The magnitude of enthalpy changes for irreversible adiabatic expansion of a gas from `1 L` to `2 l` is `Delta H_(1)` and for reversible adiabatic expansion for the same expansion is `Delta H_(2)`. Then

Options

A. `DeltaH_(1) gt DeltaH_(2)`
B. `DeltaH_(1) lt DeltaH_(2)`
C. `DeltaH_(1)=Delta H_(2)`, enthalpy being a state function `(Delta H_(1)=Delta H_(2))`
D. `Delta H_(1)=Delta E_(1) & Delta H_(2)=Delta E_(2)` where `Delta E_(1) & DeltaE_(2)` are magnitudes of change in internal energy of gas in these expansions respectively.

Explanation

To solve the problem, we need to analyze the enthalpy changes during the irreversible and reversible adiabatic expansions of a gas. Here’s the step-by-step solution:

### Step 1: Understand the Processes
- **Irreversible Adiabatic Expansion**: This process occurs without heat exchange with the surroundings and does not follow the equilibrium conditions.
- **Reversible Adiabatic Expansion**: This process also occurs without heat exchange, but it follows equilibrium conditions throughout the expansion.

### Step 2: Work Done in Each Process
- In reversible processes, the work done is maximized compared to irreversible processes. Thus, the magnitude of work done in the reversible process is greater than that in the irreversible process.

### Step 3: Temperature Changes
- Since the work done is greater in the reversible process, the final temperature after expansion in the reversible process will be lower than that in the irreversible process. This means:
- \( T_{final, reversible} < T_{final, irreversible} \)

### Step 4: Change in Temperature
- The change in temperature (\( \Delta T \)) can be defined as:
- \( \Delta T = T_{final} - T_{initial} \)
- Since the final temperature is lower in the reversible process, the change in temperature for the reversible process will be greater than that for the irreversible process:
- \( \Delta T_{reversible} > \Delta T_{irreversible} \)

### Step 5: Relationship Between Enthalpy Change and Temperature Change
- The change in enthalpy (\( \Delta H \)) is directly proportional to the change in temperature (\( \Delta T \)):
- \( \Delta H \propto \Delta T \)
- Therefore, since \( \Delta T_{reversible} > \Delta T_{irreversible} \), it follows that:
- \( \Delta H_{reversible} > \Delta H_{irreversible} \)

### Step 6: Conclusion
- From the above reasoning, we conclude that:
- \( \Delta H_2 > \Delta H_1 \)
- Thus, the correct order is \( \Delta H_2 > \Delta H_1 \).

### Final Answer
The correct order is \( \Delta H_2 > \Delta H_1 \).

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