One mole of a non-ideal gas undergoes a change of state from (2,0 atm, 3.0 L, 100 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy, `DeltaU=30.0` Latm. The change on enthalpy of process in (L - atm) is

Answer

To find the change in enthalpy (ΔH) for the given process of a non-ideal gas, we will use the relationship between enthalpy, internal energy, and the change in pressure-volume work. The formula we will use is:

\[ \Delta H = \Delta U + \Delta (PV) \]

### Step-by-step Solution:

1. **Identify Given Values:**
- Initial state: \( P_1 = 2 \, \text{atm}, V_1 = 3.0 \, \text{L}, T_1 = 100 \, \text{K} \)
- Final state: \( P_2 = 4.0 \, \text{atm}, V_2 = 5.0 \, \text{L}, T_2 = 250 \, \text{K} \)
- Change in internal energy: \( \Delta U = 30.0 \, \text{L-atm} \)

2. **Calculate the Change in PV:**
- Calculate \( PV \) for the initial and final states:
\[ PV_1 = P_1 \times V_1 = 2 \, \text{atm} \times 3.0 \, \text{L} = 6.0 \, \text{L-atm} \]
\[ PV_2 = P_2 \times V_2 = 4.0 \, \text{atm} \times 5.0 \, \text{L} = 20.0 \, \text{L-atm} \]

3. **Calculate the Change in PV:**
- Now, find \( \Delta (PV) \):
\[ \Delta (PV) = PV_2 - PV_1 = 20.0 \, \text{L-atm} - 6.0 \, \text{L-atm} = 14.0 \, \text{L-atm} \]

4. **Substitute Values into the Enthalpy Equation:**
- Now substitute \( \Delta U \) and \( \Delta (PV) \) into the enthalpy equation:
\[ \Delta H = \Delta U + \Delta (PV) = 30.0 \, \text{L-atm} + 14.0 \, \text{L-atm} \]

5. **Calculate ΔH:**
- Finally, calculate \( \Delta H \):
\[ \Delta H = 30.0 + 14.0 = 44.0 \, \text{L-atm} \]

### Final Answer:
\[ \Delta H = 44.0 \, \text{L-atm} \]

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