Zeroth Law of Thermodynamics
Statement
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.
Significance — Defines Temperature
This seemingly simple, almost obvious law is what gives temperature its rigorous physical meaning — it establishes that temperature is a well-defined, transitive property that can be reliably measured and compared, providing the foundation for using a thermometer at all. Without this law, there would be no logical basis for saying two separate objects are "at the same temperature" just because a thermometer reads the same value for both.
Solved Example
A thermometer (system C) shows the same reading when placed in contact with both a cup of water (system A) and a metal block (system B), though A and B were never placed in direct contact with each other. Explain why we can conclude A and B are at the same temperature, using the zeroth law.
The thermometer reaching the same reading with both A and C, and separately with both B and C, means A is in thermal equilibrium with C, and B is in thermal equilibrium with C. By the zeroth law, this directly implies A and B must also be in thermal equilibrium with each other — meaning they are at the same temperature, even without ever being placed in direct contact.
Q. Why is the zeroth law considered logically necessary before temperature can be meaningfully defined or measured?
Ans: Without the zeroth law's guarantee of transitivity, there would be no logical justification for concluding that two objects are at the same temperature just because each separately matches a thermometer's reading — the law establishes that "being in thermal equilibrium with" is a transitive relationship, which is precisely the property needed to define a consistent, comparable temperature scale using any reference thermometer.
Q. Two rooms, A and B, are each found to be in thermal equilibrium with a reference thermometer showing 25°C, but at different times. Are rooms A and B in thermal equilibrium with each other?
Ans: Yes, according to the zeroth law — since both A and B are separately in thermal equilibrium with the same reference system (the thermometer, both showing 25°C), they must also be in thermal equilibrium with each other, meaning they are at the same temperature.
Heat, Work & Internal Energy
Heat vs Work vs Internal Energy
Heat (Q) is energy transferred between systems due to a temperature difference. Work (W) is energy transferred through some mechanical process (like expansion/compression). Internal energy (U) is the total kinetic and potential energy of all the molecules within a system — unlike heat and work (which are path-dependent, describing a process), internal energy is a state function, depending only on the current state of the system.
Solved Example
Explain why heat and work are described as "energy in transit," while internal energy is not.
Heat and work only have meaning during a process — they describe energy actively flowing into or out of a system, and it makes no sense to say a system "contains" a certain amount of heat or work at some fixed moment. Internal energy, by contrast, is a property the system genuinely possesses at any given instant, regardless of how it got there — it can be assigned a definite value for a system in a specific state, unlike heat or work, which are only defined for a transition between states.
Q. Why is internal energy classified as a state function, while heat and work are not?
Ans: Internal energy depends only on the current state of the system (e.g., its temperature, for an ideal gas), regardless of the path taken to reach that state — two different processes ending at the same final state will show the same change in internal energy. Heat and work, however, can differ depending on the specific path/process taken between the same initial and final states, so their values are path-dependent rather than determined solely by the initial and final states.
Q. For an ideal gas, internal energy depends only on which physical quantity?
Ans: Temperature. For an ideal gas, internal energy depends solely on temperature (since ideal gas molecules are assumed to have no intermolecular potential energy, only kinetic energy, which depends only on temperature) — it is independent of the gas's volume or pressure.
First Law of Thermodynamics
Statement
The heat supplied to a system equals the sum of the increase in its internal energy and the work done by the system on its surroundings — essentially a statement of energy conservation applied to thermodynamic systems.
Solved Example
A gas absorbs 500 J of heat and does 200 J of work on its surroundings while expanding. Calculate the change in its internal energy.
$\Delta U = \Delta Q - \Delta W = 500 - 200 = 300\ \text{J}$
Q. A gas releases 100 J of heat to its surroundings while 50 J of work is done on it (compression). Calculate the change in its internal energy.
Ans: ΔQ = −100 J (released). ΔW = −50 J (work done on the gas, so work done by the gas is negative). ΔU = ΔQ − ΔW = −100 − (−50) = −50 J.
Q. Why is the first law of thermodynamics considered a specific application of the more general law of conservation of energy?
Ans: It simply accounts for all the ways energy can enter, leave, or be stored within a thermodynamic system — as heat transferred, work done, or a change in internal energy — asserting that these must always balance out consistently, exactly matching the broader principle that energy can be transformed from one form to another but never created or destroyed.
Isothermal Process
Definition
An isothermal process occurs at constant temperature. For an ideal gas, since internal energy depends only on temperature, $\Delta U = 0$ throughout an isothermal process — meaning, by the first law, all heat absorbed is entirely converted into work done by the gas: $\Delta Q = \Delta W$.
Work Done in an Isothermal Process
$$W = nRT\ln\left(\dfrac{V_f}{V_i}\right)$$
Practical Requirement
A truly isothermal process requires the gas to be in excellent thermal contact with a large heat reservoir (to instantly absorb or release heat and keep temperature constant), and must occur very slowly (quasi-statically), so the gas has time to remain in thermal equilibrium with its surroundings throughout the process.
Solved Example
2 moles of an ideal gas expand isothermally at 300 K from a volume of 10 L to 20 L. Calculate the work done by the gas. (R = 8.314 J/mol K)
$W = nRT\ln(V_f/V_i) = (2)(8.314)(300)\ln(20/10) = 4988.4 \times \ln(2) = 4988.4 \times 0.693 \approx 3457\ \text{J}$
Q. Why must an isothermal process occur very slowly to be considered ideal/genuine?
Ans: Maintaining a genuinely constant temperature requires the gas to continuously exchange heat with its surroundings fast enough to offset any temperature changes that would otherwise occur due to compression or expansion. If the process happens too quickly, there isn't enough time for this heat exchange to keep pace, and the gas's temperature would fluctuate rather than remaining truly constant — so a slow, quasi-static process is necessary to maintain the isothermal condition throughout.
Q. In an isothermal process for an ideal gas, why does all the heat absorbed convert entirely into work, with none going towards increasing internal energy?
Ans: For an ideal gas, internal energy depends only on temperature. Since temperature remains constant throughout an isothermal process (by definition), the change in internal energy (ΔU) is exactly zero — by the first law (ΔQ = ΔU + ΔW), with ΔU = 0, all the heat absorbed (ΔQ) must be entirely converted into work done by the gas (ΔW).
Adiabatic Process
Definition
An adiabatic process occurs with no heat exchange with the surroundings ($\Delta Q = 0$) — either because the system is perfectly insulated, or because the process happens too quickly for any significant heat transfer to occur. By the first law, all work done comes entirely at the expense of (or contributes entirely to) internal energy: $\Delta U = -\Delta W$.
Solved Example
Explain why a gas cools down when it expands adiabatically, using the first law of thermodynamics.
In an adiabatic expansion, no heat enters the system ($\Delta Q = 0$), but the gas still does positive work on its surroundings as it expands ($\Delta W > 0$). By the first law, $\Delta U = \Delta Q - \Delta W = 0 - \Delta W = -\Delta W$, which is negative — meaning internal energy decreases. Since internal energy for an ideal gas depends only on temperature, a decrease in internal energy directly corresponds to a decrease in temperature — the gas does work using its own stored internal energy, since no external heat is available to replace it, causing it to cool.
Q. Why does a bicycle pump feel warm to the touch after rapidly compressing air inside it?
Ans: Rapid compression doesn't allow enough time for significant heat exchange with the surroundings, making the process approximately adiabatic. Since work is done ON the gas during compression (rather than by it), and no heat escapes fast enough, this work goes entirely into increasing the gas's internal energy — raising its temperature, which is why the pump (and the compressed air inside) feels noticeably warmer after quick, repeated compressions.
Q. Why is an adiabatic P-V curve steeper than an isothermal P-V curve at the same point?
Ans: The adiabatic relation PV^γ = constant involves the exponent γ (always greater than 1 for real gases), making pressure fall off more rapidly with increasing volume compared to the isothermal relation PV = constant (equivalent to an exponent of 1). This mathematically steeper fall in pressure per unit volume increase is why the adiabatic curve is consistently steeper than the isothermal curve wherever they might intersect.
Isochoric & Isobaric Processes
Isochoric (Constant Volume) Process
Since volume doesn't change, no work is done by or on the gas ($\Delta W = 0$, since $W = P\Delta V$ and $\Delta V = 0$). By the first law, all heat supplied goes entirely into increasing internal energy: $\Delta Q = \Delta U$.
Isobaric (Constant Pressure) Process
Volume can change, so work is done: $\Delta W = P\Delta V$. Heat supplied is shared between increasing internal energy and doing this work: $\Delta Q = \Delta U + P\Delta V$.
Solved Example
A gas is heated at constant volume, absorbing 300 J of heat. Calculate the change in its internal energy and the work done.
Since volume is constant, $\Delta W = 0$. By the first law: $\Delta U = \Delta Q - \Delta W = 300 - 0 = 300\ \text{J}$
Solved Example
2 moles of gas are heated at constant pressure of $1\times10^5$ Pa, causing its volume to increase from 0.02 m³ to 0.03 m³, while absorbing 600 J of heat. Calculate the change in internal energy.
$\Delta W = P\Delta V = (1\times10^5)(0.03-0.02) = (1\times10^5)(0.01) = 1000\ \text{J}$
$\Delta U = \Delta Q - \Delta W = 600 - 1000 = -400\ \text{J}$
Q. Why is no work done by a gas undergoing an isochoric (constant volume) process, regardless of how much heat is absorbed?
Ans: Work done by an expanding/compressing gas is given by W = PΔV, which directly depends on the change in volume. Since volume remains fixed throughout an isochoric process, ΔV = 0, making the work done exactly zero regardless of how much pressure or heat is involved.
Q. Why does an isobaric process generally require more heat than an isochoric process to produce the same rise in temperature?
Ans: In an isochoric process, all supplied heat goes directly into raising internal energy (and hence temperature). In an isobaric process, the gas also expands and does work against its surroundings at constant pressure, so some of the supplied heat is "used up" doing this extra work rather than solely raising internal energy — meaning more total heat must be supplied to achieve the same temperature rise as would occur at constant volume, which is exactly why Cp > Cv.
Relating Molar Heat Capacities
For an ideal gas, the molar heat capacity at constant volume ($C_v$) and constant pressure ($C_p$) are related by $C_p - C_v = R$ (the universal gas constant) — this reflects the extra heat needed in an isobaric process specifically to perform expansion work, beyond what's needed just to raise internal energy (temperature) alone.
Solved Example
For a monatomic ideal gas, Cv = (3/2)R. Calculate Cp and the ratio γ = Cp/Cv for this gas.
$C_p = C_v + R = \tfrac{3}{2}R + R = \tfrac{5}{2}R$
$\gamma = \dfrac{C_p}{C_v} = \dfrac{5/2\,R}{3/2\,R} = \dfrac{5}{3} \approx 1.67$
Q. Why is Cp always greater than Cv for any ideal gas, regardless of the specific gas involved?
Ans: At constant volume, all supplied heat contributes solely to raising internal energy (temperature). At constant pressure, supplied heat must both raise internal energy by the same amount AND additionally provide the energy for the gas to expand and do work against the constant external pressure — since this second process always requires extra heat input for the identical temperature rise, Cp must always exceed Cv, with their difference exactly equal to R for an ideal gas.
Second Law of Thermodynamics
Kelvin-Planck Statement
It is impossible to construct a device that operates in a cycle and produces no effect other than extracting heat from a single reservoir and converting it completely into work — some heat must always be rejected to a colder reservoir.
Clausius Statement
It is impossible for heat to spontaneously flow from a colder body to a hotter body without external work being done — heat naturally flows from hot to cold, and reversing this direction always requires an external energy input (as in a refrigerator).
Why the Second Law Matters — Direction of Natural Processes
The first law alone doesn't forbid many processes that never actually happen in nature (like heat spontaneously flowing from cold to hot, or a broken cup spontaneously reassembling) — these would technically conserve energy but never occur naturally. The second law provides the crucial additional restriction on which energy-conserving processes are actually physically possible, establishing a preferred, irreversible direction for natural processes (increasing entropy).
Solved Example
Explain why a ship's engine cannot simply extract heat from the ocean and convert all of it into useful work to propel the ship, even though the ocean contains an enormous amount of thermal energy.
According to the Kelvin-Planck statement of the second law, no cyclic device can convert heat from a single reservoir (here, the ocean) completely into work without rejecting some heat to a colder reservoir. Since there's no naturally colder reservoir readily available for the ship's engine to reject heat into (the ocean is essentially the only reservoir at hand, at a fairly uniform temperature), such a scheme — however appealing given the ocean's vast energy content — is fundamentally forbidden by the second law, regardless of engineering cleverness.
Q. Does the first law of thermodynamics forbid heat from flowing spontaneously from a cold object to a hot object? If not, what does forbid it?
Ans: No, the first law (energy conservation) alone does not forbid this — energy could technically still be conserved even if heat flowed from cold to hot. It is specifically the second law of thermodynamics (Clausius statement) that forbids this spontaneous direction of heat flow without external work being done, establishing the natural, one-way direction of heat transfer that we observe in everyday experience.
Q. Why is the Kelvin-Planck statement of the second law significant for understanding the fundamental limitations of engines?
Ans: It establishes that no heat engine can ever be 100% efficient — some fraction of the heat absorbed from the hot reservoir must always be rejected to a cold reservoir rather than being fully converted to useful work, setting a fundamental theoretical upper limit on any engine's efficiency, regardless of how well-engineered or friction-free it might be.
Heat Engines & Carnot Engine
Heat Engine — Basic Concept
A heat engine absorbs heat ($Q_1$) from a hot reservoir, converts part of it into useful work ($W$), and rejects the remaining heat ($Q_2$) to a cold reservoir, operating in a repeating cycle.
Efficiency of a Heat Engine
$$\eta = \dfrac{W}{Q_1} = \dfrac{Q_1-Q_2}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$$
Carnot Engine — The Theoretical Ideal
The Carnot engine is an idealised, perfectly reversible heat engine operating between two fixed temperatures, providing the maximum possible efficiency theoretically achievable by any heat engine operating between those same two temperatures.
$$\eta_{Carnot} = 1 - \dfrac{T_2}{T_1}$$
where $T_1$, $T_2$ are the absolute (Kelvin) temperatures of the hot and cold reservoirs respectively.
Solved Example
A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate its maximum possible efficiency.
$\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.4 = 40\%$
Q. Why can no real heat engine achieve 100% efficiency, according to the Carnot efficiency formula?
Ans: η_Carnot = 1 − T₂/T₁ can only equal 1 (100%) if T₂ = 0 K — but achieving a cold reservoir at absolute zero is physically impossible (a separate consequence of the third law of thermodynamics). Since the cold reservoir always has some non-zero absolute temperature in practice, the theoretical maximum efficiency is always strictly less than 100%, and real engines (with additional practical losses like friction) perform even further below this already-limited Carnot maximum.
Q. A heat engine absorbs 800 J of heat from the hot reservoir and rejects 500 J to the cold reservoir in each cycle. Calculate its efficiency.
Ans: η = 1 − Q₂/Q₁ = 1 − 500/800 = 1 − 0.625 = 0.375 = 37.5%.
Refrigerators & Heat Pumps
Refrigerator — Reverse Heat Engine
A refrigerator operates like a heat engine running in reverse — using external work input to extract heat from a cold region (inside the fridge) and reject a larger amount of heat to a hotter region (the room), moving heat in the "unnatural" direction with the help of supplied work, consistent with the Clausius statement of the second law.
Solved Example
A refrigerator extracts 200 J of heat from its interior while 50 J of work is supplied to it. Calculate its coefficient of performance.
$\text{COP} = \dfrac{Q_2}{W} = \dfrac{200}{50} = 4$
Q. Why can the coefficient of performance of a refrigerator be greater than 1, while the efficiency of a heat engine is always less than 1?
Ans: Efficiency (for an engine) is defined as useful work output divided by heat input, and since only part of the heat input can ever become work (second law), this ratio is always less than 1. COP, however, is defined as heat extracted (the "useful" quantity for a refrigerator) divided by the smaller quantity of work input — since a well-designed refrigerator can move considerably more heat than the amount of work supplied to drive that process, this ratio can indeed exceed 1.
Q. What is a heat pump, and how does it differ in application (though not in basic operating principle) from a refrigerator?
Ans: A heat pump operates on the exact same principle as a refrigerator (using work input to move heat from a cold region to a hot region), but its intended application is reversed — while a refrigerator's purpose is to cool its interior (with heat rejected to the room as an unwanted by-product), a heat pump's purpose is specifically to warm a room or space (extracting heat from a colder outside environment and delivering it, along with the work input, as useful heating to the target warm space).
Reversible & Irreversible Processes
Reversible Process
A reversible process can be exactly reversed, restoring both the system and its surroundings to their original states, with no net change anywhere in the universe. This requires the process to occur infinitely slowly (quasi-statically), through a continuous sequence of equilibrium states, with no friction, turbulence, or other dissipative effects.
Irreversible Process
An irreversible process cannot be undone without leaving some permanent change in the surroundings — nearly all real, naturally occurring processes are irreversible, due to the presence of friction, unrestrained expansion, finite temperature differences during heat transfer, or other dissipative/non-equilibrium effects.
Reversible Processes Are an Idealisation
A truly reversible process is a theoretical idealisation, useful for establishing the maximum possible efficiency limits (like the Carnot engine) — no real, finite-speed process can be perfectly reversible, since real processes always involve at least some dissipation (friction, finite temperature gradients) and occur over a finite (not infinitely slow) timescale.
Solved Example
Explain why the free (unrestrained) expansion of a gas into a vacuum is considered an irreversible process.
When a gas expands freely into a vacuum, it does so suddenly and chaotically, without passing through a well-defined sequence of equilibrium states, and without doing any useful work in the process (since there's no opposing pressure to do work against). To reverse this — compressing the gas back into its original smaller volume — would require external work to be done on it, and this compression process would generate heat/entropy elsewhere (e.g., in the compressing mechanism), leaving a permanent change in the surroundings — meaning the original free expansion cannot be undone without such a lasting external change, making it irreversible.
Q. Why is friction always associated with irreversibility in a thermodynamic process?
Ans: Friction converts organised, useful mechanical energy into disorganised heat energy (dissipation), and this heat, once generated and dispersed into the surroundings, cannot be spontaneously and completely gathered back up and converted entirely back into the original organised mechanical form — this one-way conversion (order into disorder) is exactly what characterises an irreversible process, since perfectly restoring the original state without any lasting change elsewhere becomes impossible once such dissipation has occurred.
Q. Why is a Carnot engine, despite being physically unrealisable in practice, still a useful theoretical concept?
Ans: Even though no real engine can achieve perfectly reversible operation, the Carnot engine establishes an absolute theoretical upper limit on efficiency for any heat engine operating between two given temperatures — this provides an invaluable benchmark against which the performance of real, practical engines can be measured and compared, helping engineers understand how much room for improvement exists relative to this fundamental thermodynamic limit.
Q. Why does melting ice at 0°C, when done extremely slowly with the surrounding temperature kept just infinitesimally above 0°C, come close to being a reversible process?
Ans: Because the process occurs with essentially zero temperature difference driving the heat transfer at any instant (surroundings only infinitesimally warmer than the ice), the system remains arbitrarily close to thermal equilibrium throughout — reversing the process (refreezing) would similarly require only an infinitesimal temperature difference in the other direction, restoring both the ice and its surroundings almost exactly to their original states with negligible net change, which is the hallmark of a nearly reversible process.
⚡ Formula Cheat-Sheet
First lawΔQ = ΔU + ΔW
Isothermal workW = nRT ln(Vf/Vi)
Adiabatic relationPV^γ = constant
IsochoricΔQ = ΔU (ΔW=0)
IsobaricΔQ = ΔU + PΔV
Heat engine efficiencyη = 1 − Q₂/Q₁
Carnot efficiencyη = 1 − T₂/T₁
Refrigerator COPCOP = Q₂/W