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 determining 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**:
- We have 2 moles of an ideal gas expanding isothermally (at constant temperature) from 2 liters to 20 liters at a temperature of 300 K.

2. **Recall the Enthalpy Formula**:
- The enthalpy (H) of a system is given by the equation:
\[ H = E + PV \]
- Where E is the internal energy, P is the pressure, and V is the volume.

3. **Isothermal Process Implications**:
- In an isothermal process for an ideal gas, the internal energy (E) depends only on temperature. Since the temperature is constant (300 K), the change in internal energy (ΔE) is zero:
\[ ΔE = 0 \]

4. **Calculate the Change in Enthalpy**:
- Since ΔE = 0, we can substitute this into the enthalpy equation:
\[ ΔH = ΔE + Δ(PV) \]
- For an ideal gas, the term PV can be expressed as:
\[ PV = nRT \]
- Since n (number of moles) and T (temperature) are constant during the isothermal expansion, the change in PV (Δ(PV)) is also zero:
\[ Δ(PV) = 0 \]

5. **Final Calculation**:
- Therefore, substituting these values into the enthalpy change equation gives:
\[ ΔH = 0 + 0 = 0 \]

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

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