X and Y are two equal size containers. X contains a gas A at a temperature `75^@ `C and Y contains a gas B at a temperature `40^@ `C. Each gas behaves as an ideal gas and specific heat at constant pressure for both gases has the same value.
It can be concluded that the

Options

A. average translational kinetic energy of a molecule in container X is more than that of a molecule in container Y.
B. average translational kinetic energy of a molecule in container Y is more than that of a molecule in container X
C. average translational kinetic energy of a molecule in container X could be more or could also be less than that of a molecule in container Y.
D. average translational kinetic energy of a molecule in container X is equal to that of a molecule in container Y.

Answer

To solve the problem, we need to analyze the information given about the two gases in containers X and Y, and how their temperatures relate to the average translational kinetic energy of the gas molecules.

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- Container X has gas A at a temperature of \(75^\circ C\).
- Container Y has gas B at a temperature of \(40^\circ C\).

2. **Convert Temperatures to Kelvin**:
- To use the ideal gas laws and kinetic theory, we need to convert the temperatures from Celsius to Kelvin.
- The conversion formula is:
\[ T(K) = T(°C) + 273 \]
- Thus, for gas A:
\[ T_A = 75 + 273 = 348 \, K \]
- For gas B:
\[ T_B = 40 + 273 = 313 \, K \]

3. **Understand the Relationship Between Temperature and Kinetic Energy**:
- According to the kinetic theory of gases, the average translational kinetic energy (\(KE\)) of a gas molecule is given by:
\[ KE = \frac{3}{2} k T \]
- Here, \(k\) is the Boltzmann constant, and \(T\) is the absolute temperature in Kelvin.

4. **Compare the Average Kinetic Energies**:
- Since both gases behave as ideal gases and have the same specific heat at constant pressure, we can directly compare their average kinetic energies based on their temperatures.
- The average kinetic energy of gas A:
\[ KE_A \propto T_A = 348 \, K \]
- The average kinetic energy of gas B:
\[ KE_B \propto T_B = 313 \, K \]

5. **Conclusion**:
- Since \(T_A > T_B\), it follows that:
\[ KE_A > KE_B \]
- Therefore, the average translational kinetic energy of the molecules in container X (gas A) is greater than that in container Y (gas B).

### Final Answer:
The correct conclusion is that the average translational kinetic energy of a molecule in container X (gas A) is greater than that of a molecule in container Y (gas B).

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