For the isothernmal expansion of an ideal gas

Options

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

Answer

To solve the question regarding the isothermal expansion of an ideal gas, we need to analyze the changes in internal energy (U) and enthalpy (H) during this process.

### Step-by-Step Solution:

1. **Understanding Isothermal Process**:
- An isothermal process is one where the temperature (T) remains constant throughout the process. For an ideal gas, this means that the internal energy (U) is dependent solely on temperature.

2. **Change in Internal Energy (ΔU)**:
- The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
where \( n \) is the number of moles, \( C_V \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.
- Since the process is isothermal, \( \Delta T = 0 \). Therefore:
\[ \Delta U = n C_V \cdot 0 = 0 \]
- This indicates that the internal energy does not change during the isothermal expansion.

3. **Change in Enthalpy (ΔH)**:
- The change in enthalpy (ΔH) is given by:
\[ \Delta H = \Delta U + \Delta(PV) \]
- For an ideal gas, we can express \( PV \) as \( nRT \). Since T is constant during an isothermal process, the term \( nRT \) remains constant.
- Thus, the change in \( PV \) is also zero:
\[ \Delta(PV) = 0 \]
- Therefore, we have:
\[ \Delta H = \Delta U + 0 = 0 \]

4. **Conclusion**:
- From the above calculations, we find that both the change in internal energy (ΔU) and the change in enthalpy (ΔH) are zero:
\[ \Delta U = 0 \quad \text{and} \quad \Delta H = 0 \]
- This means that both U and H remain constant during the isothermal expansion of an ideal gas.

5. **Final Answer**:
- The correct option is that both U and H are unaltered (remain constant).