An insulated vessel contains 1mole of a liquid, molar volume `100mL` at 1bar. When liquid is steeply passed to `100`bar, volume decreases to `99mL`. Find `DeltaH` and `DeltaU` for the process.

Explanation

To solve the problem, we need to find the changes in enthalpy (ΔH) and internal energy (ΔU) for the given process. We will follow these steps:

### Step 1: Identify the Given Data
- Number of moles (n) = 1 mole
- Initial pressure (P1) = 1 bar
- Final pressure (P2) = 100 bar
- Initial volume (V1) = 100 mL = 0.1 L
- Final volume (V2) = 99 mL = 0.099 L

### Step 2: Understand the Process
The process is adiabatic, which means no heat is exchanged with the surroundings (Q = 0). Therefore, the change in internal energy (ΔU) is equal to the work done (W) on the system.

### Step 3: Calculate Work Done (W)
For an adiabatic process, the work done can be calculated using the formula:
\[ W = -P \Delta V \]
Where:
- \( \Delta V = V2 - V1 = 99 \, \text{mL} - 100 \, \text{mL} = -1 \, \text{mL} \)

Now, we need to convert the pressure from bar to mL (since 1 bar = 100 mL):
- \( P2 = 100 \, \text{bar} = 100 \, \text{bar} \times 1 \, \text{mL/bar} = 100 \, \text{mL} \)

Thus, the work done is:
\[ W = -P2 \Delta V = -100 \, \text{bar} \times (-1 \, \text{mL}) = 100 \, \text{bar mL} \]

### Step 4: Calculate Change in Internal Energy (ΔU)
Since \( Q = 0 \):
\[ \Delta U = Q + W = 0 + 100 \, \text{bar mL} = 100 \, \text{bar mL} \]

### Step 5: Calculate Change in Enthalpy (ΔH)
The change in enthalpy can be calculated using the formula:
\[ \Delta H = \Delta U + \Delta (PV) \]
Where:
\[ \Delta (PV) = P2V2 - P1V1 \]

Calculating \( \Delta (PV) \):
- \( P2V2 = 100 \, \text{bar} \times 99 \, \text{mL} = 9900 \, \text{bar mL} \)
- \( P1V1 = 1 \, \text{bar} \times 100 \, \text{mL} = 100 \, \text{bar mL} \)

Now substituting these values:
\[ \Delta (PV) = 9900 \, \text{bar mL} - 100 \, \text{bar mL} = 9800 \, \text{bar mL} \]

Now substituting back into the enthalpy equation:
\[ \Delta H = \Delta U + \Delta (PV) = 100 \, \text{bar mL} + 9800 \, \text{bar mL} = 9900 \, \text{bar mL} \]

### Final Results
- \( \Delta U = 100 \, \text{bar mL} \)
- \( \Delta H = 9900 \, \text{bar mL} \)