Internal Energy as a State Function
Every thermodynamic system stores energy — chemical, electrical, mechanical, the translational, vibrational and rotational motion of its molecules, and the energy of its electrons and nuclei. The sum of all these contributions is the internal energy, denoted $U$. NCERT introduces $U$ as the quantity that represents the total energy of a chemical system and that may change when heat flows across the boundary, when work is done, or when matter is exchanged.
The decisive property of $U$ is that it is a state function: its value depends only on the present state of the system, not on the path used to reach that state. Joule established this experimentally between 1840 and 1850. He took an insulated (adiabatic) vessel of water and raised its temperature in two completely different ways — once by churning it with paddles (mechanical work) and once with an immersion heater (electrical work). A given amount of work, no matter how it was delivered, produced the same change of state as measured by the same temperature rise.
That path-independence lets us define internal energy through the adiabatic work required to move between two states: $\Delta U = U_2 - U_1 = w_{ad}$. Because the result is the same for every adiabatic route, $U$ qualifies as a state function, standing alongside the other familiar state functions $p$, $V$ and $T$.
We cannot assign an absolute value to the internal energy of a system; we can only measure changes, $\Delta U$. This distinguishes the thermodynamic property energy from a purely mechanical property such as volume, whose absolute value is unambiguous.
Statement and the Master Equation
In the general case, the state of a system is changed both by doing work on it and by transferring heat. NCERT writes the change in internal energy for this combined case as the central equation of the chapter:
$$\Delta U = q + w \qquad (5.1)$$
For a given change between two fixed states, $q$ and $w$ can each take many values depending on how the change is carried out — they are not state functions. Yet their sum, $q + w = \Delta U$, depends only on the initial and final states. This is the mathematical statement of the first law of thermodynamics. In words, the NCERT and NIOS texts state it identically:
The energy of an isolated system is constant. Energy can neither be created nor destroyed — it only changes from one form to another.
An immediate consequence follows for an isolated system. If neither heat nor work crosses the boundary, $q = 0$ and $w = 0$, so $\Delta U = 0$: the internal energy of an isolated system is conserved. The first law is therefore the law of conservation of energy, made specific to a thermodynamic system.
IUPAC Sign Convention
Because $\Delta U = q + w$ is an equation in signed quantities, getting the signs of $q$ and $w$ right is the single most important skill in applying the first law. NCERT and NIOS both adopt the IUPAC convention for chemical thermodynamics, in which anything that increases the internal energy of the system carries a positive sign.
| Quantity | Positive (+) | Negative (−) | Effect on U |
|---|---|---|---|
| Heat, q | Heat absorbed by the system from surroundings | Heat released by the system to surroundings | +q raises U; −q lowers U |
| Work, w | Work done on the system (compression) | Work done by the system (expansion) | +w raises U; −w lowers U |
As a concrete illustration drawn from NIOS, suppose a change absorbs 50 kJ of heat and the system spends 30 kJ doing work on the surroundings. Then $q = +50\ \text{kJ}$ (heat in) and $w = -30\ \text{kJ}$ (work out), so $\Delta U = (+50) + (-30) = +20\ \text{kJ}$. The system gains 20 kJ of internal energy while its surroundings lose the same 20 kJ.
The old physics convention vs. the IUPAC convention
Older physics books write the first law as $\Delta U = q - w$, where $w$ is the work done by the system. NCERT Chemistry uses the IUPAC form $\Delta U = q + w$, where $w$ is the work done on the system. Both are correct — they merely define $w$ with opposite signs. Mixing the two is the most common source of wrong answers.
In NEET Chemistry always use $\Delta U = q + w$ with $w = -p_{ext}\,\Delta V$. For an expansion ($\Delta V > 0$) this makes $w$ negative, exactly as the convention demands.
Energy Bookkeeping: q and w into the System
The first law is best pictured as an accounting balance for the energy stored inside the system. Two arrows can deposit energy into the store $U$ — a heat arrow $q$ and a work arrow $w$ — and the convention fixes which direction counts as positive.
The pressure–volume work that a gas does is captured by $w = -p_{ext}\,\Delta V$. The negative sign is what reconciles the mechanics with the IUPAC convention: in a compression $\Delta V = (V_f - V_i)$ is negative, so $w$ comes out positive (work done on the gas), while in an expansion $\Delta V$ is positive, so $w$ is negative (work done by the gas). Substituting this into the first law gives the working form $\Delta U = q - p_{ext}\,\Delta V$, the starting point for analysing each process.
The terms $q$, $w$ and $U$ are defined in detail in Work, Heat and Internal Energy, including reversible vs. irreversible $pV$-work.
First-Law Form for the Four Processes
The power of $\Delta U = q + w$ is that, by recognising which quantity vanishes in a given process, you collapse the master equation into a simple, memorable form. The four standard processes for an ideal gas each set one term to zero.
| Process | Condition | What is zero | First-law form |
|---|---|---|---|
| Isothermal | $T$ constant ($\Delta T = 0$) | $\Delta U = 0$ (ideal gas) | $q = -w$ (all heat in → work out) |
| Adiabatic | No heat exchange ($q = 0$) | $q = 0$ | $\Delta U = w_{ad}$ |
| Isochoric | $V$ constant ($\Delta V = 0$) | $w = 0$ | $\Delta U = q_V$ |
| Isobaric | $p$ constant | neither (use $w = -p\Delta V$) | $\Delta U = q_p - p\Delta V$; $\;q_p = \Delta H$ |
Isothermal process — ΔU = 0
For an ideal gas the internal energy depends only on temperature. Holding $T$ constant means $\Delta T = 0$, so $\Delta U = 0$. The first law then gives $q = -w$: every joule of heat the gas absorbs is delivered as work to the surroundings. For a reversible isothermal expansion of $n$ moles, NCERT derives the work term explicitly as $w_{rev} = -nRT\ln\dfrac{V_f}{V_i} = -2.303\,nRT\log\dfrac{V_f}{V_i}$, and hence $q = -w = 2.303\,nRT\log\dfrac{V_f}{V_i}$.
Adiabatic process — q = 0
An adiabatic process exchanges no heat with the surroundings; the boundary is an adiabatic wall. Setting $q = 0$ in the first law leaves $\Delta U = w_{ad}$. Work done on the gas raises both its internal energy and temperature, while work done by the gas in expansion lowers them — which is exactly why a compressed gas heats up and an expanding one cools.
Isochoric process — w = 0
At constant volume $\Delta V = 0$, so the $pV$-work $w = -p_{ext}\,\Delta V$ vanishes. The first law reduces to $\Delta U = q_V$, where the subscript $V$ records that the heat was exchanged at constant volume. This is the principle behind the bomb calorimeter, in which the heat measured at fixed volume is the internal-energy change of the reaction.
Isobaric process and free expansion
At constant pressure the gas does work $w = -p\,\Delta V$, so $\Delta U = q_p - p\Delta V$; here $q_p$ equals the enthalpy change $\Delta H$, developed in the next section. A special limiting case is free expansion — expansion into a vacuum where $p_{ext} = 0$. No work is done ($w = 0$), and Joule found experimentally that for an ideal gas $q = 0$ as well, so $\Delta U = 0$ and the temperature does not change.
Relation to Enthalpy at Constant Pressure
Most chemical reactions are carried out in open vessels at constant atmospheric pressure, not at constant volume. The first law tells us that at constant volume the measured heat is $\Delta U$ ($\Delta U = q_V$), but it does not directly give the heat at constant pressure. That gap is exactly why enthalpy is defined.
Enthalpy is the state function $H = U + pV$, so $\Delta H = \Delta U + \Delta(pV)$. At constant pressure $\Delta H = \Delta U + p\,\Delta V$. Substituting the constant-pressure first law $\Delta U = q_p - p\,\Delta V$ gives a clean cancellation:
$$\Delta H = (q_p - p\,\Delta V) + p\,\Delta V = q_p$$
Thus $\Delta H = q_p$: the heat exchanged at constant pressure is the change in enthalpy. This pairs neatly with the constant-volume result $\Delta U = q_V$. For reactions involving gases, the two are linked through $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas — the relation NEET tested verbatim in 2023.
Insulated container does not mean ΔU = 0
A "well-insulated" or "thermos-flask" container only forces $q = 0$ (adiabatic). It does not set $\Delta U = 0$. With $q = 0$, the first law becomes $\Delta U = w = -p_{ext}\,\Delta V$, so an irreversible expansion against a finite external pressure gives a nonzero, negative $\Delta U$. Only an isothermal change of an ideal gas guarantees $\Delta U = 0$.
Read the wording carefully: "insulated" → $q = 0$; "isothermal / constant temperature" → $\Delta U = 0$ (ideal gas). They are not the same.
Worked Examples
A system absorbs 701 J of heat from the surroundings and does 394 J of work on the surroundings. Calculate the change in internal energy. (NIOS §9.7-type)
Heat is absorbed, so $q = +701\ \text{J}$. Work is done by the system, so it is negative: $w = -394\ \text{J}$. Applying the first law,
$\Delta U = q + w = (+701) + (-394) = +307\ \text{J}.$
The internal energy increases by 307 J. The gas could not keep all 701 J of the heat it absorbed because it spent 394 J of it pushing back the surroundings.
A gas expands 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. Find $\Delta U$ in joules. (NEET 2017)
The container is insulated, so the process is adiabatic and $q = 0$. The first law reduces to $\Delta U = w = -p_{ext}\,\Delta V$.
$\Delta V = 4.50 - 2.50 = 2.00\ \text{L}, \quad w = -2.5 \times 2.00 = -5.0\ \text{L atm}.$
Converting with $1\ \text{L atm} = 101.3\ \text{J} \approx 101\ \text{J}$:
$\Delta U = w = -5.0 \times 101 = -505\ \text{J}.$
The internal energy falls by 505 J — the gas cooled, since it did expansion work with no heat supply. Answer: $-505\ \text{J}$.
Two litres of an ideal gas at 10 atm expands isothermally at 25°C into a vacuum until its volume is 10 L. How much heat is absorbed and how much work is done? (NCERT Problem 5.2)
Expansion into a vacuum means $p_{ext} = 0$, so the work is $w = -p_{ext}\,\Delta V = 0$. With no opposing pressure, no work is done. Because the expansion is isothermal for an ideal gas, $\Delta U = 0$, and the first law gives $q = -w = 0$.
Both $q$ and $w$ are zero: no heat is absorbed and no work is done during free expansion of an ideal gas.
First Law in One Screen
- First law: $\Delta U = q + w$ — the energy of an isolated system is constant (conservation of energy).
- $U$ is a state function ($\Delta U$ is path-independent); $q$ and $w$ are not.
- IUPAC signs: $+q$ = heat into system, $+w$ = work done on system; use $w = -p_{ext}\,\Delta V$.
- Isothermal (ideal gas): $\Delta U = 0$, so $q = -w$.
- Adiabatic: $q = 0$, so $\Delta U = w_{ad}$. Isochoric: $w = 0$, so $\Delta U = q_V$.
- Isobaric: $q_p = \Delta H$; and $\Delta H = \Delta U + \Delta n_g RT$ for gases.
- Free / adiabatic free expansion of an ideal gas: $q = 0$, $w = 0$, $\Delta T = 0$, $\Delta U = 0$.