Question

During an isothermal expansion of an ideal gas:

• A. The temperature of the gas increases
• B. The internal energy of the gas remains constant
• C. The gas absorbs heat without doing work
• D. The pressure of the gas remains constant

Hint:

In an isothermal process for an ideal gas: constant temperature → constant internal energy → heat added = work done.

Answer 1

The Correct Option is B

Step 1: In an isothermal process, the temperature of the system remains constant throughout the process. For an ideal gas, internal energy \( U \) depends only on temperature, not on volume or pressure.
Step 2: Therefore, if the temperature is constant during the expansion, the internal energy of the gas does not change: \[ \Delta U = 0 \] Step 3: From the first law of thermodynamics: \[ \Delta Q = \Delta U + W \] Since \( \Delta U = 0 \), it follows that: \[ \Delta Q = W \] This means that the heat absorbed by the gas is entirely used to do work during the expansion.
Why the other options are incorrect:

• (A) Temperature remains constant in isothermal processes.
• (C) The gas does do work; heat is not absorbed without work.
• (D) Pressure changes during expansion to maintain constant temperature.