A system containing real gas changes it's state from state `-1` to state -`2`.
State -1`( 2atm, 3L,300K)`
State-2 `(5 atm, 4L,500K)`
If change in internal energy `= 30 L atm` then calculate change in enthalpy.

Options

A. `44 L atm`
B. `35 L atm`
C. `40 L atm`
D. None of these

Answer

To calculate the change in enthalpy (ΔH) for the given system, we can use the following relationship:

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

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

### Step 1: Identify the given values
From the problem statement, we have:
- State 1: \( P_1 = 2 \, \text{atm}, V_1 = 3 \, \text{L}, T_1 = 300 \, \text{K} \)
- State 2: \( P_2 = 5 \, \text{atm}, V_2 = 4 \, \text{L}, T_2 = 500 \, \text{K} \)
- Change in internal energy \( \Delta U = 30 \, \text{L atm} \)

### Step 2: Calculate Δ(PV)
To calculate Δ(PV), we need to find the values of \( PV \) at both states.

\[ PV_1 = P_1 \times V_1 = 2 \, \text{atm} \times 3 \, \text{L} = 6 \, \text{L atm} \]

\[ PV_2 = P_2 \times V_2 = 5 \, \text{atm} \times 4 \, \text{L} = 20 \, \text{L atm} \]

Now, we can calculate the change in \( PV \):

\[ \Delta(PV) = PV_2 - PV_1 = 20 \, \text{L atm} - 6 \, \text{L atm} = 14 \, \text{L atm} \]

### Step 3: Substitute values into the enthalpy equation
Now we can substitute the values of \( \Delta U \) and \( \Delta(PV) \) into the enthalpy equation:

\[ \Delta H = \Delta U + \Delta(PV) = 30 \, \text{L atm} + 14 \, \text{L atm} \]

### Step 4: Calculate ΔH
\[ \Delta H = 30 + 14 = 44 \, \text{L atm} \]

### Final Answer
The change in enthalpy \( \Delta H \) is \( 44 \, \text{L atm} \).