The internal energy of an ideal gas depends only on its temperature. Which statement is true of an isothermal (constant-temperature) expansion of an ideal gas against a constant external pressure? Explain. a. \(\Delta E\) is positive. b. \(w\) is positive. c. \(q\) is positive. d. \(\Delta E\) is negative.

The concept of internal energy is a cornerstone in understanding thermodynamics and it specifically refers to the total energy contained within a system due to the kinetic and potential energies of the molecules. In an ideal gas, internal energy is a function of temperature, which means that if there's no change in temperature, the internal energy of the gas remains constant. During an isothermal expansion of an ideal gas, where the temperature is kept constant, the gas maintains constant internal energy, thus \( \Delta E = 0 \).

Even as the gas expands and the volume changes, the speed of the gas particles and their collisions—which dictate the internal energy—do not change. This is crucial in understanding why, despite the expansion, no internal energy is gained or lost in isothermal processes for an ideal gas. Students often struggle with the concept that even though volume changes, if the temperature does not, the internal energy stays the same. This is because temperature is a measure of the average kinetic energy of the particles, and not directly related to the volume of the gas.

As a principle conservation law in physics, the first law of thermodynamics essentially states that energy cannot be created or destroyed, only transformed or transferred. Formally, it is expressed as \( \Delta E = q + w \), linking the change in internal energy of a system (\( \Delta E \)) to the heat added to the system (\( q \)) and the work done by the system (\( w \)).

In an isothermal expansion, while the internal energy remains unchanged (\( \Delta E = 0 \)), work is being done by the gas as it expands, which is accounted for as negative by our chosen sign convention. To keep the internal energy unchanged, heat must therefore flow into the gas (\( q > 0 \)) to compensate for this work done. This fundamental thermodynamic law can be confusing for students because it is necessary to pay attention to the sign convention (work done by the system is negative), and to understand that in the absence of a change in temperature, any work done must be exactly balanced by heat exchange.

The 'work done by gas' is an expression of energy transfer resulting from a gas expanding or contracting within its surroundings. When a gas expands, it applies a force over a distance against external pressure, and that's the work done by the gas. According to the sign conventions in thermodynamics, work done by a system (in this case, an expanding gas) is considered negative. Therefore, during an isothermal expansion against a constant external pressure, \( w \) is negative.

This negativeness can perplex students as they often equate expansion with positive work, not realizing that it's the perspective of the system doing the work that dictates the sign. In our scenario, a negative \( w \) implies that the gas is losing energy by doing work on its surroundings. Yet, since the process is isothermal, this energy loss is offset by an equivalent heat gain (\( q > 0 \)), resulting in no net change in internal energy.