One mole of a non-ideal gas undergoes a change of state `( 2.0 atm, 3.0L, 95 K) rarr (4.0 atm, 5.0L, 245 K)` with a change in internal energy , `Delta U= 30.0 L. atm`. Calculate change in enthalpy of the process in `L. atm`.

Explanation

To calculate the change in enthalpy (ΔH) for the process, we can use the following formula:

\[ \Delta H = \Delta U + nR\Delta T \]

Where:
- ΔH is the change in enthalpy
- ΔU is the change in internal energy
- n is the number of moles of gas
- R is the universal gas constant
- ΔT is the change in temperature

### Step-by-Step Solution:

1. **Identify the given values**:
- Change in internal energy (ΔU) = 30.0 L·atm
- Number of moles (n) = 1 mole (since it is given that we have one mole of gas)
- Initial temperature (T1) = 95 K
- Final temperature (T2) = 245 K
- Universal gas constant (R) = 0.0821 L·atm/(K·mol)

2. **Calculate the change in temperature (ΔT)**:
\[ \Delta T = T2 - T1 = 245 \, K - 95 \, K = 150 \, K \]

3. **Substitute the values into the enthalpy change formula**:
\[ \Delta H = \Delta U + nR\Delta T \]
\[ \Delta H = 30.0 \, L·atm + (1 \, mol)(0.0821 \, L·atm/(K·mol))(150 \, K) \]

4. **Calculate nRΔT**:
\[ nR\Delta T = 1 \cdot 0.0821 \cdot 150 = 12.315 \, L·atm \]

5. **Add ΔU and nRΔT to find ΔH**:
\[ \Delta H = 30.0 \, L·atm + 12.315 \, L·atm = 42.315 \, L·atm \]

6. **Round the final answer**:
\[ \Delta H \approx 42.3 \, L·atm \]

### Final Answer:
The change in enthalpy (ΔH) of the process is approximately **42.3 L·atm**.