For the isothermal expansion of an ideal gas

Options

A. E and H increases
B. E increases but H decreases
C. H increases but E decreases
D. E and H are unaltered

Answer

**Step-by-Step Solution:**

1. **Understanding Isothermal Process:**
- An isothermal process is defined as a thermodynamic process that occurs at a constant temperature. For an ideal gas undergoing isothermal expansion, the temperature (T) remains constant throughout the process.

2. **Change in Internal Energy (ΔE):**
- The change in internal energy (ΔE) of an ideal gas can be expressed using the formula:
\[ \Delta E = nC_v \Delta T \]
- Here, \(n\) is the number of moles, \(C_v\) is the molar specific heat at constant volume, and \(\Delta T\) is the change in temperature. Since the process is isothermal, \(\Delta T = 0\).

3. **Calculating ΔE:**
- Substituting \(\Delta T = 0\) into the equation:
\[ \Delta E = nC_v \cdot 0 = 0 \]
- Therefore, the change in internal energy (ΔE) during isothermal expansion is zero.

4. **Change in Enthalpy (ΔH):**
- The change in enthalpy (ΔH) can be expressed as:
\[ \Delta H = \Delta E + \Delta (PV) \]
- For an ideal gas, we know that \(PV = nRT\). Since the temperature is constant, the product \(PV\) does not change during the isothermal process.

5. **Calculating ΔH:**
- Since both ΔE and Δ(PV) are zero in this case:
\[ \Delta H = 0 + 0 = 0 \]
- Thus, the change in enthalpy (ΔH) is also zero.

6. **Conclusion:**
- In the case of isothermal expansion of an ideal gas, both the change in internal energy (ΔE) and the change in enthalpy (ΔH) are zero. This means that neither the internal energy nor the enthalpy of the system changes during the process.