First Law of Thermodynamics

The law as energy conservation

The internal energy $U$ of a system can change in only two ways: heat can flow across its boundary, and work can be done across that boundary. The first law of thermodynamics is the bookkeeping statement that ties these three quantities together. Following NCERT, let $\Delta Q$ be the heat supplied to the system by the surroundings, $\Delta W$ the work done by the system on the surroundings, and $\Delta U$ the change in internal energy. Conservation of energy then demands

$$ \Delta Q = \Delta U + \Delta W $$

Read in words: the energy $\Delta Q$ supplied to the system goes partly to increase the internal energy $\Delta U$ and the rest into work $\Delta W$ done on the environment. This is nothing more profound than the conservation of energy applied to a system once heat transfer to and from the surroundings is included. It is not derived from a deeper law within thermodynamics; it is the conservation principle itself, extended to count heat as a genuine mode of energy transfer.

The same content is often written in the rearranged form $\Delta U = \Delta Q - \Delta W$, which reads "the change in internal energy equals the heat added minus the work done by the system." Both forms are identical algebra; NEET problems use whichever is convenient. If the gas in a cylinder is pushed against a constant external pressure $P$, the work it does is $\Delta W = P\,\Delta V$, and the law specialises to $\Delta Q = \Delta U + P\,\Delta V$.

The energy-flow picture

Picture a fixed mass of gas in a cylinder fitted with a movable piston. There are exactly two ways to change its internal energy, and the first law accounts for both. Put the cylinder in contact with a hotter body and heat flows in — that is $\Delta Q$. Let the gas push the piston outward and it does work on the surroundings — that is $\Delta W$. Whatever heat enters and does not leave as work is stored as a rise in internal energy.

NCERT illustrates the split with a concrete number. To convert 1 g of water at atmospheric pressure from liquid to vapour requires $\Delta Q = 2256\,\text{J}$ (the latent heat). The volume jumps from $1\,\text{cm}^3$ to $1671\,\text{cm}^3$, so the work done against atmospheric pressure is

$$ \Delta W = P\,(V_g - V_l) = 1.013\times10^5 \times 1670\times10^{-6} \approx 169.2\,\text{J} $$

The first law then fixes the rest as internal energy: $\Delta U = 2256 - 169.2 = 2086.8\,\text{J}$. Most of the heat supplied goes into the internal energy of the new vapour phase; only a small slice is spent doing expansion work. This single example shows the law doing exactly what it is for — splitting one measured heat input into a stored part and a work part.

Sign conventions

The first law is only as reliable as the signs you feed it. NCERT and NIOS agree on a single convention, summarised below. The cardinal rule: $\Delta W$ in $\Delta Q = \Delta U + \Delta W$ is the work done by the system, which is positive when the gas expands.

Two equivalent equations follow from the convention, and confusing them is the most common slip. The NCERT/expansion form treats work done by the system as positive: $\Delta Q = \Delta U + \Delta W$. The NIOS form writes $\Delta U = \Delta Q - \Delta W$ from the same convention — identical content. As long as $\Delta W$ means "work done by the system" throughout a problem, the two never disagree.

Path-dependence and state functions

A system can move from an initial state $(P_1, V_1)$ to a final state $(P_2, V_2)$ along infinitely many paths. NCERT notes that since $U$ is a state variable, $\Delta U$ depends only on the initial and final states — never on the route. Heat $\Delta Q$ and work $\Delta W$, by contrast, are not state variables: each depends on the path chosen. The first law makes a sharp claim about their combination: although $\Delta Q$ and $\Delta W$ each vary path to path, the difference $\Delta Q - \Delta W$ is always equal to the same $\Delta U$, hence always path-independent.

This is the deepest content of the law beyond mere accounting. It tells you that "heat content" and "work content" of a state are meaningless phrases — a gas in a given state does not "contain" a certain amount of heat. But "internal energy of a state" is perfectly meaningful, because $U$ is fixed by the state alone.

Cp, Cv and the Mayer relation

For a gas, the heat needed to raise the temperature by one kelvin depends on the conditions under which heating occurs, because work may or may not be done. This gives a gas two distinct molar specific heats, defined in NCERT §11.6: $C_v$ at constant volume and $C_p$ at constant pressure. The first law links them in a remarkably clean relation.

Start from the law for one mole, $\Delta Q = \Delta U + P\,\Delta V$.

Constant volume. Here $\Delta V = 0$, so no work is done and all heat raises internal energy:

$$ C_v = \left(\frac{\Delta Q}{\Delta T}\right)_V = \frac{\Delta U}{\Delta T} $$

Because the internal energy of an ideal gas depends only on temperature, this $\Delta U/\Delta T$ is the same in every process — the subscript can be dropped.

Constant pressure. Now the gas also expands and does work, so

$$ C_p = \left(\frac{\Delta Q}{\Delta T}\right)_P = \frac{\Delta U}{\Delta T} + P\frac{\Delta V}{\Delta T} = C_v + P\frac{\Delta V}{\Delta T} $$

For one mole of an ideal gas $PV = RT$, so at constant pressure $P\,\Delta V = R\,\Delta T$, giving $P\,\dfrac{\Delta V}{\Delta T} = R$. Substituting yields Mayer's relation:

$$ \boxed{\,C_p - C_v = R\,} $$

The physics is in that extra term $R$. At constant volume every joule of heat becomes internal energy. At constant pressure, the same temperature rise needs the same internal-energy increase plus the work $R\,\Delta T$ the gas spends pushing back the surroundings. So $C_p$ exceeds $C_v$ by exactly $R$ — the gas always asks for more heat when it is allowed to expand.

The ratio γ = Cp/Cv

The dimensionless ratio of the two specific heats is given its own symbol:

$$ \gamma = \frac{C_p}{C_v} $$

Because $C_p = C_v + R$, the ratio always exceeds 1. It governs adiabatic behaviour through $PV^\gamma = \text{const}$, and it encodes the molecular structure of the gas through the degrees of freedom $f$, since $C_v = \tfrac{f}{2}R$ and therefore $\gamma = 1 + \tfrac{2}{f}$. The standard NEET values follow directly.

Note the pattern verified by experiment in NCERT: for solids, where $\Delta V$ is negligible, $\Delta Q \approx \Delta U$ and the molar heat capacity approaches $3R$ (the Dulong–Petit value) at ordinary temperatures.

The law applied per process

The single equation $\Delta Q = \Delta U + \Delta W$ takes a distinct, simpler form in each standard process, because each process kills one of the three terms or freezes one variable. The table collects the per-process specialisations; the full quantitative treatment of each — work integrals, $PV^\gamma$ curves, indicator diagrams — is developed in the dedicated thermodynamic processes page.

The limitation of the first law

The first law is a powerful constraint, but it is silent on one crucial matter: direction. It says energy is conserved in any process, yet it does not say which processes actually happen. NCERT puts it vividly — a book lying on a table never spontaneously cools the table and hops upward, even though such an event would conserve energy perfectly. The first law would permit it; nature forbids it.

NIOS lists the same gaps explicitly. The first law does not prohibit heat flowing from a cold body to a hot one, and it sets no ceiling on how much heat can be converted into work. Both of these are real, observed restrictions in nature that the first law cannot express. Supplying the missing arrow of direction — and capping the efficiency of heat engines — is the job of the second law of thermodynamics.