Thermodynamics: common doubts, answered
The questions students ask most often about Thermodynamics, each with a short answer. For the full chapter, read the Thermodynamics notes.
System, surroundings and state functions
Read this section in the notes →What is the difference between open, closed and isolated systems?
An open system exchanges both energy and matter with its surroundings, like reactants in an uncovered beaker. A closed system exchanges energy but not matter, like the same reaction inside a sealed metal flask that can still gain or lose heat. An isolated system exchanges neither energy nor matter, a situation that a sealed, insulated vessel such as a thermos flask comes close to.
Why are heat and work not state functions when internal energy is?
A state function depends only on the present condition of a system, not on the route taken to reach it. The amounts of heat and work exchanged do depend on the route: one change can be carried out with more heat and less work, or the other way round. Only their sum, ΔU = q + w, is fixed by the initial and final states.
Work, heat and the first law
Read this section in the notes →Why is work negative when a gas expands in chemistry?
Because NCERT chemistry counts work from the system's point of view: w counts as positive when the surroundings do work on the system and negative when the system pushes on the surroundings. With ΔU = q + w and w = −pₑₓΔV, an expanding gas has positive ΔV, so w is negative as it spends energy pushing back the surroundings. Some physics books use ΔU = q − w instead; never mix the two conventions.
Pressure-volume work
Read this section in the notes →Why is no work done in the free expansion of a gas?
Expansion work is w = −pₑₓΔV, and a gas expanding into a vacuum faces zero external pressure, so w = 0 however much the volume grows. For an ideal gas expanding freely there is also no heat exchanged and no change in internal energy. With nothing to push against, the gas uses no energy to expand.
Why is ΔU zero in the isothermal expansion of an ideal gas?
The internal energy of an ideal gas depends only on its temperature, because its molecules exert no forces on one another. In an isothermal process the temperature does not change, so ΔU = 0, and the first law then gives q = −w. Whatever work the expanding gas does is exactly made up by heat drawn in from the surroundings.
Why does a reversible expansion give the maximum work?
In a reversible expansion the external pressure is kept only infinitesimally lower than the gas pressure at every stage, so the gas always pushes against the largest pressure it can still overcome. That draws out the most work possible, w_rev = −2.303 nRT log(V_f/V_i) for an ideal gas at constant temperature. Expanding against a smaller fixed pressure yields less work for the same volume change.
Enthalpy, heat capacity and extensive properties
Read this section in the notes →What is the difference between ΔH and ΔU and when are they equal?
ΔU is the heat change at constant volume, while ΔH is the heat change at constant pressure, which also accounts for expansion or compression work. For reactions with gases, ΔH = ΔU + Δn_gRT, where Δn_g is moles of gaseous products minus gaseous reactants, ignoring liquids and solids. When Δn_g = 0, as in H₂ + Cl₂ → 2HCl, the two are equal.
Why is C_p greater than C_v for a gas?
At constant volume every bit of heat supplied goes into raising the gas's internal energy and temperature. At constant pressure the gas also expands and does work on its surroundings, so part of the heat is used up there and more heat is needed for the same temperature rise. For one mole of an ideal gas this extra heat is exactly R, so C_p − C_v = R.
What is the difference between extensive and intensive properties?
An extensive property depends on how much matter is present, such as mass, volume, internal energy, enthalpy and heat capacity. An intensive property does not, such as temperature, pressure, density and molar heat capacity. The ratio of two extensive properties is intensive, which is why heat capacity is extensive but molar and specific heat capacities are intensive.
Calorimetry
Read this section in the notes →Why does a bomb calorimeter measure ΔU and not ΔH?
A bomb calorimeter is a sealed steel container whose volume cannot change, so the reaction does no expansion work and the heat released equals ΔU. Heat measured at constant pressure, in a vessel open to the atmosphere, gives ΔH instead, because the system is free to expand or contract. A bomb calorimeter value can be converted using ΔH = ΔU + Δn_gRT.
Reaction enthalpy, standard states and thermochemical equations
Read this section in the notes →Why is the standard enthalpy of formation of an element zero?
Standard enthalpy of formation is the enthalpy change for making one mole of a substance from its elements in their reference states. Forming an element in its reference state from itself involves no change, so the value is zero by definition. The reference state is the most stable form at 1 bar, so ΔfH° is zero for graphite but not for diamond.
Hess's law
Read this section in the notes →Why does Hess's law work?
Hess's law works because enthalpy is a state function: its change depends only on where a reaction starts and finishes, not on the steps in between. So the overall enthalpy change is the same in one step or many. This lets us find values that are hard to measure, such as for C + ½O₂ → CO, by combining equations whose enthalpies are easy to measure.
Enthalpies of combustion, atomisation, bonds, lattices, solution and dilution
Read this section in the notes →Why is reaction enthalpy from bond enthalpies calculated as reactants minus products?
Bond enthalpy is the energy needed to break a bond, so breaking the reactants' bonds absorbs energy and forming the products' bonds releases it. That gives ΔᵣH = Σ bonds broken − Σ bonds formed, the reverse of the formation-enthalpy method, which uses products minus reactants. The bond method is reliable only for reactions in which every species is a gas.
Why does NaCl dissolve in water even though dissolving absorbs a little heat?
Because dissolving raises entropy enough to make ΔG negative. The enthalpy of solution balances two terms, ΔsolH = ΔlatticeH + ΔhydH: energy to pull the ions apart against energy released as water surrounds them. For NaCl the lattice term slightly outweighs hydration, so dissolving is mildly endothermic, about +4 kJ mol⁻¹, but the ions scattering through the water raise entropy enough for TΔS to outweigh it.
Spontaneity and entropy
Read this section in the notes →Is every exothermic reaction spontaneous?
No. At constant temperature and pressure spontaneity is decided by ΔG, not by ΔH alone. An exothermic reaction that lowers entropy has ΔG = ΔH − TΔS negative only at low temperatures; when it is hot enough, the −TΔS term turns positive and wins. Many endothermic changes, such as ice melting above 273 K, are spontaneous because they bring a large rise in entropy.
Does spontaneous mean that a reaction happens quickly?
No. A spontaneous process is one that can proceed by itself in a given direction; thermodynamics says nothing about how fast it goes. Diamond changing into graphite is spontaneous under ordinary conditions yet far too slow to notice, and hydrogen and oxygen can sit together unreacted until a spark sets them off. Speed is a question for chemical kinetics.
Why is entropy change defined as ΔS = q_rev/T?
Entropy measures how spread out the energy and particles of a system are, and dividing heat by temperature reflects how much difference that heat makes. Heat added to a cold system raises its disorder a lot, while the same heat added to an already hot, disordered system changes it relatively little. For any spontaneous process, the combined entropy of system and surroundings increases.
What does the third law of thermodynamics say?
It says a perfect crystal of a pure substance has an entropy that falls towards zero as it is cooled towards 0 K, since its particles are then locked into a single possible arrangement. This fixed zero point lets absolute entropies be found by measuring heat capacities upward from near 0 K. Solutions and glassy solids keep some disorder even at 0 K, so their entropy does not reach zero.
Gibbs energy and spontaneity
Read this section in the notes →How do the signs of ΔH and ΔS decide whether a reaction is spontaneous?
Use ΔG = ΔH − TΔS. With ΔH negative and ΔS positive, ΔG is negative at every temperature; with ΔH positive and ΔS negative, it is never negative. With both negative, the reaction is spontaneous only at low temperatures, because −TΔS grows as T rises. With both positive, it is spontaneous only at high temperatures, where TΔS outweighs ΔH.
How do you calculate the temperature above which a reaction becomes spontaneous?
Set ΔG = 0 and solve for T = ΔH/ΔS, with both in the same energy unit. For ΔH = +178 kJ mol⁻¹ and ΔS = +160 J K⁻¹ mol⁻¹, first convert ΔS to 0.160 kJ K⁻¹ mol⁻¹, which gives T ≈ 1112 K; above that temperature ΔG is negative. Forgetting the joule to kilojoule conversion gives an answer that is off by a factor of 1000.
Gibbs energy and equilibrium
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