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 find the change in enthalpy (\( \Delta H \)) for the given process involving a non-ideal gas, we can use the following formula:

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

Where:
- \( \Delta U \) is the change in internal energy.
- \( \Delta (PV) \) is the change in the product of pressure and volume.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial state: \( P_1 = 1 \, \text{atm} \), \( V_1 = 3 \, \text{L} \), \( T_1 = 200 \, \text{K} \)
- Final state: \( P_2 = 4 \, \text{atm} \), \( V_2 = 5 \, \text{L} \)
- Change in internal energy: \( \Delta U = 40 \, \text{L-atm} \)

2. **Calculate \( \Delta (PV) \):**
- Calculate \( P_1 V_1 \):
\[ P_1 V_1 = 1 \, \text{atm} \times 3 \, \text{L} = 3 \, \text{L-atm} \]
- Calculate \( P_2 V_2 \):
\[ P_2 V_2 = 4 \, \text{atm} \times 5 \, \text{L} = 20 \, \text{L-atm} \]
- Now, find \( \Delta (PV) \):
\[ \Delta (PV) = P_2 V_2 - P_1 V_1 = 20 \, \text{L-atm} - 3 \, \text{L-atm} = 17 \, \text{L-atm} \]

3. **Substitute values into the enthalpy change formula:**
\[ \Delta H = \Delta U + \Delta (PV) = 40 \, \text{L-atm} + 17 \, \text{L-atm} \]

4. **Calculate \( \Delta H \):**
\[ \Delta H = 57 \, \text{L-atm} \]

### Final Answer:
The change in enthalpy of the process is \( \Delta H = 57 \, \text{L-atm} \).

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