Two moles of an ideal gas is expanded isothermally and reversibly from 2 litre to 20 litre at 300 K. The enthalpy change (in kJ) for the process is

Options

A. 22.4 kJ
B. `-22.4 kJ`
C. `9.6 kJ`
D. `0 kJ`

Answer

To solve the problem of finding the enthalpy change (ΔH) for the isothermal and reversible expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- The gas is expanded isothermally (constant temperature) and reversibly from a volume of 2 liters to 20 liters at a temperature of 300 K.
- For an ideal gas undergoing an isothermal process, the temperature remains constant.

2. **Recall the Enthalpy Change Formula**:
- The change in enthalpy (ΔH) for an ideal gas can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]
- Where ΔU is the change in internal energy and Δ(PV) is the change in pressure-volume work.

3. **Evaluate the Change in Internal Energy (ΔU)**:
- For an ideal gas, the internal energy (U) depends only on temperature. Since the process is isothermal (constant temperature), the change in internal energy (ΔU) is zero:
\[ \Delta U = 0 \]

4. **Evaluate the Change in PV**:
- The term Δ(PV) can be expressed as:
\[ \Delta(PV) = nR\Delta T \]
- Since the process is isothermal, the change in temperature (ΔT) is also zero:
\[ \Delta T = 0 \]
- Therefore, Δ(PV) is also zero:
\[ \Delta(PV) = nR \cdot 0 = 0 \]

5. **Calculate the Enthalpy Change (ΔH)**:
- Now substituting the values into the enthalpy change formula:
\[ \Delta H = \Delta U + \Delta(PV) = 0 + 0 = 0 \]

6. **Final Answer**:
- The enthalpy change (ΔH) for the process is:
\[ \Delta H = 0 \text{ kJ} \]

### Conclusion:
The enthalpy change for the isothermal expansion of the ideal gas is 0 kJ.