Chemistry · Thermodynamics · NEET
Joule took water inside a container with adiabatic (thermally insulated) walls, so no heat could enter or leave. Falling weights turned paddle wheels that stirred the water. This stirring did mechanical work ON the water. He measured the temperature before (state A) and after (state B). The temperature rose (T_B > T_A). So work alone, with zero heat, changed the state of the water.
Work and heat are two different ways to add energy to a system. Stirring puts in energy as WORK, not as heat. That energy stays inside the water and shows up as a higher temperature. This is important for NEET: it proves you can change a system's internal energy using only work, with q = 0.
Joule did the same experiment in many different ways: stirring, passing electric current, compressing, or rubbing. As long as the walls were adiabatic (q = 0), the SAME amount of work always produced the SAME temperature change. The result did not depend on the method or 'path'. Only the start state and end state mattered.
Because adiabatic work is the same for any path between two fixed states, we can call it a property of the states themselves. We define this fixed quantity as the change in internal energy: ΔU = w(adiabatic). A quantity that depends only on the initial and final state (not the path) is a state function. So U is a state function.
The full first law is ΔU = q + w. In an adiabatic process, no heat flows, so q = 0, which leaves ΔU = w. If the walls were diathermic (heat can pass), then q would not be zero and ΔU = w would be wrong. For NEET, always check first whether the container is insulated before writing ΔU = w.
This is the biggest trap. Work in general is a PATH function — different paths give different work. Joule's point is special: ONLY when q = 0 (adiabatic) does the work become path-independent, because it then equals ΔU, and ΔU is a state function. Do not say 'work is always path-independent' — that is wrong.
The correct option for free expansion of an ideal gas under adiabatic condition is:
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the proof that internal energy is a state function and that ΔU = w when q = 0. NEET questions on adiabatic processes, free expansion, and 'well insulated container' problems all rely on this single idea.
It measures WORK. The falling weights do mechanical work on the water. No heat is added because the walls are adiabatic. The temperature rise happens purely from work input.
ΔU = w(adiabatic), which is just the first law ΔU = q + w with q = 0. This defines internal energy change as the adiabatic work between two states.
No. In a normal process work depends on the path (single-step vs multi-step give different work). Work becomes path-independent only in the special adiabatic case, because then it equals the state function ΔU.
