Chemistry · Thermodynamics · NEET
Start with the first law of thermodynamics: ΔU = q + w. The only work here is pressure-volume (expansion) work, given by w = -p_ext·ΔV. At constant volume the volume stays the same, so ΔV = 0. That makes w = -p_ext·(0) = 0. Now substitute w = 0 into the first law: ΔU = q + 0 = q. Since this heat is measured at constant volume, we write it as qV. So ΔU = qV. This is why heat added at constant volume goes fully into internal energy.
The gas does mechanical (expansion) work only when it pushes its boundary outward or gets pushed inward, that is, when volume changes. Work is w = -p_ext·ΔV. If ΔV = 0, the boundary does not move, so no push-pull happens and w = 0. Think of a sealed steel container (like a bomb calorimeter): the walls cannot move, so the gas cannot do or receive expansion work.
The subscript V tells you the heat is exchanged while the volume is held constant (an isochoric process). It is a label, not a multiplication. qV means 'heat at constant volume'. Similarly qP means 'heat at constant pressure'. NEET often uses this subscript to hint which formula to apply.
qV equals ΔU only when volume is constant AND the only work is expansion (pressure-volume) work. If there is other work, like electrical work, or if volume changes, then qV = ΔU no longer holds. For normal NEET reactions in a sealed rigid vessel with only PV work, qV = ΔU is correct.
ΔU (internal energy change) depends only on the start and end states, so it is a state function. Heat q normally depends on the path taken, so it is a path function. But at constant volume, work is forced to zero, so the first law makes qV take exactly one fixed value, ΔU. In this special condition the path is pinned down, so qV behaves like ΔU. It does not mean heat is always a state function.
qV is measured using a bomb calorimeter. A sealed steel vessel (the bomb) sits in a water bath. Because the steel walls cannot move, volume is constant, so no work is done. The heat released by the reaction warms the surrounding water, and the temperature rise gives the heat, which equals ΔU. This is why the bomb calorimeter measures ΔU directly, not ΔH.
Because it is the bridge between measured heat and internal energy. Bomb calorimeters give qV, and qV = ΔU lets you find ΔU. Then you convert ΔU to ΔH using ΔH = ΔU + Δn_g·RT for constant-pressure reactions. Many NEET numericals start from qV = ΔU and end at ΔH, so this step is the foundation.
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:
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.
Yes. Even if gas moles change during the reaction, as long as the reaction runs in a sealed rigid vessel the total volume is fixed, so ΔV = 0, w = 0, and qV = ΔU still holds. The gas cannot expand because the walls do not move.
qV = ΔU is heat at constant volume and equals the internal energy change. qP = ΔH is heat at constant pressure and equals the enthalpy change. Most lab reactions in open flasks are at constant pressure (qP = ΔH); bomb calorimeter reactions are at constant volume (qV = ΔU). They are linked by ΔH = ΔU + Δn_g·RT.
Yes. If the process releases heat (exothermic) at constant volume, qV is negative, so ΔU is negative too. If heat is absorbed (endothermic), qV and ΔU are positive. The sign of qV directly tells you the sign of ΔU.
Yes. It is in Class 11 Chemistry, Chapter Thermodynamics, under 'Applications of the First Law'. NEET regularly tests the first law ΔU = q + w and the constant-volume result qV = ΔU, often combined with bomb calorimeter or ΔH = ΔU + Δn_g·RT problems.
