`2` moles of ideal gas is expanded isothermally & reversibly from `1` litre to `10` litre. Find the enthalpy changes in `KJ mol^(-1)`.

Options

A. 0
B. `11.7`
C. `-11.7`
D. 25

Explanation

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

### Step 1: Understand the relationship between enthalpy (H), internal energy (U), pressure (P), and volume (V).
The enthalpy (H) is defined as:
\[ H = U + PV \]

### Step 2: Write the expression for the change in enthalpy (ΔH).
The change in enthalpy can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]

### Step 3: Use the ideal gas law to express PV.
For an ideal gas, we know that:
\[ PV = nRT \]
where:
- n = number of moles of gas
- R = ideal gas constant
- T = temperature in Kelvin

### Step 4: Analyze the isothermal process.
In an isothermal process, the temperature (T) remains constant. Therefore, the change in temperature (ΔT) is zero:
\[ \Delta T = 0 \]

### Step 5: Determine the change in internal energy (ΔU).
For an ideal gas, the change in internal energy (ΔU) is given by:
\[ \Delta U = nC_V \Delta T \]
Since ΔT = 0, we have:
\[ \Delta U = 0 \]

### Step 6: Determine the change in PV (Δ(PV)).
Since temperature is constant, the product PV also changes, but we need to evaluate it:
\[ \Delta(PV) = nR\Delta T + \Delta(nRT) \]
Again, since ΔT = 0, we find that:
\[ \Delta(PV) = 0 \]

### Step 7: Substitute the values into the ΔH equation.
Now substituting the values we found:
\[ \Delta H = \Delta U + \Delta(PV) = 0 + 0 = 0 \]

### Step 8: Conclusion.
Thus, the enthalpy change (ΔH) for the isothermal expansion of the gas is:
\[ \Delta H = 0 \, \text{KJ mol}^{-1} \]

### Final Answer:
The enthalpy change for the process is **0 KJ mol^(-1)**.

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