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:

### Part (a): Calculate the average kinetic energy of translation of the molecules of an ideal gas at 0°C and at 100°C.

1. **Understanding the Formula**: The average kinetic energy (KE) of translation of a molecule in an ideal gas is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is Boltzmann's constant and \( T \) is the absolute temperature in Kelvin.

2. **Convert Celsius to Kelvin**:
- For \( 0°C \):
\[ T = 0 + 273 = 273 \, K \]
- For \( 100°C \):
\[ T = 100 + 273 = 373 \, K \]

3. **Calculate KE at 0°C**:
\[ KE_{0°C} = \frac{3}{2} \times (1.38 \times 10^{-23} \, J/K) \times 273 \, K \]
\[ KE_{0°C} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 273 \]
\[ KE_{0°C} \approx 5.65 \times 10^{-21} \, J \]

4. **Calculate KE at 100°C**:
\[ KE_{100°C} = \frac{3}{2} \times (1.38 \times 10^{-23} \, J/K) \times 373 \, K \]
\[ KE_{100°C} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 373 \]
\[ KE_{100°C} \approx 7.72 \times 10^{-21} \, J \]

### Part (b): Calculate the corresponding energies per mole of the gas.

1. **Using Avogadro's Number**: The energy per mole can be calculated by multiplying the energy per molecule by Avogadro's number \( N_A = 6.02 \times 10^{23} \).

2. **Calculate Energy per Mole at 0°C**:
\[ KE_{0°C, \text{ per mole}} = KE_{0°C} \times N_A \]
\[ KE_{0°C, \text{ per mole}} = (5.65 \times 10^{-21} \, J) \times (6.02 \times 10^{23}) \]
\[ KE_{0°C, \text{ per mole}} \approx 3401 \, J \]

3. **Calculate Energy per Mole at 100°C**:
\[ KE_{100°C, \text{ per mole}} = KE_{100°C} \times N_A \]
\[ KE_{100°C, \text{ per mole}} = (7.72 \times 10^{-21} \, J) \times (6.02 \times 10^{23}) \]
\[ KE_{100°C, \text{ per mole}} \approx 4647 \, J \]

### Final Answers:
- Average kinetic energy at \( 0°C \): \( 5.65 \times 10^{-21} \, J \) per molecule
- Average kinetic energy at \( 100°C \): \( 7.72 \times 10^{-21} \, J \) per molecule
- Energy per mole at \( 0°C \): \( 3401 \, J \)
- Energy per mole at \( 100°C \): \( 4647 \, J \)