This MCQ module is based on: Thermal Equilibrium Zeroth Law
Thermal Equilibrium Zeroth Law
This assessment will be based on: Thermal Equilibrium Zeroth Law
Upload images, PDFs, or Word documents to include their content in assessment generation.
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.
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.
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:
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.
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.
- \(Q > 0\) — 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 \(P_\text{ext}\):
- \(W > 0\) — system does work on surroundings (gas expands).
- \(W < 0\) — work is done on the system (gas compressed).
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 P | Heat Q absorbed in a process |
| Volume V | Work 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:
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.
\(= 2\times 12.47 \times 100 = \boxed{2494\text{ J}}\).
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.
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?
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.
- Take a thick (~1 cm) rubber band. Touch it to your lips to feel its initial temperature.
- Quickly stretch it as much as you can. Immediately touch the stretched part to your lips.
- Hold it stretched for a minute (let it cool). Then release suddenly and touch to your lips again.
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
Q1. L1 Remember State the Zeroth Law of Thermodynamics.
Q2. L2 Understand Distinguish heat from internal energy.
Q3. L3 Apply Find ΔU in the scenario above.
Q4. L3 Apply Compute the temperature rise of the gas (C_v = 3R/2).
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?
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.
Assertion (A): Heat and work are state functions.
Reason (R): They are both forms of energy.
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.