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 calculate the change in enthalpy (ΔH) for a reversible isothermal process at a temperature of T = 300 K, where the volume is increased from Vi = 1 L to Vf = 10 L, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Enthalpy (ΔH)**:
The change in enthalpy (ΔH) for a system can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]
where ΔU is the change in internal energy and Δ(PV) is the change in the product of pressure and volume.

2. **Use the Ideal Gas Law**:
For an ideal gas, the relationship between pressure (P), volume (V), and temperature (T) is given by:
\[ PV = nRT \]
where n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

3. **Determine Δ(PV)**:
Since we are considering a reversible isothermal process, we can express Δ(PV) as:
\[ \Delta(PV) = nR(T_f - T_i) \]
However, since the process is isothermal, the temperature does not change (T_f = T_i = T). Therefore, ΔT = 0, and thus:
\[ \Delta(PV) = 0 \]

4. **Calculate ΔU for an Isothermal Process**:
For an ideal gas undergoing an isothermal process, the change in internal energy (ΔU) is also zero because internal energy for an ideal gas depends only on temperature. Since there is no change in temperature:
\[ \Delta U = 0 \]

5. **Substitute Values into the Enthalpy Equation**:
Now substituting the values of ΔU and Δ(PV) into the enthalpy equation:
\[ \Delta H = \Delta U + \Delta(PV) = 0 + 0 = 0 \]

6. **Conclusion**:
Therefore, the change in enthalpy (ΔH) for this isothermal process is:
\[ \Delta H = 0 \]

### Final Answer:
\[ \Delta H = 0 \text{ kJ} \]