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Thermal Equilibrium Zeroth Law

🎓 Class 11 Physics CBSE Theory Ch 11 – Thermodynamics ⏱ ~14 min
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Thermal Equilibrium Zeroth Law

11.1 Introduction — From Heat to Thermodynamics

Chapter 10 dealt with heat as a flow that depends on temperature differences, and with how heat causes phase change and expansion. Thermodynamics asks deeper questions: What is heat? How is it related to work? Can we convert heat completely into work (a steam engine)? Can we extract heat from a cold body to warm a hot one (a refrigerator)? The answers form three quantitative laws — the Zeroth, First and Second Laws of Thermodynamics — among the most far-reaching laws of physics.

Mechanics describes the motion of a single object given the forces. Thermodynamics describes systems containing ~10²³ particles using only a handful of macroscopic variables: pressure P, volume V, temperature T, internal energy U, entropy S. These are state variables: they specify the equilibrium state of the system.

11.2 Thermal Equilibrium

Two systems in thermal contact are said to be in thermal equilibrium when no net heat flows between them. At that point both systems share a common value of a single macroscopic variable — temperature.

An adiabatic wall is one that prevents any heat exchange (an idealised, perfectly insulating partition). A diathermic wall is one that allows heat to pass freely (a thin metal sheet, for example). These two idealised walls give us laboratory control over whether systems can exchange thermal energy.

A (T₁) B (T₂) adiabatic wall No heat flow A (T₁) B (T₂) diathermic wall Heat flows ⇒ T₁ = T₂
Fig 11.1: An adiabatic wall blocks heat flow; a diathermic wall allows it, eventually equalising temperatures.

11.3 Zeroth Law of Thermodynamics

Suppose three systems A, B and C are arranged so that A is in thermal equilibrium with C, and B is in thermal equilibrium with C (separately). The experimental observation is that A and B will then also be in thermal equilibrium when brought into contact. This empirical fact is elevated to a law:

Zeroth Law of Thermodynamics: Two systems each in thermal equilibrium with a third system are also in thermal equilibrium with each other.

The Zeroth law guarantees that temperature is well-defined: a thermometer (which is just our system C) gives the same reading whenever it is in equilibrium with any system at the same "hotness". Without the Zeroth law, the very notion of measuring temperature with an instrument would be meaningless.

A B C therefore A ↔ B A ↔ C B ↔ C
Fig 11.2: Zeroth Law. If A is in thermal equilibrium with C, and B with C, then A and B are also in equilibrium — they share a common temperature.
The "Zeroth" law was named after the First and Second laws had already been formalised — but logically it precedes them, so it was placed before "First". It was first stated formally by R.H. Fowler in the 1930s.

11.4 Heat, Internal Energy and Work

11.4.1 Internal Energy U

The total energy stored within a system due to the random translational, rotational and vibrational motions of its molecules and the inter-molecular potential energies is called the internal energy \(U\). Internal energy is a state function: it depends only on the current state of the system (P, V, T) and not on how the system got there.

For an ideal gas of \(\mu\) moles, \(U = U(T)\) only — depending only on temperature, not on volume or pressure. The molar internal energy is \(U = \mu C_v T\) (a useful result we will use repeatedly).

11.4.2 Heat Q (Energy in Transit)

Heat is the energy transferred between system and surroundings only because of a temperature difference. Heat is not stored in a body — once absorbed, it becomes internal energy. Heat depends on the path of the process; it is therefore a path function, not a state function.

Sign convention for Q:
  • \(Q > 0\) — heat added to the system.
  • \(Q < 0\) — heat released by the system.

11.4.3 Work W

Mechanical work can also transfer energy between the system and surroundings. For a gas in a cylinder fitted with a frictionless piston, when the gas expands by an infinitesimal volume \(dV\) against external pressure \(P_\text{ext}\):

\[\delta W = P_\text{ext}\, dV \qquad \text{(work done BY the system)}\] For a quasi-static process, \(P_\text{ext}\) equals the gas pressure \(P\), so the total work for expansion from \(V_1\) to \(V_2\) is: \[W = \int_{V_1}^{V_2} P\, dV\]
Sign convention for W:
  • \(W > 0\) — system does work on surroundings (gas expands).
  • \(W < 0\) — work is done on the system (gas compressed).
Gas (P, V, T) cylinder piston dx (expansion) cross-section A
Fig 11.3: Gas in a cylinder with a frictionless piston. As the gas expands by dx, it does work δW = P·A·dx = P·dV on the surroundings.

11.5 State Variables vs Path Variables

This is one of the trickiest distinctions in thermodynamics. The key idea:

State variables (functions of state)Path variables
Pressure PHeat Q absorbed in a process
Volume VWork W done in a process
Temperature T
Internal energy U
Entropy S

State variables are determined uniquely once the equilibrium state is known. Heat and work, however, depend on how the system was taken from one state to another: an isothermal expansion versus an adiabatic expansion can give different Q and W between the same end states, but exactly the same ΔU.

11.6 Equation of State

For a thermodynamic system there is a relation between the state variables, called the equation of state. For an ideal gas, the equation of state is just the familiar:

PV = µRT

For real gases at moderate pressure, the van der Waals equation is more accurate: \(\left(P + \dfrac{a}{V^2}\right)(V - b) = \mu RT\). For solids and liquids, more complex forms are needed, but the conceptual structure (relating P, V, T) is the same.

Worked Examples

Example 11.1: Internal energy change of an ideal gas

Two moles of an ideal monatomic gas (\(C_v = \tfrac{3}{2}R\)) are heated from 27 °C to 127 °C. Find the change in internal energy. R = 8.314 J/mol·K.

\(\Delta U = \mu C_v \Delta T = 2 \times \tfrac{3}{2}\times 8.314 \times (127-27)\)
\(= 2\times 12.47 \times 100 = \boxed{2494\text{ J}}\).
Note: ΔU depends only on the temperature change — not on the path or process.

Example 11.2: Work in isobaric expansion

A gas at constant pressure of 2.0 × 10⁵ Pa expands from 0.020 m³ to 0.080 m³. Compute the work done by the gas.

At constant P: \(W = P\,\Delta V = 2.0\times 10^5 \times (0.080 - 0.020) = 1.2\times 10^4\) J = 12 kJ (positive — work done BY the gas).

Example 11.3: Heat input vs work output

In an experiment, 800 J of heat are added to a gas, and the gas does 300 J of work on the surroundings. By how much does the internal energy of the gas change?

First law (which we will formally state in Part 2): \(\Delta U = Q - W = 800 - 300 = \boxed{500\text{ J}}\) — internal energy increases by 500 J. Temperature must rise (for ideal gas).

Interactive: Energy Bookkeeping for a Gas L3 Apply

Choose how much heat Q is added and how much work W the gas does. The applet computes ΔU and predicts whether the gas warms or cools.

800
300
Activity — Rubber Band Stretch & Snap L4 Analyse
Predict: When you suddenly stretch a thick rubber band and immediately touch it to your lips, does it feel warmer or cooler? When you let it suddenly snap back?
  1. Take a thick (~1 cm) rubber band. Touch it to your lips to feel its initial temperature.
  2. Quickly stretch it as much as you can. Immediately touch the stretched part to your lips.
  3. Hold it stretched for a minute (let it cool). Then release suddenly and touch to your lips again.
Observation: A freshly stretched band feels noticeably warmer; a freshly snapped band feels cooler.
Explanation: Quick stretching is approximately adiabatic (no time for heat exchange). Work is done on the polymer chains, increasing internal energy and raising temperature. The reverse happens on snapping. This rubber-band thermodynamics is even the basis for some heat-engine prototypes!

Competency-Based Questions

A piston-cylinder arrangement holds 3 moles of an ideal monatomic gas at 27 °C. The gas is heated such that 600 J of heat enter the gas while 200 J of work is done by the gas on the surroundings.

Q1. L1 Remember State the Zeroth Law of Thermodynamics.

If two systems A and B are each separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other. The law guarantees the existence of temperature as a well-defined property.

Q2. L2 Understand Distinguish heat from internal energy.

Internal energy is energy stored within the system (state function). Heat is energy in transit between system and surroundings due to a temperature difference (path function). After absorption, heat ceases to exist as heat — it becomes internal energy.

Q3. L3 Apply Find ΔU in the scenario above.

ΔU = Q − W = 600 − 200 = 400 J (internal energy increases).

Q4. L3 Apply Compute the temperature rise of the gas (C_v = 3R/2).

ΔU = µC_v ΔT, so ΔT = ΔU/(µC_v) = 400/(3 × 1.5 × 8.314) = 400/37.4 = 10.7 K. Final T = 37.7 °C.

Q5. L5 Evaluate A frictionless piston is replaced by one with friction. The same 600 J of heat is added but only 100 J of useful work emerges. Where did the rest go?

Friction dissipates 100 J as heat back into the gas (and a little to the cylinder walls). Effective W (useful) = 100 J; ΔU = 500 J. Practical heat engines lose ~10–20 % of their potential output to friction and dissipation in this way.

Assertion-Reason Questions

Assertion (A): The internal energy of an ideal gas depends only on its absolute temperature.

Reason (R): Ideal-gas molecules have no inter-molecular potential energy and only kinetic energy, which is set by T.

  • A. Both A and R are true, and R is the correct explanation of A.
  • B. Both A and R are true, but R is NOT the correct explanation of A.
  • C. A is true, but R is false.
  • D. A is false, but R is true.
Answer: A. Joule's free-expansion experiment confirmed that an ideal gas does not change temperature on free expansion, exactly because U depends only on T.

Assertion (A): Heat and work are state functions.

Reason (R): They are both forms of energy.

  • A. Both A and R are true, and R is the correct explanation of A.
  • B. Both A and R are true, but R is NOT the correct explanation of A.
  • C. A is true, but R is false.
  • D. A is false, but R is true.
Answer: D. A is false. Heat and work are path functions, not state functions — they depend on how the process is carried out. R is correct on its own.

Assertion (A): Temperature is a thermodynamic state variable.

Reason (R): The Zeroth law guarantees that the temperature of a system is uniquely defined when it is in thermal equilibrium.

  • A. Both A and R are true, and R is the correct explanation of A.
  • B. Both A and R are true, but R is NOT the correct explanation of A.
  • C. A is true, but R is false.
  • D. A is false, but R is true.
Answer: A. The Zeroth law is exactly what makes temperature a well-defined state variable.

Did You Know?

Frequently Asked Questions - Thermal Equilibrium Zeroth Law

What is the main concept covered in Thermal Equilibrium Zeroth Law?
In NCERT Class 11 Physics Chapter 11 (Thermodynamics), "Thermal Equilibrium Zeroth Law" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Thermal Equilibrium Zeroth Law useful in real-life applications?
Real-life applications of Thermal Equilibrium Zeroth Law from NCERT Class 11 Physics Chapter 11 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Thermal Equilibrium Zeroth Law?
Key formulas in Thermal Equilibrium Zeroth Law (NCERT Class 11 Physics Chapter 11 Thermodynamics) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 11?
NCERT Class 11 Physics Chapter 11 (Thermodynamics) is structured so each part builds on the previous one. Thermal Equilibrium Zeroth Law connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Thermal Equilibrium Zeroth Law?
CBSE board questions from Thermal Equilibrium Zeroth Law typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Thermal Equilibrium Zeroth Law lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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