One mole of non - ideal gas undergoes a change of state (2.0 atm , 3 .0 L , 95 K `rarr` (4.0 atm , 5.0 L , 245 K) with a change in internal energy , `DeltaU = 30.0L` atm . The change in enthalpy `(DeltaH)` of the process in L atm is

Options

A. `40.0`
B. `22.0`
C. `44.0`
D. `48.0`

Answer

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial state (P1, V1, T1):
- P1 = 2.0 atm
- V1 = 3.0 L
- T1 = 95 K
- Final state (P2, V2, T2):
- P2 = 4.0 atm
- V2 = 5.0 L
- T2 = 245 K
- Change in internal energy (ΔU) = 30.0 L·atm

2. **Calculate Δ(PV):**
- We can calculate the change in PV using the formula:
\[ \Delta(PV) = P2 \cdot V2 - P1 \cdot V1 \]
- Substitute the values:
\[ \Delta(PV) = (4.0 \, \text{atm} \cdot 5.0 \, \text{L}) - (2.0 \, \text{atm} \cdot 3.0 \, \text{L}) \]
- Calculate:
\[ \Delta(PV) = (20.0 \, \text{L} \cdot \text{atm}) - (6.0 \, \text{L} \cdot \text{atm}) = 14.0 \, \text{L} \cdot \text{atm} \]

3. **Use the Relationship Between ΔH and ΔU:**
- The relationship is given by:
\[ \Delta H = \Delta U + \Delta(PV) \]
- Substitute the values:
\[ \Delta H = 30.0 \, \text{L} \cdot \text{atm} + 14.0 \, \text{L} \cdot \text{atm} \]
- Calculate:
\[ \Delta H = 44.0 \, \text{L} \cdot \text{atm} \]

4. **Final Answer:**
- The change in enthalpy (ΔH) of the process is **44.0 L·atm**.