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 solve the problem, we need to calculate the change in enthalpy (ΔH) for the process involving the ideal gas. We will use the formula:

\[ \Delta H = \Delta U + (P_2V_2 - P_1V_1) \]

Where:
- \(\Delta H\) = change in enthalpy
- \(\Delta U\) = change in internal energy
- \(P_1, V_1\) = initial pressure and volume
- \(P_2, V_2\) = final pressure and volume

### Step 1: Identify the given values
- Initial state: \(V_1 = 4 \, \text{L}\), \(P_1 = 3 \, \text{atm}\)
- Final state: \(V_2 = 6 \, \text{L}\), \(P_2 = 5 \, \text{atm}\)
- Change in internal energy: \(\Delta U = 45 \, \text{L-atm}\)

### Step 2: Calculate \(P_1V_1\) and \(P_2V_2\)
- Calculate \(P_1V_1\):
\[ P_1V_1 = 3 \, \text{atm} \times 4 \, \text{L} = 12 \, \text{L-atm} \]

- Calculate \(P_2V_2\):
\[ P_2V_2 = 5 \, \text{atm} \times 6 \, \text{L} = 30 \, \text{L-atm} \]

### Step 3: Substitute values into the enthalpy equation
Now substitute the values into the enthalpy equation:
\[ \Delta H = \Delta U + (P_2V_2 - P_1V_1) \]
\[ \Delta H = 45 \, \text{L-atm} + (30 \, \text{L-atm} - 12 \, \text{L-atm}) \]

### Step 4: Simplify the expression
Calculate the difference and then add to \(\Delta U\):
\[ \Delta H = 45 \, \text{L-atm} + (30 - 12) \, \text{L-atm} \]
\[ \Delta H = 45 \, \text{L-atm} + 18 \, \text{L-atm} \]
\[ \Delta H = 63 \, \text{L-atm} \]

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