Thermodynamics: theory | Physics Circle

CONCEPT

Gas Laws & Ideal Gas Equation

In one line
A gas is described by four quantities: P, V, T and amount n. The gas laws link them, and together they give PV = nRT.

SYSTEM

13.1.1 Thermodynamic system

System: the part (usually a gas) we study; everything outside is the surroundings.
Open: exchanges energy and matter. Closed: energy only. Isolated: neither.
State variables: P, V, T, U, n. They fix the state of the system.

Open, closed and isolated systems

ZEROTH LAW

13.1.2 Zeroth law of thermodynamics

If A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other.
This law defines temperature: bodies in equilibrium have the same T.

BOYLE’S LAW

13.1.3 Boyle’s law (T constant)

PV = constant · P₁V₁ = P₂V₂

Also Pρ = constant, since ρ ∝ 1V.
P–V graph: rectangular hyperbola. PV–P graph: horizontal line.
Volume falls by x% ⇒ pressure rises by 100x100 − x% (not x%).

CHARLES’S LAW

13.1.4 Charles’s law (P constant)

V ∝ T · V₁T₁ = V₂T₂

In °C: V_t = V₀(1 + t273.15).
V–t line meets the t-axis at −273.15°C. Always use kelvin in the formula.

GAY-LUSSAC’S LAW

13.1.5 Gay-Lussac’s law (V constant)

P ∝ T · P₁T₁ = P₂T₂

∝ TTPGay-Lussac: P ∝ T

Boyle, Charles and Gay-Lussac graphs

AVOGADRO AND DALTON

13.1.6 Avogadro’s and Dalton’s laws

Avogadro: equal volumes of all gases at the same P and T have equal numbers of molecules.
1 mole of any gas at STP occupies 22.4 L and has N_A = 6.023 × 10²³ molecules.
Dalton: total pressure of non-reacting gases = sum of partial pressures.

P = P₁ + P₂ + P₃ + …

IDEAL GAS EQUATION

13.1.7 Ideal gas equation

PV = nRT

Amount of gas	Equation	Constant
n moles	PV = nRT	R = 8.31 J mol⁻¹ K⁻¹
N molecules	PV = Nk_BT	k_B = RN_A = 1.38 × 10⁻²³ J K⁻¹
mass m	PV = mMRT	M = molar mass

R ≈ 8.31 J mol⁻¹ K⁻¹ ≈ 2 cal mol⁻¹ K⁻¹.
PV has the unit of energy (N m = J).

DENSITY AND NUMBER

13.1.8 Density and number of molecules

ρ = PMRT · ρP ∝ 1T

N ∝ PVT · n_v = NV = Pk_BT

Same P, V, T ⇒ same number of molecules, whatever the gas.
Gas leaking from a fixed vessel: mass left ∝ PT.
Two vessels joined, total internal energy kept: P(V₁ + V₂) = P₁V₁ + P₂V₂ (same T) and n is conserved.

Practise PYQs on this topicGas Laws & Ideal Gas Equation · 86 questions with solutions

CONCEPT

Degrees of Freedom & Internal Energy

In one line
Degrees of freedom count the independent ways a molecule can move. Each one stores energy, so they decide the internal energy of a gas.

DEGREES OF FREEDOM

13.2.1 Degrees of freedom

Degrees of freedom

A = number of atoms, B = number of fixed bonds (constraints).
Translational f is always 3 (motion along x, y, z).
Rotation about the axis of a linear molecule is not counted (negligible moment of inertia).

Monoatomic, diatomic and non-linear triatomic molecules

Gas	Example	f (room temperature)
monoatomic	He, Ne, Ar	3 (trans)
diatomic	H₂, O₂, N₂	5 (3 trans + 2 rot)
triatomic linear	CO₂	5 (rigid)
triatomic non-linear	H₂O, CH₄	6 (3 trans + 3 rot)
diatomic, vibrating	high T	7 (+2 vibrational)

One vibrational mode adds 2 degrees of freedom (kinetic + potential energy).

INTERNAL ENERGY

13.2.2 Internal energy

U = kinetic energy + potential energy of the molecules.
Ideal gas: no intermolecular force ⇒ U is wholly kinetic and depends on T only.
Each degree of freedom carries 12k_BT per molecule (law of equipartition, proved in Kinetic Theory of Gases).

U = f2nRT · ΔU = f2nRΔT = nC_vΔT

U = f2PV

U is a state function: ΔU depends only on the initial and final states, not on the path.
For a cyclic process ΔU = 0.

MIXTURES

13.2.3 Internal energy of a mixture

U = f₁2n₁RT + f₂2n₂RT

f_mix = n₁f₁ + n₂f₂n₁ + n₂ · T_mix = n₁f₁T₁ + n₂f₂T₂n₁f₁ + n₂f₂

MOVING CONTAINER STOPPED

13.2.4 Moving container stopped suddenly

12Mv² = f2RΔT ⇒ ΔT = Mv²fR = (γ − 1)Mv²2R

M = molar mass. Insulated container: no heat is lost.
Monoatomic: ΔT = Mv² ÷ 3R. Diatomic: ΔT = Mv² ÷ 5R.

Practise PYQs on this topicDegrees of Freedom & Internal Energy · 45 questions with solutions

CONCEPT

Specific Heats of Gases

In one line
A gas has two main specific heats: at constant volume (C_v) and at constant pressure (C_p). C_p is larger because some heat goes into work.

MOLAR SPECIFIC HEATS

13.3.1 Molar specific heats

C_v = 1n(ΔQΔT)_V · C_p = 1n(ΔQΔT)_P

Specific heat per unit mass: c = C ÷ M.
Constant volume: all heat raises U. Constant pressure: heat raises U and does work PΔV.

MAYER’S RELATION

13.3.2 Mayer’s relation

C_p − C_v = R

Per unit mass: c_p − c_v = RM.
So C_p > C_v always.
Using cal: R ≈ 2 cal mol⁻¹ K⁻¹; then J = R (in J)R (in cal).

IN TERMS OF f

13.3.3 Specific heats in terms of f

C_v = f2R · C_p = (f2 + 1)R

γ = C_pC_v = 1 + 2f

C_v = Rγ − 1 · C_p = γRγ − 1 · f = 2γ − 1

Gas	f	C_v	C_p	γ
monoatomic	3	32R	52R	53 = 1.67
diatomic (rigid)	5	52R	72R	75 = 1.4
diatomic (vibrating)	7	72R	92R	97 = 1.29
polyatomic (non-linear)	6	3R	4R	43 = 1.33

More degrees of freedom ⇒ C_v and C_p rise, γ falls towards 1.
γ is always greater than 1.
RC_v = γ − 1 (asked often).

MIXTURE OF GASES

13.3.4 Specific heat of a mixture

C_v,mix = n₁C_v1 + n₂C_v2n₁ + n₂ · C_p,mix = C_v,mix + R

γ_mix = C_p,mixC_v,mix · n₁ + n₂γ_mix − 1 = n₁γ₁ − 1 + n₂γ₂ − 1

Never average γ directly; average C_v first.

HEAT CAPACITY OF A PROCESS

13.3.5 Heat capacity of a process

Process	Molar heat capacity
isochoric	C_v
isobaric	C_p
isothermal (ΔT = 0)	∞
adiabatic (ΔQ = 0)	0
polytropic PV^x = const	C_v + R1 − x

Practise PYQs on this topicSpecific Heats of Gases - Cp, Cv, γ & Mayer's Relation · 65 questions with solutions

CONCEPT

Heat, Work & First Law

In one line
Heat given to a gas is shared between raising its internal energy and doing work: ΔQ = ΔU + ΔW.

HEAT AND WORK

13.4.1 Heat and work

Heat flows because of a temperature difference.
Work done by a gas against a piston:

dW = P dV · W = ∫_V₁^V₂ P dV

Work = area under the P–V curve

Heat and work are path functions; U is a state function.

FIRST LAW

13.4.2 First law of thermodynamics

ΔQ = ΔU + ΔW

Quantity	+	−
ΔQ	heat given to the system	heat taken out
ΔW	work done by the gas (expansion)	work done on the gas (compression)
ΔU	temperature rises	temperature falls

All three in the same unit (J; 1 cal = 4.2 J).
If W is the work done on the gas: ΔQ = ΔU − W.
As rates: dQdt = dUdt + dWdt (power in, power out).
Limitation: it does not tell the direction of heat flow (that is the second law).

CHANGE OF STATE

13.4.3 Change of state

ΔU = mL − PΔV

Boiling: volume rises a lot ⇒ part of mL goes into work.
Melting of ice: volume falls ⇒ work is done on the system by the atmosphere.

Practise PYQs on this topicHeat, Work & First Law of Thermodynamics · 24 questions with solutions

CONCEPT

Isobaric & Isochoric Processes

In one line
A thermodynamic process changes the state (P, V, T) of a gas. In an isobaric process P stays fixed; in an isochoric process V stays fixed.

TYPES OF PROCESSES

13.5.1 Types of processes

Process	Constant	Key result
isobaric	P	V ∝ T
isochoric	V	W = 0
isothermal	T	ΔU = 0
adiabatic	Q (no heat flow)	ΔU = −W
cyclic	returns to start	ΔU = 0, Q = W

Isobaric and isochoric lines on a P–V graph

ISOBARIC

13.5.2 Isobaric process

W = PΔV = nRΔT · ΔU = nC_vΔT · Q = nC_pΔT

ΔQ : ΔU : ΔW = C_p : C_v : R = γ : 1 : (γ − 1)

Charles’s law holds: V ∝ T.
Fraction of heat used as work = RC_p = 1 − 1γ.
Monoatomic: 40% work, 60% to U. Diatomic: 27 work, 57 to U.
ΔU = PΔVγ − 1.

Gas	ΔQ : ΔU : ΔW
monoatomic	5 : 3 : 2
diatomic	7 : 5 : 2

ISOCHORIC

13.5.3 Isochoric process

W = 0 · Q = ΔU = nC_vΔT

Gay-Lussac’s law holds: P ∝ T.
All heat supplied raises the internal energy.
On a P–V graph: a vertical line.

SAME HEAT, TWO CYLINDERS

13.5.4 Same heat, two cylinders

ΔT_P = ΔT_Vγ

Free piston (constant P) heats up less, because part of the heat does work.

Practise PYQs on this topicThermodynamic Processes · 255 questions with solutions

CONCEPT

Isothermal & Adiabatic Processes

In one line
Isothermal: temperature constant, heat flows freely. Adiabatic: no heat flow, so the temperature changes.

ISOTHERMAL

13.6.1 Isothermal process

PV = constant · ΔU = 0 · Q = W

W = nRT lnV₂V₁ = 2.303 nRT logV₂V₁ = nRT lnP₁P₂

Isotherms: higher T lies farther from the axes

Walls perfectly conducting; process very slow.
Examples: melting of ice, boiling of water.
Molar heat capacity is infinite; v_rms does not change.
Slope of P–V curve = −PV. Isothermal bulk modulus = P.

ADIABATIC

13.6.2 Adiabatic process

PV^γ = const · TV^{γ − 1} = const

T^γP^{γ − 1} = const

W = nR(T₁ − T₂)γ − 1 = P₁V₁ − P₂V₂γ − 1 · ΔU = −W

Sudden compression or expansion; bursting of a tyre; sound waves in air.
Expansion cools the gas; compression heats it.
P ∝ T^{γ/(γ − 1)}: monoatomic 2.5, diatomic 3.5.
Molar heat capacity is zero. Adiabatic bulk modulus = γP.

Gas	P ∝	T ∝
monoatomic (γ = 53)	V⁻⁵³	V⁻²³
diatomic (γ = 75)	V⁻⁷⁵	V⁻²⁵
polyatomic (γ = 43)	V⁻⁴³	V⁻¹³

COMPARISON

13.6.3 Isothermal vs adiabatic

Through the same point the adiabatic curve is steeper

(slope)_adiabatic = γ × (slope)_isothermal

Equal expansion: W_iso > W_adi, final P_iso > P_adi.
Equal compression: W_adi > W_iso, final P_adi > P_iso.
Expanded to 2V: P_isoP_adi = 2^{γ − 1}.

Practise PYQs on this topicThermodynamic Processes · 255 questions with solutions

CONCEPT

Polytropic & Other Processes

In one line
Every process of the form PV^x = constant is polytropic. Isobaric, isothermal, adiabatic and isochoric are its special cases.

POLYTROPIC PROCESS

13.7.1 Polytropic process

W = P₁V₁ − P₂V₂x − 1 = nR(T₁ − T₂)x − 1

C = C_v + R1 − x = Rγ − 1 − Rx − 1

x	Process	C
0	isobaric	C_p
1	isothermal	∞
γ	adiabatic	0
∞	isochoric	C_v

C = 0 ⇒ x = γ (adiabatic).
Monoatomic gas with PV³ = constant: C = 32R − R2 = R.

READING T FROM THE LAW

13.7.2 Reading T from the process law

PV² = constant ⇒ T ∝ 1V: the gas cools on expansion.
VP² = constant ⇒ P ∝ V⁻¹² ⇒ T ∝ √V.
P ∝ V (straight line through origin) ⇒ T ∝ V².

ADIABATIC EXPONENTS

13.7.3 Adiabatic exponents

Relation	Exponent
P ∝ T^x	γγ − 1
TV^x = const	γ − 1
P ∝ V^−x	γ

Practise PYQs on this topicThermodynamic Processes · 255 questions with solutions

CONCEPT

P-V Diagrams & Cyclic Processes

In one line
On a P–V diagram, the area under the curve is the work. Work depends on the path; internal energy does not.

WORK FROM A P-V GRAPH

13.8.1 Work from a P-V graph

Expansion (V increases): W positive. Compression: W negative.
Straight-line path: W = area of the trapezium.

trapezium: W = 12(P₁ + P₂)(V₂ − V₁)

Same end points, different work

TWO PATHS, SAME END POINTS

13.8.2 Two paths, same end points

ΔU = Q₁ − W₁ = Q₂ − W₂

Find ΔU from the path whose Q and W are known.
Then Q or W on the other path follows.
Change in U from the end points: ΔU = f2(P₂V₂ − P₁V₁).

CYCLIC PROCESS

13.8.3 Cyclic process

ΔU = 0 · Q_net = W_net = area enclosed

Clockwise cycle: positive work; anticlockwise: negative

Clockwise on P–V: W positive (engine). Anticlockwise: W negative (refrigerator).
Rectangle: W = ΔP × ΔV; circle: W = πr_Pr_V.
Same cycle on a V–P graph (axes swapped): the sign is reversed.

FREE EXPANSION

13.8.4 Free expansion

Sudden and uncontrolled; intermediate states are not equilibrium states, so it cannot be drawn on a P–V diagram.
Ideal gas: T does not change.

GRAPHS OF THE FOUR PROCESSES

13.8.5 Graphs of the four processes

Same four processes on three kinds of graph

Process	P–V	P–T	V–T
isobaric	horizontal	horizontal	line through origin
isochoric	vertical	line through origin	horizontal
isothermal	hyperbola	vertical	vertical
adiabatic	steeper hyperbola	P ∝ T^γ/(γ−1)	V ∝ T^−1γ−1

Practise PYQs on this topicP-V Diagrams & Cyclic Processes · 92 questions with solutions

CONCEPT

Heat Engines, Second Law & Refrigerators

In one line
A heat engine runs in a cycle: it takes heat Q₁ from a hot source, turns part of it into work and rejects Q₂ to a cold sink. No engine can turn all heat into work.

HEAT ENGINE

13.9.1 Heat engine

Heat engine

W = Q₁ − Q₂ · η = WQ₁ = 1 − Q₂Q₁

Working substance: steam, petrol vapour, an ideal gas.
Real engines: steam ~ 8–15%, petrol ~ 40%, diesel ~ 50–55%.

SECOND LAW

13.9.2 Second law of thermodynamics

Kelvin–Planck: no engine working in a cycle can convert all the heat taken from a source into work. So η < 1.
Clausius: heat cannot flow from a colder body to a hotter body by itself, without external work.

CARNOT ENGINE

13.9.3 Carnot engine

Carnot cycle: two isotherms, two adiabats

A → B isothermal expansion at T₁ (heat Q₁ in).
B → C adiabatic expansion (T falls to T₂).
C → D isothermal compression at T₂ (heat Q₂ out).
D → A adiabatic compression (back to T₁).

Q₂Q₁ = T₂T₁ · η = 1 − T₂T₁

Use kelvin. η depends only on T₁ and T₂, not on the working substance.
η = 1 only if T₂ = 0 K (impossible).
Carnot theorem: no engine between the same two temperatures beats a Carnot engine.
Sink lowered by ΔT at fixed T₁: Δη = ΔTT₁.

REFRIGERATOR

13.9.4 Refrigerator (heat pump)

Refrigerator

β = Q₂W = Q₂Q₁ − Q₂ = T₂T₁ − T₂

β = 1 − ηη · η = 11 + β

T₂ = inside (cold), T₁ = room (hot).
β can be greater than 1. β = 0 when T₂ = 0 K.
Heat released to the room Q₁ = Q₂ + W; Q₂ = βQ₁1 + β.
Power needed = Q₂ per second ÷ β.

Practise PYQs on this topicHeat Engines, Refrigerators & Second Law · 95 questions with solutions

EXAM-DAY CARD

Just see this before the exam

PV = nRT

gas law

U = (f2)nRT

internal energy

C_p − C_v = R

Mayer

ΔQ = ΔU + ΔW

first law

PV^γ = const

adiabatic

η = 1 − T₂T₁

Carnot

FORMULA SHEET

Whole chapter on one page

Gas laws

PV = nRT, PV = Nk_BT

ρ = PMRT

P₁V₁T₁ = P₂V₂T₂

P = P₁ + P₂ + …

Internal energy

f = 3A − B

U = (f2)nRT

ΔU = nC_vΔT

stopped: ΔT = Mv²fR

Specific heats

C_p − C_v = R

C_v = Rγ − 1

γ = 1 + 2f

C = C_v + R1 − x

First law

ΔQ = ΔU + ΔW

W = ∫P dV = area

isobaric: W = nRΔT

isochoric: W = 0

Iso / adiabatic

W_iso = nRT ln(V₂V₁)

PV^γ, TV^γ−1 = const

W_adi = nR(T₁ − T₂)γ − 1

slope ratio = γ

Engines

η = 1 − Q₂Q₁ = 1 − T₂T₁

β = T₂T₁ − T₂

β = (1 − η)/η

cycle: Q = W, ΔU = 0

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using °C in PV = nRT or η = 1 − T₂T₁

Averaging γ of a mixture directly

Isothermal: thinking Q = 0

Cycle sign: ignoring the direction

CORRECT ✓

Always convert to kelvin

Average C_v by moles, then γ

Isothermal: ΔU = 0, Q = W

Clockwise on P–V: W > 0