One mole of an ideal gas undergoes change of state from (4 L, 3 atm) to (6 L, 5 atm). If the change in internal energy is 45 L-atm then the change of enthalpy for the process is

Options

A. 27 L-atm
B. 45 L-atm
C. 55 L-atm
D. 63 L-atm

Answer

To find the change in enthalpy (ΔH) for the process, we can use the relationship between change in enthalpy, change in internal energy (ΔU), and the work done by the system. The formula we will use is:

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

Where:
- ΔH = change in enthalpy
- ΔU = change in internal energy
- Δ(PV) = change in pressure-volume product

### Step 1: Identify the given values
- Change in internal energy (ΔU) = 45 L-atm
- Initial state: P1 = 3 atm, V1 = 4 L
- Final state: P2 = 5 atm, V2 = 6 L

### Step 2: Calculate the initial and final PV values
- Initial PV (P1V1):
\[ P1V1 = 3 \, \text{atm} \times 4 \, \text{L} = 12 \, \text{L-atm} \]

- Final PV (P2V2):
\[ P2V2 = 5 \, \text{atm} \times 6 \, \text{L} = 30 \, \text{L-atm} \]

### Step 3: Calculate the change in PV (Δ(PV))
\[ \Delta(PV) = P2V2 - P1V1 = 30 \, \text{L-atm} - 12 \, \text{L-atm} = 18 \, \text{L-atm} \]

### Step 4: Substitute values into the enthalpy change equation
Now we can substitute the values into the enthalpy change formula:
\[ \Delta H = \Delta U + \Delta(PV) = 45 \, \text{L-atm} + 18 \, \text{L-atm} = 63 \, \text{L-atm} \]

### Final Answer
The change in enthalpy (ΔH) for the process is:
\[ \Delta H = 63 \, \text{L-atm} \]

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