First Law of Thermodynamics

Internal energy is everything the system already owns. $U$ is the sum of every form of energy stored inside a system — translational, rotational and vibrational motion of molecules, plus the potential energy locked in chemical bonds and intermolecular attractions. It is a state function: its value depends only on the current state (temperature, pressure, composition), never on the path taken to reach it. So $\Delta U$ for any process is fixed once the initial and final states are fixed, even though the heat and work exchanged along the way can differ with the path.

The law itself is conservation of energy, restricted to a closed system. Energy can transform from one form to another — heat into work, work into internal energy — but the total can neither be created nor destroyed. For a closed system exchanging only heat and work with its surroundings,

where: $\Delta U$ is the change in internal energy of the system, $q$ is the heat absorbed by the system, and $w$ is the work done on the system. This is the NCERT/IUPAC sign convention used throughout CBSE Class 11 Chemistry, and it is worth stating the signs explicitly before doing any calculation:

• $q > 0$ when heat flows into the system (endothermic); $q < 0$ when heat flows out (exothermic).
• $w > 0$ when the surroundings do work on the system, such as compression; $w < 0$ when the system does work on the surroundings, such as expansion.

Expansion work against a constant external pressure. The commonest work term in this chapter is mechanical (pressure-volume) work done when a gas changes volume against a resisting external pressure $P_{ext}$. If the volume changes from $V_1$ to $V_2$, the work done on the system is

where: $P_{ext}$ is the constant external pressure resisting the gas, and $\Delta V = V_2 - V_1$ is the change in volume. The minus sign is not a bookkeeping accident — when the gas expands ($\Delta V > 0$), it is pushing the surroundings back and spending its own energy to do so, so the system loses energy and $w$ must come out negative in this convention. When the gas is compressed ($\Delta V < 0$), the surroundings push the gas, energy flows in, and $w$ is positive. The figure below shows the energy bookkeeping for a gas that both absorbs heat and expands, so that the two effects — one adding energy, one removing it — combine into a smaller net change in internal energy.

Special cases worth memorising. Several process types collapse this general equation into simpler forms that examiners test directly:

• Isochoric (constant volume, $\Delta V = 0$): $w = 0$, so $\Delta U = q_v$, the heat exchanged at constant volume.
• Free expansion (into a vacuum, $P_{ext} = 0$): $w = -P_{ext}\Delta V = 0$ even though $\Delta V \ne 0$, since there is nothing to push against.
• Adiabatic (no heat exchange, $q = 0$): $\Delta U = w$.
• Isothermal expansion of an ideal gas: since $U$ depends only on temperature, $\Delta U = 0$, forcing $q = -w$ — every joule absorbed converts, joule for joule, into work done by the gas.

Problem: A gas absorbs 500 J of heat and expands against a constant external pressure of $1.0\times10^5\ \text{Pa}$, its volume increasing from $2.0\ \text{L}$ to $5.0\ \text{L}$. Calculate the work done and the change in internal energy of the gas.

Step 1. Find the volume change. $\Delta V = 5.0 - 2.0 = 3.0\ \text{L} = 3.0\times10^{-3}\ \text{m}^3$.

Step 2. Compute the work done on the system. $w = -P_{ext}\Delta V = -(1.0\times10^5\ \text{Pa})(3.0\times10^{-3}\ \text{m}^3) = -300\ \text{J}$. The negative sign confirms the gas does work on the surroundings as it pushes outward.

Step 3. Apply the first law. With $q = +500\ \text{J}$ absorbed, $\Delta U = q + w = 500 + (-300) = 200\ \text{J}$. The internal energy of the gas increases by 200 J.

Competitive stretch

Problem: The same gas is now compressed back from $5.0\ \text{L}$ to $2.0\ \text{L}$ against the same external pressure, and its internal energy falls by $50\ \text{J}$. Find $q$ for this stage.

Step 1. The volume change reverses: $\Delta V = 2.0 - 5.0 = -3.0\times10^{-3}\ \text{m}^3$, so $w = -P_{ext}\Delta V = +300\ \text{J}$ — positive, since the surroundings now do work on the gas.

Step 2. With $\Delta U = -50\ \text{J}$, the first law gives $q = \Delta U - w = -50 - 300 = -350\ \text{J}$. The gas releases 350 J of heat, even though 300 J of work was done on it, because internal energy still fell overall.

• Sign confusion: NCERT uses $\Delta U = q + w$ with $w$ as work done on the system, while some physics texts use $\Delta U = q - w$ with $w$ done by the system. Always use the NCERT convention here.
• Forgetting that $w = -P_{ext}\Delta V$ uses the external pressure, not the gas's own pressure — the two are equal only for a reversible, quasi-static process.
• Assuming $\Delta U = 0$ for every isothermal process — true only for an ideal gas, since only then does $U$ depend on temperature alone.
• Treating $q$ and $w$ individually as fixed by the initial and final states — only their sum $\Delta U$ is path-independent.
• Skipping the litre-to-cubic-metre conversion, which leaves the work term wrong by a factor of $10^3$.

Ask the tutor: Want to practice more sign-convention problems, or see how the first law connects to enthalpy and Hess's law in the next topic — ask the AI tutor for a guided problem set.