There is `1 mol` liquid (molar volume `100 ml`) in an adiabatic container initial, pressure being `1` bar Now the pressure is steeply increased to `100` bar, and the volume decreased by `1 ml` under constant pressure of `100` bar. Calculate `Delta H` and `Delta E`. [Given `1 "bar"=10^(5)N//m^(2)`]

Explanation

To solve the problem, we need to calculate the change in enthalpy (ΔH) and the change in internal energy (ΔU) for the given system. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Initial and Final Conditions
- **Initial Volume (V_initial)** = 100 ml = 0.1 L
- **Final Volume (V_final)** = 100 ml - 1 ml = 99 ml = 0.099 L
- **Number of Moles (n)** = 1 mol
- **Initial Pressure (P_initial)** = 1 bar = 10^5 N/m²
- **Final Pressure (P_final)** = 100 bar = 10^6 N/m²

### Step 2: Calculate Change in Internal Energy (ΔU)
For an adiabatic process, the change in internal energy (ΔU) can be calculated using the work done (W) on the system. The work done during a volume change at constant pressure can be expressed as:

\[ W = -P_{ext} \Delta V \]

Where:
- \( \Delta V = V_{final} - V_{initial} = 99 \, \text{ml} - 100 \, \text{ml} = -1 \, \text{ml} = -0.001 \, \text{L} \)

Substituting the values into the equation:

\[ W = -100 \, \text{bar} \times (-0.001 \, \text{L}) = 100 \, \text{bar} \times 0.001 \, \text{L} = 0.1 \, \text{bar L} \]

Since \( 1 \, \text{bar L} = 10^5 \, \text{J} \):

\[ W = 0.1 \, \text{bar L} = 0.1 \times 10^5 \, \text{J} = 10000 \, \text{J} \]

Thus, the change in internal energy is:

\[ \Delta U = W = 10000 \, \text{J} \]

### Step 3: Calculate Change in Enthalpy (ΔH)
The change in enthalpy can be calculated using the formula:

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

Where:

\[ \Delta (PV) = P_{final} \cdot V_{final} - P_{initial} \cdot V_{initial} \]

Calculating \( \Delta (PV) \):

\[ \Delta (PV) = (100 \, \text{bar} \cdot 99 \, \text{ml}) - (1 \, \text{bar} \cdot 100 \, \text{ml}) \]

Converting ml to L:

\[ \Delta (PV) = (100 \cdot 99 \times 10^{-3} \, \text{L}) - (1 \cdot 100 \times 10^{-3} \, \text{L}) = (9.9 \, \text{bar L}) - (0.1 \, \text{bar L}) = 9.8 \, \text{bar L} \]

Now converting \( 9.8 \, \text{bar L} \) to Joules:

\[ \Delta (PV) = 9.8 \times 10^5 \, \text{J} \]

Now substituting back to find ΔH:

\[ \Delta H = \Delta U + \Delta (PV) = 10000 \, \text{J} + 9.8 \times 10^5 \, \text{J} = 10000 \, \text{J} + 980000 \, \text{J} = 990000 \, \text{J} \]

### Final Results
- **ΔU = 10000 J**
- **ΔH = 990000 J**