Thermodynamics and Kinetic Theory of Gases

1The first law and internal energy

The first law is conservation of energy written for a gas: Q = ΔU + W. Two sign conventions carry the whole thing — Q counts positive when heat enters the gas, and W counts positive when the gas expands and pushes something outward. Whatever heat is not spent as work stays behind as internal energy.

The one line worth memorising is ΔU = nC_VΔT. Internal energy of an ideal gas depends only on temperature, so that expression holds on every path — isobaric, adiabatic, a jagged path nobody named — not just at constant volume. Q and W depend on how you got there; ΔU does not.

This is the physics convention. If a source writes ΔU = Q + w, its w means work done on the gas and equals −W. Never mix the two halfway through a solution. Here expansion against nonzero external pressure has W > 0, free expansion has W = 0, and compression has W < 0; heat supplied has Q > 0, heat rejected Q < 0; and for an ideal gas ΔU has the sign of T₂ − T₁.

Figure. Heat entering splits exactly two ways: part leaves again as work when the piston moves out, and the rest stays behind as internal energy. Nothing else can happen to it — that is the whole of the first law.

How it works

1. Heat inQ > 0 when heat flows into the gas, negative when the gas loses heat.
2. Work outW > 0 when the gas expands against nonzero external pressure. Free expansion has W = 0; compression has W < 0.
3. What is leftThe remainder shows up as temperature: ΔU = nC_VΔT, on any path at all.
4. Check the signsExpansion against pressure, heating, and heat supplied point to W > 0, ΔU > 0, and Q > 0 respectively; free expansion has W = 0.

2Work is the area under the curve

For a quasistatic path, W = ∫P dV means that on a P–V diagram the work done by the gas is the area under the path. In general W = ∫P_ext dV, so free expansion into vacuum has W = 0 even while the volume increases.

Because it is an area under a path, W is path-dependent — two routes joining the same two states enclose different areas and cost different work, even though ΔU is identical for both. Close the path into a loop and ΔU vanishes entirely, so the net heat equals the net work equals the enclosed area: positive going clockwise (an engine), negative going anticlockwise (a refrigerator).

Figure. The two horizontal legs do all the work; the vertical legs do none. What survives is the rectangle itself, and going round it clockwise means the expansion happened at the higher pressure — so the gas comes out ahead.

Reading a P–V diagram

1. Area underFor a quasistatic path, the area under the P–V curve is the work done by the gas.
2. Direction signsMoving right (expansion) counts positive, moving left counts negative.
3. Close the loopRound a cycle ΔU = 0, so net heat = net work = the area inside the loop.

3Isothermal processes

Hold the temperature fixed and PV = constant: the path is a hyperbola, one of a family with the hotter isotherms further from the origin. Doing this in practice means good thermal contact with a reservoir and a change slow enough for the gas to keep up.

Because T never changes, ΔU = 0 and the first law gives Q = W. For a reversible quasistatic ideal-gas path, their common value is nRT ln(V₂/V₁). Every joule of heat absorbed comes straight back out as work along that reversible path; on compression every joule of work put in is rejected as heat.

Figure. Every point on one hyperbola is the same temperature, so moving along it changes P and V but not U. The work is the area under the arc from V₁ to V₂ — and because ΔU is zero, that same area is the heat that had to be supplied.

How it works

1. Fix TReservoir contact plus a slow change keeps the gas at the reservoir's temperature.
2. ΔU = 0An ideal gas's internal energy tracks T alone, so it does not move.
3. Q = WFor a reversible quasistatic ideal-gas path, Q = W = nRT ln(V₂/V₁), positive on expansion.

4Adiabatic processes

Adiabatic means Q = 0 — either the walls are insulating, or the change is too fast for heat to go anywhere. The first law then reads ΔU = −W. An adiabatic expansion that does positive work cools an ideal gas, while free expansion does no work and leaves its temperature unchanged. Adiabatic compression that has work done on the gas heats it.

Three equivalent relations describe a reversible adiabatic path, and picking the right one saves the integration: PV^{\gamma} = constant, TV^{\gamma-1} = constant, and TP^{(1-\gamma)/\gamma} = constant, with γ = C_P/C_V fixed by the molecule's degrees of freedom. Through any given state the adiabat is steeper than the isotherm — slope −γP/V against −P/V — by exactly the factor γ.

The bulk modulus is B=-V\,dP/dV. Therefore an ideal gas has isothermal bulk modulus B_T=P, but under adiabatic compression B_S=\gamma P. The gas is effectively stiffer when heat cannot escape; this is why the speed of sound is v=\sqrt{B_S/\rho}=\sqrt{\gamma P/\rho} rather than \sqrt{P/\rho}.

Figure. Both curves pass through the same state, but the adiabat leaves it more steeply in both directions. Expand along it and the pressure collapses faster because the gas is cooling as well as spreading out.

How it works

1. Set Q = 0Insulated, or fast enough that heat has no time to flow either way.
2. Pick the pairUse PV^γ for a P–V question, TV^(γ−1) for a T–V one. No integral needed.
3. Work from ΔUW = −ΔU = nC_V(T₁ − T₂), also written nR(T₁ − T₂)/(γ − 1).
4. Differentiate for stiffnessFrom PV^γ = constant, dP/dV = −γP/V and hence the adiabatic bulk modulus is B_S = γP.

5Isobaric, isochoric and polytropic processes

An isobaric process holds pressure fixed. The gas moves along a horizontal line on a P–V graph, does W=P(V_2-V_1)=nR\Delta T, changes internal energy by nC_V\Delta T, and exchanges Q=nC_P\Delta T: heat is absorbed for \Delta T>0 and rejected for \Delta T<0. At fixed pressure, Charles's law gives V/T= constant, so V against kelvin temperature is a straight line through the origin.

An isochoric process holds volume fixed. Its P–V path is vertical, so dV=0 and W = 0; every joule of heat changes internal energy, Q=\Delta U=nC_V\Delta T. At fixed volume, P/T= constant, so P against kelvin temperature is a straight line through the origin.

A quasistatic polytropic path obeys PV^N= constant. It contains the familiar ideal-gas paths as special cases: N = 0 is isobaric, N = 1 is reversible isothermal, N = γ is reversible adiabatic, and the limiting case N → ∞ approaches an isochore. For N ≠ 1, W=(P_2V_2-P_1V_1)/(1-N); then use \Delta U=nC_V\Delta T and Q=\Delta U+W rather than trying to memorise a separate heat formula.

Figure. On P–V axes the fingerprints are immediate: isobaric is horizontal, isochoric vertical, isothermal a rectangular hyperbola. One panel carries the three rules the multi-graph table names.

Identify the process before calculating

1. Read what is fixedP fixed means isobaric, V fixed is isochoric, T fixed is isothermal, and Q = 0 is adiabatic.
2. Read the graphOn P–V axes: horizontal isobar, vertical isochore, rectangular-hyperbola isotherm, and a steeper falling adiabat.
3. Use the first lawCompute W from the path and ΔU from temperature; Q follows from Q = ΔU + W with no new convention.

6Kinetic theory: where pressure comes from

Treat the gas as point molecules in ceaseless random motion, colliding elastically and feeling no forces in between. A molecule rebounding off a wall reverses v_x and hands the wall 2mv_x of momentum; add up every such bounce per second over the whole wall and the force per unit area comes out as P = (1/3)Nm⟨v²⟩/V, or equivalently P = (1/3)ρv_rms².

Set that beside the experimental PV = nRT and the two can only agree if (1/2)m⟨v²⟩ = (3/2)k_BT. Temperature is nothing but the average translational kinetic energy of a molecule, measured in different units — and notice that the molecular mass has dropped out.

Figure. One molecule striking the right-hand wall and rebounding reverses its x-velocity, so the wall absorbs 2mv_x. Pressure is nothing more than the time-average of billions of such kicks per square metre.

From one bounce to a pressure

1. One moleculeAn elastic bounce off the wall reverses v_x, delivering 2mv_x of momentum.
2. Add them upAveraging over all molecules and directions gives P = (1/3)Nm⟨v²⟩/V.
3. Match PV = nRTComparing the two forces (1/2)m⟨v²⟩ = (3/2)k_BT per molecule.

7Molecular speeds

The molecules do not share one speed; they are spread over the Maxwell distribution, and three points on it get named. In descending order: v_rms = √(3RT/M), then v_avg = √(8RT/πM), then the most probable speed v_mp = √(2RT/M) at the peak — in the fixed ratio √3 : √(8/π) : √2, roughly 1.73 : 1.60 : 1.41.

All three scale the same way, as √(T/M). At a given temperature the lighter gas is faster, v₁/v₂ = √(M₂/M₁); for a given gas, doubling any characteristic speed costs four times the kelvin temperature. The molecular forms are v_{mp}=\sqrt{2k_BT/m}, v_{avg}=\sqrt{8k_BT/(\pi m)} and v_{rms}=\sqrt{3k_BT/m}. Heating also flattens and broadens the curve, sliding the peak to the right while keeping its total area fixed.

Figure. Heating does not slide the curve bodily to the right — it lowers and widens it as well, because the same number of molecules must spread over a wider range of speeds. The long tail on the right is what makes evaporation and chemical reactions possible at all.

Two scalings to keep

1. Same T, lighter gasSpeeds go as 1/√M — hydrogen beats oxygen by a factor of 4.
2. Same gas, hotterSpeeds go as √T, so 4× the kelvin temperature to double the speed.
3. Which speed?v_rms for anything energetic; v_mp only when asked for the peak of the curve.

8Graham's law of effusion and diffusion

Graham's law says that at the same temperature and pressure the rate at which a gas effuses through a tiny hole, or approximately diffuses, is inversely proportional to the square root of its molar mass: r\propto1/\sqrt{M}. Hence r_1/r_2=\sqrt{M_2/M_1}. At equal pressure and temperature density is proportional to molar mass, so the same ratio may be written r_1/r_2=\sqrt{\rho_2/\rho_1}.

A time is the reciprocal of a rate. For equal amounts passing under the same conditions, t_1/t_2=r_2/r_1=\sqrt{M_1/M_2}. The law has the same mass dependence as every Maxwell speed because the faster molecules reach and cross the opening more often; do not reverse the ratio when the question switches from rate to time.

Figure. With oxygen's rate set to 1, helium's rate is √(32/4) = √8 ≈ 2.83. The bars compare rates under identical temperature, pressure and opening conditions.

Rate ratio without a sign error

1. Put the asked gas on topFor r₁/r₂, put the other gas's molar mass M₂ in the numerator under the square root.
2. Check lighter means fasterIf gas 1 is lighter, the computed r₁/r₂ must exceed 1.
3. Invert for timeFor equal transferred amounts, t₁/t₂ = r₂/r₁; the faster gas takes less time.

9Degrees of freedom and specific heats

Equipartition says every active quadratic energy term carries (1/2)k_BT per molecule, i.e. (1/2)RT per mole. There are three translational terms for every molecule. A rigid linear molecule adds two rotational terms; a rigid non-linear molecule adds three. Each active vibrational mode contributes two terms — one kinetic and one potential — and therefore contributes k_BT per molecule rather than half of it.

From the total number f of active quadratic terms everything drops out: molar C_V=(f/2)R, then Mayer's relation C_P=C_V+R, and \gamma=C_P/C_V=1+2/f. The extra R in C_P is not extra molecular energy storage — it is the P\Delta V=R\Delta T of work one mole must do while expanding at constant pressure.

At ordinary JEE temperatures, vibrational modes of common diatomic gases are usually treated as frozen, so f = 5, C_V=5R/2, C_P=7R/2 and γ = 7/5. At higher temperature vibration becomes active and the heat capacities rise while γ falls. If the question asks for specific heat per unit mass rather than molar heat capacity, divide by molar mass: c_V=C_V/M and c_P=C_P/M.

Figure. Equipartition counts active quadratic terms: 3 translational for every gas, plus 2 rotations for a rigid linear molecule (f = 5), plus a third rotation when the molecule is non-linear (f = 6). The excluded bond-axis spin never earns a term.

How it works

1. Count f3 for a monatom, 5 for a rigid linear molecule, 6 for a rigid non-linear molecule; add 2 for each active vibrational mode.
2. C_V = (f/2)RInternal energy per mole is (f/2)RT, so heating at fixed V costs (f/2)R.
3. Add R for C_PAt constant pressure the gas also does RΔT of work, so C_P = C_V + R.

10Heat engines and the Carnot limit

An engine works in cycles, taking Q_H from a hot reservoir, dumping Q_C into a cold one and delivering the difference as work: W = Q_H − Q_C, with efficiency η = W/Q_H = 1 − Q_C/Q_H. The second law, in Kelvin's phrasing, forbids Q_C from being zero — no cycle turns heat entirely into work, so η < 1 always.

Carnot's theorem sharpens that into a number. Between two reservoirs no engine beats η = 1 − T_C/T_H, achieved by the reversible cycle of two isotherms and two adiabats, and the ceiling depends on the two temperatures alone — not on the gas, the design, or the engineering budget. Temperatures must be in kelvin. Run the same cycle backwards and it becomes a refrigerator, moving heat uphill with COP = Q_C/W = T_C/(T_H − T_C).

Figure. Heat enters only along AB, at T_H, and leaves only along CD, at T_C; the two adiabatic legs BC and DA exchange none. That is why the efficiency can depend on nothing but the two temperatures.

How it works

1. Split the heatEvery joule from the source either leaves as work or is dumped into the sink.
2. Kelvin onlyη = 1 − T_C/T_H is meaningless in Celsius — convert first, every time.
3. Reverse itDriven backwards, the engine pumps heat: COP = T_C/(T_H − T_C).