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 determine which option is correct for the irreversible expansion of an ideal gas under isothermal conditions, we need to analyze the changes in internal energy and entropy.

1. Internal Energy Change (\(\Delta U\)) for an Ideal Gas:
   • For an isothermal process, the temperature remains constant by definition. In ideal gases, the internal energy (\(U\)) is a function of temperature only. Therefore, for an isothermal process of an ideal gas, the change in internal energy is zero: \(\Delta U = 0\).
2. Total Entropy Change (\(\Delta S_{\text{total}}\)) in an Irreversible Process:
   • The second law of thermodynamics states that the total entropy change, which includes both the system and the surroundings, must increase in an irreversible process. This means \(\Delta S_{\text{total}} \neq 0\); it must be greater than zero.

Given these analyses:

The correct option is: \(\Delta U = 0\) and \(\Delta S_{\text{total}} \neq 0\).

Let's briefly consider why the other options are incorrect:

• \(\Delta U \neq 0, \Delta S_{\text{total}} = 0\): This is incorrect because, for an isothermal process involving an ideal gas, \(\Delta U = 0\) by virtue of constant temperature. Additionally, the total entropy change cannot be zero for an irreversible process.
• \(\Delta U = 0, \Delta S_{\text{total}} = 0\): While the internal energy change is correct, the statement of zero total entropy change is incorrect for an irreversible process.
• \(\Delta U \neq 0, \Delta S_{\text{total}} \neq 0\): This contradicts the zero internal energy change characteristic of an isothermal ideal gas process.

Thus, the nature of an isothermal expansion in an irreversible process of an ideal gas dictates that \(\Delta U = 0\) and \(\Delta S_{\text{total}} \neq 0\).