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 process, we can use the formula:

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

Where:
- ΔH is the change in enthalpy
- ΔU is the change in internal energy
- Δ(PV) is the change in the product of pressure and volume

### Step 1: Calculate ΔU
We are given that:
\[ \Delta U = 30.0 \, \text{L atm} \]

### Step 2: Calculate Δ(PV)
To find Δ(PV), we need to calculate the initial and final values of PV.

- **Initial State (P1, V1)**:
- P1 = 2.0 atm
- V1 = 3.0 L
- Therefore, \( P1 \cdot V1 = 2.0 \, \text{atm} \cdot 3.0 \, \text{L} = 6.0 \, \text{L atm} \)

- **Final State (P2, V2)**:
- P2 = 4.0 atm
- V2 = 5.0 L
- Therefore, \( P2 \cdot V2 = 4.0 \, \text{atm} \cdot 5.0 \, \text{L} = 20.0 \, \text{L atm} \)

Now, we can calculate Δ(PV):
\[ \Delta (PV) = P2 \cdot V2 - P1 \cdot V1 = 20.0 \, \text{L atm} - 6.0 \, \text{L atm} = 14.0 \, \text{L atm} \]

### Step 3: Substitute values into the ΔH equation
Now we can substitute the values of ΔU and Δ(PV) into the enthalpy change equation:
\[ \Delta H = \Delta U + \Delta (PV) = 30.0 \, \text{L atm} + 14.0 \, \text{L atm} = 44.0 \, \text{L atm} \]

### Final Answer
Thus, the change in enthalpy (ΔH) for the process is:
\[ \Delta H = 44.0 \, \text{L atm} \]

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