One mole of an ideal gas undergoes a change of state (2.0) atm, 3.0 L) to (2.0 atm, 7.0 L) with a change in internal energy `(DeltaU) = 30` L-atm. The change in enthalpy `(DeltaH)` of the process in L-atm :

Options

A. 22
B. 38
C. 25
D. None of these

Explanation

To find the change in enthalpy (ΔH) for the given process, we can use the relationship between the change in enthalpy, change in internal energy, and the work done by the system. The formula we will use is:

\[ \Delta H = \Delta U + P \Delta V \]

Where:
- ΔH = Change in enthalpy
- ΔU = Change in internal energy
- P = Pressure (constant in this case)
- ΔV = Change in volume

### Step-by-step solution:

1. **Identify the given values**:
- Change in internal energy (ΔU) = 30 L-atm
- Initial volume (V1) = 3.0 L
- Final volume (V2) = 7.0 L
- Pressure (P) = 2.0 atm

2. **Calculate the change in volume (ΔV)**:
\[ \Delta V = V2 - V1 = 7.0 \, \text{L} - 3.0 \, \text{L} = 4.0 \, \text{L} \]

3. **Substitute the values into the enthalpy formula**:
\[ \Delta H = \Delta U + P \Delta V \]
\[ \Delta H = 30 \, \text{L-atm} + (2.0 \, \text{atm} \times 4.0 \, \text{L}) \]

4. **Calculate the pressure-volume work (PΔV)**:
\[ P \Delta V = 2.0 \, \text{atm} \times 4.0 \, \text{L} = 8.0 \, \text{L-atm} \]

5. **Final calculation of ΔH**:
\[ \Delta H = 30 \, \text{L-atm} + 8.0 \, \text{L-atm} = 38.0 \, \text{L-atm} \]

Thus, the change in enthalpy (ΔH) of the process is **38.0 L-atm**.