Question

For irreversible expansion of an ideal gas under isothermal conditions, which of the following is correct?

• A. \( \Delta U > 0, \, q > 0 \)
• B. \( \Delta U = 0, \, q = -w \)
• C. \( \Delta U = 0, \, q = w \)
• D. \( \Delta U < 0, \, q < 0 \)

Hint:

In isothermal processes for ideal gases, the internal energy (\( \Delta U \)) remains constant (\( \Delta U = 0 \)). The heat absorbed by the system (\( q \)) is always equal to the negative of the work done by the system (\( w \)).

Answer 1

The Correct Option is B

For an irreversible expansion of an ideal gas under isothermal conditions, the temperature of the gas remains constant throughout the process. This condition plays a crucial role in determining the behavior of the internal energy (\( \Delta U \)) of the gas.
Step 1: First Law of Thermodynamics. The First Law of Thermodynamics is given by the equation: \[ \Delta U = q + w, \] where: - \( \Delta U \) is the change in the internal energy of the system, - \( q \) is the heat exchanged by the system, - \( w \) is the work done by the system on the surroundings.
Step 2: Isothermal Process for an Ideal Gas. In an **isothermal process** for an ideal gas, the temperature is constant. Since the internal energy of an ideal gas depends only on temperature (and not on volume or pressure), the **change in internal energy** (\( \Delta U \)) for an isothermal process is zero: \[ \Delta U = 0. \] Thus, applying this to the First Law of Thermodynamics: \[ 0 = q + w. \] Rearranging this equation: \[ q = -w. \] This implies that the heat absorbed by the system is equal in magnitude to the work done by the system, but with the opposite sign.
Step 3: Work and Heat Direction. During the expansion of the gas: - The work (\( w \)) done by the gas is positive because the gas is expanding and pushing against external pressure. - To maintain a constant temperature during this expansion, the gas must absorb heat (\( q \)) equal in magnitude but opposite in sign to the work done, so \( q = -w \).
Conclusion: Thus, in the case of an irreversible isothermal expansion of an ideal gas: \[ \Delta U = 0, \quad q = -w. \] Therefore, the correct answer is \( \mathbf{(2)} \).