Work, Heat & Internal Energy

Internal Energy as a State Function

When a chemical system loses or gains energy, we describe it through a single quantity that sums every form of energy it holds — chemical, electrical, mechanical and all others. In thermodynamics this total is called the internal energy, denoted $U$. According to NCERT §5.1.4, the internal energy of a system can change when heat passes into or out of it, when work is done on or by it, or when matter enters or leaves it.

To see why $U$ behaves as a state function, consider an adiabatic system — water enclosed in a thermos flask whose walls forbid heat exchange. Take the gas or water in state A at temperature $T_A$ with internal energy $U_A$. We can raise it to state B at $T_B$ in two distinct ways: by doing 1 kJ of mechanical work (churning with paddles), or by doing 1 kJ of electrical work (an immersion rod). Joule's experiments between 1840 and 1850 established that the same amount of adiabatic work produces the same change of state — the same temperature rise — irrespective of the path.

This path-independence is precisely what defines a state function. The adiabatic work needed to move between two states equals the difference in internal energy:

$$\Delta U = U_2 - U_1 = w_\text{ad}$$

Other familiar state functions are volume $V$, pressure $p$ and temperature $T$. If a system is taken from $25\,^\circ\text{C}$ to $35\,^\circ\text{C}$, the change is $+10\,^\circ\text{C}$ whether it is heated directly or cooled first and then warmed. Because state functions depend only on the present state and not on history, their changes are route-independent — a property we exploit constantly in the first law of thermodynamics.

Heat and Work: Two Currencies of Energy

A closed system has only two channels through which energy crosses its boundary. The first is work; the second is heat, $q$ — the energy exchanged as a result of a temperature difference between system and surroundings.

Suppose we reproduce the same change of state as before, but now through thermally conducting walls instead of an adiabatic wall. Water at $T_A$ in a copper container is enclosed in a large reservoir at $T_B$. Energy now flows as heat, and the heat absorbed can be measured from the temperature difference $T_B - T_A$. When no work is done at constant volume, the entire internal-energy change is accounted for by heat:

$$\Delta U = q \quad (\text{at constant volume, } w = 0)$$

Combining both channels gives the general statement. For a change of state brought about by both work and heat:

$$\Delta U = q + w \tag{5.1}$$

Here is the deep point: $q$ and $w$ individually depend on how the change is carried out — they are path functions — yet their sum $q + w = \Delta U$ depends only on the initial and final states. This equation is the mathematical statement of the first law of thermodynamics, the law of conservation of energy: the energy of an isolated system is constant.

The IUPAC Sign Convention

Signs in thermodynamics are not arbitrary — they encode the direction in which energy flows. NCERT chemistry follows the IUPAC convention throughout, and getting it wrong is the single most common error in NEET numericals.

In words: energy added to the system is positive, energy leaving is negative. The internal energy of the system increases when work is done on it ($w_\text{ad}$ positive) and decreases when the system does work. Likewise, $q$ is positive when heat is transferred from surroundings to system and negative when heat flows the other way.

Pressure-Volume Work

In chemistry the work that matters most is mechanical pressure-volume work — the work associated with a change in volume against an external pressure. Consider one mole of an ideal gas in a cylinder fitted with a frictionless piston of cross-sectional area $A$. Let the external pressure be $p_\text{ext}$ and let the piston move a distance $l$.

Force on the piston is $p_\text{ext} \cdot A$, and the volume change is $\Delta V = l \times A = (V_f - V_i)$. The work done on the system is force times distance:

$$w = p_\text{ext} \cdot A \cdot l = -p_\text{ext}\,\Delta V = -p_\text{ext}(V_f - V_i) \tag{5.2}$$

The negative sign is what makes the result obey the IUPAC convention. During compression, $(V_f - V_i)$ is negative, so $-p_\text{ext}\Delta V$ is positive — correctly signalling that work is done on the gas. During expansion, $\Delta V$ is positive, so $w$ is negative — work is done by the gas. The sign emerges automatically from the geometry.

If the external pressure is not constant but changes in a series of finite steps, the work done is summed over all steps as $-\sum p\,\Delta V$. In the limit of infinitesimal steps — where $p_\text{ext}$ at each stage is always infinitesimally different from the gas pressure — the sum becomes an integral, which is the route to reversible work.

Reversible vs Irreversible Work

A reversible process is one carried out so that it could, at any moment, be reversed by an infinitesimal change. It proceeds infinitely slowly through a continuous series of near-equilibrium states, with the system and surroundings always in near balance. Any process that is not reversible — for instance expansion against a single fixed external pressure — is an irreversible process.

For a reversible change we set $p_\text{ext} = (p_\text{in} \pm \mathrm{d}p)$, and since $\mathrm{d}p \cdot \mathrm{d}V$ is negligibly small, the work integrates as:

$$w_\text{rev} = -\int_{V_i}^{V_f} p_\text{in}\,\mathrm{d}V \tag{5.4}$$

Substituting the ideal gas relation $p = nRT/V$ and holding $T$ constant (isothermal) gives the reversible isothermal work expression:

$$w_\text{rev} = -nRT\ln\frac{V_f}{V_i} = -2.303\,nRT\log\frac{V_f}{V_i} \tag{5.5}$$

Worked Examples

The clearest way to see the difference between irreversible and reversible work is to expand the same gas between the same two volumes by both routes and compare the magnitudes.

Free and Isothermal Expansion

Free expansion is the expansion of a gas into a vacuum, where $p_\text{ext} = 0$. Putting $p_\text{ext} = 0$ into $w = -p_\text{ext}\Delta V$ immediately gives $w = 0$ — and this holds whether the process is reversible or irreversible, because there is simply no opposing pressure to do work against.

For the isothermal free expansion of an ideal gas, Joule showed experimentally that $q = 0$ as well. The first law then forces:

$$\Delta U = q + w = 0 + 0 = 0$$

For an isothermal change that is not free, the temperature is constant so $\Delta U = 0$ still holds for an ideal gas, but now heat and work are non-zero and exactly cancel. NCERT §5.2.1 summarises both cases:

The pattern to internalise: in an isothermal expansion all the heat absorbed is paid out as work done by the gas; in an adiabatic change, with $q = 0$, the work comes entirely at the expense of internal energy. These special cases are the workhorses of NEET numericals on this chapter.

Why work and heat are path functions

A subtle but examinable idea hides inside equation 5.1. Although $\Delta U$ between two fixed states is fixed, the split of that change between $q$ and $w$ is not. Take a gas from state $i$ to state $f$: do it reversibly and you obtain one pair of $(q, w)$ values; do it irreversibly against a low constant pressure and you obtain a different pair — yet $q + w = \Delta U$ comes out identical both times.

This is exactly why $q$ and $w$ are called path functions: their individual values depend on the route, not merely on the endpoints. The two worked examples above demonstrate this concretely — the same $2\ \text{L} \to 10\ \text{L}$ expansion delivered $8\ \text{L atm}$ of work irreversibly but about $39\ \text{L atm}$ reversibly, and the heat absorbed adjusted to match in each case so that $\Delta U$ stayed zero. State functions ($U$, $H$, $V$, $p$, $T$) are written as differences; path functions ($q$, $w$) are reported only for a specified process.