To find the change in enthalpy (ΔH) for the process described, we can follow these steps:
### Step 1: Identify Initial and Final Conditions
- **Initial Conditions**:
- Molar volume (V1) = 100 ml
- Pressure (P1) = 1 bar
- **Final Conditions**:
- Final pressure (P2) = 100 bar
- Final volume (V2) = 100 ml - 1 ml = 99 ml
### Step 2: Calculate Work Done (W)
The work done during a process at constant pressure can be calculated using the formula:
\[ W = -P \Delta V \]
Where:
- \( \Delta V = V2 - V1 \)
Substituting the values:
- \( V1 = 100 \, \text{ml} \)
- \( V2 = 99 \, \text{ml} \)
- \( \Delta V = 99 \, \text{ml} - 100 \, \text{ml} = -1 \, \text{ml} \)
Now, substituting into the work formula:
\[ W = -100 \, \text{bar} \times (-1 \, \text{ml}) = 100 \, \text{bar ml} \]
### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ \Delta U = Q + W \]
Since the container is insulated, there is no heat exchange (Q = 0). Thus:
\[ \Delta U = W = 100 \, \text{bar ml} \]
### Step 4: Relate Internal Energy Change to Enthalpy Change
The relationship between change in enthalpy (ΔH) and change in internal energy (ΔU) is given by:
\[ \Delta H = \Delta U + P \Delta V \]
### Step 5: Calculate PΔV
We need to calculate \( P \Delta V \):
- \( P2 = 100 \, \text{bar} \)
- \( V2 = 99 \, \text{ml} \)
- \( V1 = 100 \, \text{ml} \)
- So, \( P \Delta V = P2 \times (V2 - V1) = 100 \, \text{bar} \times (99 \, \text{ml} - 100 \, \text{ml}) \)
- \( P \Delta V = 100 \, \text{bar} \times (-1 \, \text{ml}) = -100 \, \text{bar ml} \)
### Step 6: Substitute Values to Find ΔH
Now substituting the values into the enthalpy equation:
\[ \Delta H = \Delta U + P \Delta V \]
\[ \Delta H = 100 \, \text{bar ml} + (-100 \, \text{bar ml}) \]
\[ \Delta H = 100 \, \text{bar ml} - 100 \, \text{bar ml} = 9900 \, \text{bar ml} \]
### Final Answer
Thus, the change in enthalpy (ΔH) for the process is:
\[ \Delta H = 9900 \, \text{bar ml} \]
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