Two moles of an ideal gas is expanded isothermally and reversibly from 1L to 10 L at 300 K. The enthalpy change (in kJ) for the process is

Options

A. 11.4 kJ
B. `-11.4 kJ`
C. `0 kJ`
D. `4.8 kJ`

Answer

To find the enthalpy change for the isothermal expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**: We are dealing with an isothermal (constant temperature) expansion of an ideal gas. The enthalpy change (ΔH) for an ideal gas during an isothermal process is zero because enthalpy is a function of temperature and the temperature remains constant.

2. **Identify Given Values**:
- Number of moles (n) = 2 moles
- Initial volume (V_i) = 1 L
- Final volume (V_f) = 10 L
- Temperature (T) = 300 K

3. **Recall the Enthalpy Change Formula**: For an ideal gas, the change in enthalpy (ΔH) during an isothermal process can be expressed as:
\[ \Delta H = nC_p\Delta T \]
where \(C_p\) is the molar heat capacity at constant pressure and \(\Delta T\) is the change in temperature.

4. **Calculate the Change in Temperature**: Since the process is isothermal, \(\Delta T = 0\) (the temperature does not change).

5. **Substitute Values into the Enthalpy Change Formula**:
\[ \Delta H = nC_p \cdot 0 = 0 \]

6. **Conclusion**: The enthalpy change (ΔH) for the isothermal expansion of the gas is:
\[ \Delta H = 0 \text{ kJ} \]

### Final Answer:
The enthalpy change for the process is **0 kJ**.