Chemistry · Thermodynamics · NEET
Expansion work in chemistry is w = -p(external) × ΔV. If the volume does not change, ΔV = 0. Anything multiplied by zero is zero, so w = 0. The gas has no space to expand into and pushes nothing, so it cannot do or receive pressure-volume work. This is the whole reason ΔU = qV works.
The small V just tells you the heat was measured while the volume was held constant. qV is not a new kind of heat; it is ordinary heat q, but with the condition 'volume fixed' attached. NCERT writes it this way so you never forget the condition that makes ΔU = qV true.
Start with the first law: ΔU = q + w. At constant volume w = 0. Substitute: ΔU = q + 0 = q. Since this q was measured at constant volume we call it qV. So ΔU = qV. It is one substitution, nothing more.
No. Internal energy U is always a state function, but heat q is not. It only happens that at constant volume the value of qV becomes equal to ΔU, so qV takes a fixed value for given start and end states. The heat is still path-dependent in general; the constant-volume condition just pins it to ΔU.
In a bomb calorimeter. The steel bomb is sealed, so its volume cannot change. Any heat released by a burning sample is measured at constant volume, so that heat equals ΔU directly. This is why bomb calorimeters give ΔU (internal energy change), not ΔH.
At constant pressure the gas can expand and do work, so some heat leaks out as work; that heat equals ΔH (enthalpy). At constant volume no work is done, so no heat leaks out as work; all of it stays as internal energy, giving ΔU. Different condition, different state function.
At a certain temperature T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then the change in internal energy of the system is:
Consider the reaction 2A(g) + B(g) -> 2D(g), with ΔU° = -10 kJ/mol and ΔS° = -44 J/K at 298 K. Identify the correct ΔG° and the spontaneity at 298 K. (R = 8.31 J/mol/K)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is true only when the volume is held constant so that no pressure-volume work is done (w = 0). Change the condition and it no longer holds; at constant pressure you use ΔH = qP instead.
An isochoric (or isovolumetric) process is one that happens at constant volume. 'Iso' means same and 'choric' relates to volume. Because volume is fixed, no expansion work is done.
No. If total volume is fixed, ΔV = 0, so w = 0 regardless of what gas is added. NCERT notes that adding an inert gas at constant volume does not change partial pressures or the reaction, so no work is involved.
Because it links the measured heat directly to internal energy change with no work correction. In bomb-calorimeter problems, the heat you read off is ΔU. To get ΔH you then add Δn_g RT.
Heat absorbed by the system is +q; heat released is -q. Work done ON the system is +w; work done BY the system is -w. This IUPAC convention is what NEET expects.
