula for average kinetic energy
The average kinetic energy (KE) of a molecule in an ideal gas is given by the formula:
\[ KE = \frac{3}{2} k T \]
where:
- \( k \) is Boltzmann's constant (\( 1.38 \times 10^{-23} \, \text{J/K} \))
- \( T \) is the absolute temperature in Kelvin.

### Step 2: Convert temperatures from Celsius to Kelvin
- For \( 0^\circ C \):
\[ T_1 = 0 + 273 = 273 \, \text{K} \]

- For \( 100^\circ C \):
\[ T_2 = 100 + 273 = 373 \, \text{K} \]

### Step 3: Calculate average kinetic energy at \( 0^\circ C \)
Using the formula:
\[ KE_1 = \frac{3}{2} k T_1 \]
Substituting the values:
\[ KE_1 = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (273 \, \text{K}) \]

Calculating:
\[ KE_1 = \frac{3}{2} \times 1.38 \times 10^{-23} \times 273 \]
\[ KE_1 = 5.65 \times 10^{-21} \, \text{J} \, \text{(per molecule)} \]

### Step 4: Calculate average kinetic energy at \( 100^\circ C \)
Using the formula:
\[ KE_2 = \frac{3}{2} k T_2 \]
Substituting the values:
\[ KE_2 = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (373 \, \text{K}) \]

Calculating:
\[ KE_2 = \frac{3}{2} \times 1.38 \times 10^{-23} \times 373 \]
\[ KE_2 = 7.72 \times 10^{-21} \, \text{J} \, \text{(per molecule)} \]

### Step 5: Calculate the corresponding energies per mole of the gas
To find the energy per mole, we multiply the kinetic energy per molecule by Avogadro's number (\( N = 6.02 \times 10^{23} \)):
- For \( 0^\circ C \):
\[ KE_{1, \text{mole}} = KE_1 \times N \]
\[ KE_{1, \text{mole}} = (5.65 \times 10^{-21} \, \text{J}) \times (6.02 \times 10^{23}) \]
Calculating:
\[ KE_{1, \text{mole}} = 3401 \, \text{J/mol} \]

- For \( 100^\circ C \):
\[ KE_{2, \text{mole}} = KE_2 \times N \]
\[ KE_{2, \text{mole}} = (7.72 \times 10^{-21} \, \text{J}) \times (6.02 \times 10^{23}) \]
Calculating:
\[ KE_{2, \text{mole}} = 4647 \, \text{J/mol} \]

### Final Answers:
(a) Average kinetic energy of translation of the molecules:
- At \( 0^\circ C \): \( 5.65 \times 10^{-21} \, \text{J} \)
- At \( 100^\circ C \): \( 7.72 \times 10^{-21} \, \text{J} \)

(b) Corresponding energies per mole of the gas:
- At \( 0^\circ C \): \( 3401 \, \text{J/mol} \)
- At \( 100^\circ C \): \( 4647 \, \text{J/mol} \)