For a reversible process at T = 300K, the volume is increased from `V_(i)=1L` to `V_(f)=10L`. Calculate `Delta H` if the process is isothermal -

Options

A. 11.47 kJ
B. 4.98 kJ
C. 0
D. `-11.47 kJ`

Answer

To solve the problem of calculating the change in enthalpy (ΔH) for an isothermal process where the volume is increased from \( V_i = 1L \) to \( V_f = 10L \) at a constant temperature of \( T = 300K \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Conditions**:
- The process is isothermal, meaning the temperature remains constant at \( T = 300K \).
- The initial volume \( V_i = 1L \) and the final volume \( V_f = 10L \).

2. **Use the Enthalpy Change Formula**:
- The change in enthalpy (ΔH) for a process can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]
- For an ideal gas, \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature.

3. **Consider the Internal Energy Change (ΔU)**:
- For an isothermal process involving an ideal gas, the change in internal energy (ΔU) is given by:
\[ \Delta U = 0 \]
- This is because the internal energy of an ideal gas depends only on temperature, which is constant in this case.

4. **Calculate the Change in PV (Δ(PV))**:
- Since \( PV = nRT \), and \( n \) and \( R \) are constants, we can express the change in \( PV \) as:
\[ \Delta (PV) = P_fV_f - P_iV_i \]
- However, since the process is isothermal and we are not given the specific values for pressure, we can simplify our calculations by noting that for isothermal expansion, the change in enthalpy can also be simplified.

5. **Final Calculation of ΔH**:
- Substituting the values into the enthalpy change formula:
\[ \Delta H = \Delta U + \Delta (PV) = 0 + (nR(T_f - T_i)) \]
- Since \( T_f = T_i = T \) (isothermal), \( (T_f - T_i) = 0 \), thus:
\[ \Delta H = 0 \]

### Conclusion:
The change in enthalpy (ΔH) for the isothermal process is:
\[ \Delta H = 0 \]