Question

For irreversible expansion of an ideal gas under isothermal condition, the correct option is

• A. ∆U≠0, ∆S_total=0
• B. ∆U=0, ∆S_total=0
• C. ∆U≠0, ∆S_total≠0
• D. ∆U=0, ∆S_total≠0

Answer 1

The Correct Option is D

To solve the question about the irreversible expansion of an ideal gas under isothermal conditions, let's consider the thermodynamic principles involved.

Concept Explanation: 

• Isothermal Process: In an isothermal process, the temperature (T) of the system remains constant. For an ideal gas, the internal energy (U) depends only on temperature. Therefore, if the temperature doesn't change, the internal energy remains constant, i.e., \(\Delta U = 0\).
• Irreversible Process: In an irreversible process, the system is not in equilibrium, and entropy is produced. This means that the total change in entropy of the universe (\( \Delta S_{\text{total}} \)) is greater than zero for irreversible processes.

Analysis of Options:

1. \(\Delta U \neq 0, \Delta S_{\text{total}} = 0\): This is incorrect because for an isothermal process involving an ideal gas, \(\Delta U = 0\). Also, \(\Delta S_{\text{total}} = 0\) cannot be correct for an irreversible process.
2. \(\Delta U = 0, \Delta S_{\text{total}} = 0\): This is incorrect. While \(\Delta U = 0\) is correct for an isothermal process, \(\Delta S_{\text{total}} = 0\) is true only for reversible processes, not irreversible ones.
3. \(\Delta U \neq 0, \Delta S_{\text{total}} \neq 0\): This is incorrect because for an isothermal process of an ideal gas, \(\Delta U\) should be zero.
4. \(\Delta U = 0, \Delta S_{\text{total}} \neq 0\): This is the correct option. In an isothermal, irreversible expansion of an ideal gas, \(\Delta U = 0\) because the temperature is constant, and \(\Delta S_{\text{total}} \neq 0\) because entropy increases in an irreversible process.

Conclusion: Therefore, the correct answer is \(\Delta U = 0, \Delta S_{\text{total}} \neq 0\), which is option D.