The macroscopic viewpoint
Thermodynamics is a macroscopic science. It describes a bulk system using a handful of directly measurable variables — pressure, volume, temperature, mass — and deliberately ignores the coordinates and velocities of the $\sim 10^{23}$ molecules inside. NCERT stresses that the laws of thermodynamics were framed in the nineteenth century, before the molecular nature of matter was firmly established, precisely because they need no molecular bookkeeping to hold.
This is what separates thermodynamics from mechanics. In mechanics we track the motion of a body as a whole. Thermodynamics is indifferent to that bulk motion: when a bullet is fired, its mechanical kinetic energy changes, but its temperature does not. Only when the bullet strikes wood and stops does its ordered kinetic energy convert into the disordered internal motion of molecules — and the temperature climbs. Temperature is a marker of internal, disordered energy, never of the system's overall velocity.
Internal energy as a state function
Internal energy $U$ is the sum of the kinetic and potential energies of all the molecules of the system, measured in the frame in which the centre of mass is at rest. It includes translational, rotational and vibrational kinetic energy plus the potential energy of intermolecular forces. It excludes the kinetic energy of the system moving as a whole — a beaker of water on a moving train carries no extra $U$ because of the train's speed.
The defining property, in NCERT's words, is that $U$ depends only on the state of the system, not on how that state was achieved. Internal energy is a thermodynamic state variable: its value is fixed once you specify $P$, $V$ and $T$, and it carries no memory of the path taken to arrive there. Take a gas from state 1 to state 2 by any route — slow, fast, via a detour through a third state — and $\Delta U = U_2 - U_1$ is identical every time.
Why ideal-gas internal energy depends only on temperature
For a general substance $U$ depends on the full state. For an ideal gas the picture simplifies dramatically. We neglect the small intermolecular forces, so the molecular potential energy is zero and $U$ is purely the kinetic energy of random molecular motion. Since that kinetic energy is set entirely by temperature, the internal energy of an ideal gas depends only on temperature — not separately on pressure or volume.
The operational consequence is the formula students reach for again and again: $\Delta U = nC_v\,\Delta T$. NCERT derives $C_v=\left(\Delta U/\Delta T\right)$ at constant volume, then drops the constant-volume subscript precisely because $U$ of an ideal gas depends only on $T$. The formula therefore holds for any process of an ideal gas, not only isochoric ones.
Heat and work — energy in transit
There are exactly two ways to change the internal energy of a closed system, illustrated by a gas in a cylinder with a movable piston. Put the cylinder in contact with a hotter body: energy flows in because of the temperature difference — this is heat. Or push the piston: energy is transferred mechanically with no temperature difference involved — this is work. Both raise $U$; reverse either and $U$ falls.
Heat is energy, but NCERT insists it is energy in transit. The distinction is not wordplay. A system stores internal energy; it does not store heat. The statement "a gas in a given state has a certain amount of heat" is, NCERT says verbatim, as meaningless as "a gas in a given state has a certain amount of work." Heat and work name the mode of transfer across the boundary, and the instant the transfer ends the energy is simply part of $U$. Statements like "200 J of heat was supplied" or "the gas did 50 J of work" are perfectly meaningful; "the gas holds 200 J of heat" is not.
State function vs path function
This is the conceptual spine of the topic, and it is best held as a side-by-side contrast rather than prose.
Internal energy $U$ — state function
- A property of the system, with a definite value in each equilibrium state.
- Depends only on the present state $(P, V, T)$, never on history.
- $\Delta U = U_2 - U_1$ is path-independent.
- Stored in the system; "the gas has internal energy $U$" is meaningful.
- For an ideal gas, $U = U(T)$ only; $\Delta U = nC_v\,\Delta T$.
- Extensive variable (doubles if the system is doubled).
Heat $Q$ and work $W$ — path functions
- Not properties of the system — modes of energy transfer.
- Defined only during a process, not in a state.
- $Q$ and $W$ each depend on the path between the two states.
- Energy in transit; "the gas has heat $Q$" is meaningless.
- Heat flows due to a temperature difference; work due to a force through a displacement.
- Only the combination $Q - W = \Delta U$ is path-independent.
The first law $\Delta Q = \Delta U + \Delta W$ ties the three together. Although $\Delta Q$ and $\Delta W$ each vary with the path, their difference $\Delta Q - \Delta W$ equals $\Delta U$ and is therefore forced to be path-independent. This is exactly why a process with $\Delta U = 0$ (an isothermal expansion of an ideal gas, or any complete cycle) gives $\Delta Q = \Delta W$.
The bookkeeping equation $\Delta Q = \Delta U + \Delta W$ is developed in full on the first law of thermodynamics.
Sign conventions for Q and W
NCERT fixes the signs so the first law reads $\Delta Q = \Delta U + \Delta W$. Memorise the table; a flipped sign is worth a full mark lost.
| Quantity | Positive when… | Negative when… |
|---|---|---|
| Heat $Q$ | heat is added to the system | heat is removed from the system |
| Work $W$ (NCERT) | work is done BY the system — the gas expands, $V$ increases | work is done ON the system — the gas is compressed, $V$ decreases |
| Internal energy $\Delta U$ | internal energy increases ($T$ of ideal gas rises) | internal energy decreases ($T$ of ideal gas falls) |
The NIOS module writes the same first law as $\Delta U = \Delta Q - \Delta W$, using $-\Delta W$ to denote work done on the system — algebraically identical to the NCERT form. Note one frequent clash: many chemistry textbooks define $W$ as work done on the system, which inverts the sign in front of the work term. For NEET physics, follow NCERT: $W$ positive means the system does work on the surroundings.
Work of expansion W = ∫P dV
Consider the gas pushing a piston of area $A$ outward by a small distance $dx$ against external pressure $P$. The force on the piston is $F = PA$, and the small work done by the gas is
$\Delta W = F\,dx = P A\,dx = P\,\Delta V,$
since $A\,dx = \Delta V$ is the small change in volume. Summing these slivers across a finite expansion from $V_1$ to $V_2$ gives the general expression
$\displaystyle W = \int_{V_1}^{V_2} P\,dV.$
If the pressure happens to be constant (an isobaric process), the integral collapses to $W = P(V_2 - V_1) = P\,\Delta V$. In an isochoric process the volume is fixed, $\Delta V = 0$, so $W = 0$ — no work is done however much heat is supplied. The actual evaluation of $W$ for isothermal, adiabatic and other paths is handled in detail on thermodynamic processes.
Work as area under the P–V curve
The integral $\int P\,dV$ has a direct graphical meaning. On a $P$–$V$ (indicator) diagram, the work done by the gas equals the area under the process curve, bounded below by the volume axis. Because the shape of the curve between two states depends on the path, so does the enclosed area — which is the geometric statement that work is a path function.
Worked applications
Two short applications show the three quantities working together through the first law.
Find the change in internal energy when 1 g of water is converted to vapour at atmospheric pressure. The latent heat is $2256~\text{J}$; the volume changes from $1~\text{cm}^3$ (liquid) to $1671~\text{cm}^3$ (vapour) at $P = 1.013\times10^5~\text{Pa}$.
Heat added: $\Delta Q = 2256~\text{J}$ (positive, supplied to the system).
Work done by the system against constant atmospheric pressure: $\Delta W = P(V_g - V_l) = 1.013\times10^5 \times (1671-1)\times10^{-6} \approx 169.2~\text{J}$ (positive, the vapour expands).
First law: $\Delta U = \Delta Q - \Delta W = 2256 - 169.2 = 2086.8~\text{J}$. Most of the supplied heat raises the internal energy; only a small fraction goes into pushing back the atmosphere.
A gas is taken from state A to state B by two routes. Along route 1 it absorbs $\Delta Q_1 = 9.35$ cal of heat; along route 2 (an adiabatic route) $22.3~\text{J}$ of work is done on it. Are the two values of $\Delta U$ related?
Adiabatic route: $\Delta Q = 0$, work done on the gas means $\Delta W = -22.3~\text{J}$, so $\Delta U = \Delta Q - \Delta W = 0 - (-22.3) = +22.3~\text{J}$.
Because $U$ is a state function, $\Delta U$ between the same A and B must again be $+22.3~\text{J}$ on route 1. With $\Delta Q_1 = 9.35\times4.19 = 39.2~\text{J}$, the work on route 1 is $\Delta W = \Delta Q - \Delta U = 39.2 - 22.3 = 16.9~\text{J}$ done by the gas — a different work value for a different path, exactly as expected.
Internal energy, heat and work in one breath
- Thermodynamics is macroscopic: a few bulk variables, no molecular bookkeeping.
- Internal energy $U$ = sum of molecular kinetic + potential energy; it is a state function, so $\Delta U$ depends only on the end states.
- For an ideal gas $U = U(T)$ only, giving $\Delta U = nC_v\,\Delta T$ for every process.
- Heat $Q$ and work $W$ are path functions — energy in transit, not stored quantities. A system never "contains" heat.
- NCERT signs: $Q>0$ heat added; $W>0$ work done by the system (expansion). First law: $\Delta Q = \Delta U + \Delta W$.
- Expansion work $W=\int P\,dV$ equals the area under the $P$–$V$ curve; $W = P\,\Delta V$ if $P$ is constant, $W=0$ if $V$ is constant.