One mole of non`-` ideal gas undergoes a change of state `(1.0 atm, 3.0L, 200 K )` to `(4.0 atm, 5.0L,250 K)` with a change in internal energy `(DeltaU)=40 L-atm`. The change in enthalpy of the process in `L-atm`,

Options

A. 43
B. 57
C. 42
D. None of these

Explanation

To solve the problem, we need to calculate the change in enthalpy (ΔH) for the given process involving a non-ideal gas. We can use the relationship between the change in enthalpy, change in internal energy, and the change in the product of pressure and volume.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial Pressure (P1) = 1.0 atm
- Initial Volume (V1) = 3.0 L
- Final Pressure (P2) = 4.0 atm
- Final Volume (V2) = 5.0 L
- Change in Internal Energy (ΔU) = 40 L-atm

2. **Use the Relationship Between ΔH, ΔU, and Δ(PV):**
The change in enthalpy (ΔH) can be expressed as:
\[ \Delta H = \Delta U + \Delta (PV) \]
where Δ(PV) is the change in the product of pressure and volume.

3. **Calculate Δ(PV):**
We can calculate Δ(PV) using the formula:
\[ \Delta (PV) = P2 \cdot V2 - P1 \cdot V1 \]
Substituting the values:
\[ \Delta (PV) = (4.0 \, \text{atm} \times 5.0 \, \text{L}) - (1.0 \, \text{atm} \times 3.0 \, \text{L}) \]
\[ \Delta (PV) = 20.0 \, \text{L-atm} - 3.0 \, \text{L-atm} = 17.0 \, \text{L-atm} \]

4. **Substitute Values into the ΔH Equation:**
Now we can substitute ΔU and Δ(PV) into the equation for ΔH:
\[ \Delta H = \Delta U + \Delta (PV) \]
\[ \Delta H = 40 \, \text{L-atm} + 17.0 \, \text{L-atm} = 57.0 \, \text{L-atm} \]

5. **Final Result:**
The change in enthalpy (ΔH) for the process is:
\[ \Delta H = 57.0 \, \text{L-atm} \]