NDA Physics · Heat and Thermodynamics

Gas Laws and the Laws of Thermodynamics

An ideal gas obeys PV = nRT; the first law (ΔU = Q − W) tracks energy bookkeeping, and named processes — isothermal, adiabatic, isochoric, isobaric — each fix one variable and decide which heat capacity applies.

All Heat and Thermodynamics notes NDA Physics strategy

Why this matters

About 4 PYQs but punching above its weight in difficulty — recent HARD problems use a custom process (P = kT, PV² = constant) and ask you to identify its nature using the ideal gas law. The recall layer is the named processes (adiabatic = no heat exchange) and the laws (second law = heat won't flow uphill on its own). The HARD layer is combining the ideal gas law PV = nRT with the given process equation to deduce what stays constant.

Concept 1 of 4

The ideal gas law

Intuition

An ideal gas links three quantities — pressure, volume, and absolute temperature — in one equation,

PV = nRT

. Fix any of them and the other two trade off: heat a gas at constant volume and its pressure rises; squeeze it at constant temperature and its pressure climbs. At constant temperature and volume, pressure tracks the NUMBER of molecules.

Definition

For $n$ moles of an ideal gas: $PV = nRT$ , with $T$ the absolute (Kelvin) temperature. Special cases (combined gas law):

Constant $T$ (Boyle's law): $PV = \text{const}$ .
Constant $P$ (Charles's law): $V \propto T$ .
Constant $V$ (Gay-Lussac's law): $P \propto T$ .
Constant $T$ and $V$ : $P \propto n$ — pressure scales with the number of molecules.

Ideal gas law and the combined gas law

PV = nRT \qquad \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Ppressure
Vvolume
nnumber of moles (or molecules)
Runiversal gas constant
Tabsolute temperature (K)

Worked example

A rigid chamber holds n argon atoms at temperature T and pressure P. The argon is replaced by n/2 carbon-dioxide molecules at the same temperature T. What is the new pressure P′?

The chamber is rigid (constant V) and the temperature is unchanged (constant T).
From $PV = nRT$ , at fixed V and T, $P \propto n$ — pressure depends only on the number of molecules.
The number of molecules is halved (n → n/2), so the pressure halves.
$P' = P/2$ .

Answer:P′ = P/2.

Practice this conceptself-check · 4 quick reps

Try it yourself

An ideal gas at 300 K and pressure P is heated at constant volume to 600 K. What is the new pressure?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
State the ideal gas law.
2.
At constant T and V, pressure is proportional to?
3.
At constant T, PV = ? (Boyle's law)
4.
Must T be in °C or K in PV = nRT?

From the bank · past-year question

Example 1Heat and ThermodynamicsMODERATE

A stainless steel chamber contains Ar gas at a temperature

T

and pressure

P

. The total number of Ar atoms in the chamber is

n

. Now Ar gas in the chamber is replaced by

\text{CO}_2

gas and the total number of

\text{CO}_2

molecules in the chamber is

n/2

at the same temperature

T

. The pressure in the chamber now is

P'

. Which one of the following relations holds true? (Both the gases behave as ideal gases)

[Q69 · Sep · 2018]

Temperature in the gas law is ALWAYS in kelvin

Using Celsius in

PV = nRT

or in

P \propto T

gives wrong ratios. Convert to kelvin first. 'Pressure doubles when temperature doubles' is only true on the absolute scale.

Concept 2 of 4

First law of thermodynamics

Intuition

The first law is energy conservation for a gas. Heat you put IN either raises the gas's internal energy or gets spent doing work as the gas expands. Nothing is lost:

\Delta U = Q - W

. If no work is done, all the heat shows up as internal energy.

Definition

First law: $\Delta U = Q - W$ . The heat $Q$ supplied to a system equals the increase in its internal energy $\Delta U$ plus the work $W$ done BY the system.

If ** $W = 0$ ** (rigid container): $\Delta U = Q$ — all heat goes to internal energy.
Internal energy of an ideal gas depends only on temperature, so $\Delta U = 0$ for any isothermal process.

(Sign convention: $Q$ positive when heat enters, $W$ positive when the gas does work by expanding.)

First law of thermodynamics

\Delta U = Q - W

\Delta Uchange in internal energy
Qheat supplied to the system
Wwork done BY the system

Worked example

A gas is held in a rigid container so that no work is done on or by it. How does the change in internal energy relate to the heat exchanged?

Rigid container → the gas cannot expand or be compressed → $W = 0$ .
First law: $\Delta U = Q - W = Q - 0$ .
So the change in internal energy equals the heat flowing in or out.

Answer:The change in internal energy equals the heat exchanged (ΔU = Q).

Practice this conceptself-check · 4 quick reps

Try it yourself

A gas absorbs 200 J of heat and does 80 J of work in expanding. What is the change in its internal energy?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
State the first law of thermodynamics.
2.
If no work is done, ΔU equals what?
3.
For an isothermal process on an ideal gas, ΔU = ?
4.
Gas absorbs 50 J and does 50 J of work. Find ΔU.

From the bank · past-year question

Example 2Heat and ThermodynamicsMODERATE

If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct?

[Q69 · Sep · 2019]

Internal energy of an ideal gas depends only on temperature

In an ISOTHERMAL process (constant T) the internal energy does not change at all (

\Delta U = 0

), so any heat absorbed is entirely converted to work. Don't assume absorbing heat always raises internal energy.

Concept 3 of 4

Named processes — isothermal, adiabatic, isochoric, isobaric

Intuition

Each named process fixes ONE thing. Isothermal holds temperature constant; adiabatic exchanges no heat; isochoric (isovolumetric) holds volume constant; isobaric holds pressure constant. For an unfamiliar process given as an equation, the trick is to combine it with

PV = nRT

and see which variable ends up constant.

Definition

Four standard processes:

Isothermal — constant temperature ( $\Delta U = 0$ ); $PV = \text{const}$ .
Adiabatic — no heat exchange with surroundings ( $Q = 0$ ); a perfectly insulated system.
Isochoric (isovolumetric) — constant volume ( $W = 0$ ); molar heat capacity $= C_V$ .
Isobaric — constant pressure; molar heat capacity $= C_P$ (and $C_P > C_V$ ).

For a process given as an unusual equation, substitute $PV = nRT$ to find what is held fixed and hence which heat capacity / relation applies. The P–V diagram below shows how the four processes look as curves from a common start.

Identify a process by substituting PV = nRT

P = kT \;\Rightarrow\; V = \frac{nR}{k} = \text{const} \;\Rightarrow\; \text{isochoric},\; C = C_V

kthe constant in the given process equation
C_Vmolar heat capacity at constant volume
C_Pmolar heat capacity at constant pressure

From one start: isobaric holds P, isochoric holds V, isothermal follows PV = const, and the adiabatic curve (no heat exchange) is steeper than the isothermal one.

Worked example

For one mole of an ideal gas a process obeys P = kT (k constant). What is its molar heat capacity C for this process?

Use the ideal gas law for one mole: $PV = RT$ .
Substitute the process condition $P = kT$ : $(kT)V = RT$ , so $V = R/k$ .
$R/k$ is a constant, so the volume is fixed — this is an isochoric (constant-volume) process.
At constant volume the molar heat capacity is $C_V$ . Hence $C = C_V$ .

Answer:C = C_V (the process is isochoric).

Practice this conceptself-check · 4 quick reps

Try it yourself

An ideal gas undergoes a process with

PV^2 = \text{constant}

. If the initial state is

(T_1, V_1)

and final

(T_2, V_2)

, find the relation between the temperatures and volumes.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which process exchanges no heat with the surroundings?
2.
Which process holds volume constant, so W = 0?
3.
Molar heat capacity at constant pressure is denoted?
4.
In an isothermal process, ΔU = ?

From the bank · past-year question

Example 3Heat and ThermodynamicsHARD

For an ideal gas, a process is described by

P = kT

, where

k

is a constant. If the molar heat capacity for this process is

C

, then which one of the following is correct?

[Q55 · Apr · 2026]

Don't guess the process — substitute PV = nRT

A process given as

P = kT

PV^2 = \text{const}

is not one of the four standard names on sight. Substitute the ideal gas law to see which variable is actually constant, then read off the heat capacity or T-V relation.

Adiabatic means no HEAT exchange, not no temperature change

An adiabatic process has

Q = 0

but the temperature usually DOES change (an adiabatic compression heats a gas). 'No heat exchange' is the definition; 'constant temperature' is isothermal — a different process.

Drill 2 more on named processes — isothermal, adiabatic, isochoric, isobaric

Concept 4 of 4

The second law and a process summary table

Intuition

The first law says energy is conserved, but it does not say which way heat flows. The SECOND law fixes the direction: heat will not flow on its own from a colder body to a hotter one — you need work (a refrigerator) to push it uphill. This table also collects the four named processes as a one-glance recall.

Definition

Second law of thermodynamics: heat cannot flow by itself from a body at lower temperature to one at higher temperature; some external work is always needed to do so (the basis of refrigerators and heat engines). The table below summarises the named processes for quick recall.

Process / law	What is held / stated	Key consequence
Isothermal	Temperature constant	$\Delta U = 0$ ; $PV = \text{const}$ ; all heat becomes work
Adiabatic	No heat exchanged (Q = 0)	Insulated; temperature still changes (compression heats the gas)
Isochoric	Volume constant (W = 0)	$\Delta U = Q$ ; molar heat capacity $C_V$ ; $P \propto T$
Isobaric	Pressure constant	Molar heat capacity $C_P$ (and $C_P > C_V$ ); $V \propto T$
Second law	Heat won't flow cold → hot unaided	External work needed to move heat uphill (refrigerator); sets the direction of natural processes NDA 2017 — 'heat cannot flow by itself from a lower to a higher temperature' is the SECOND law of thermodynamics.

The first law is energy bookkeeping (ΔU = Q − W); the second law sets the one-way direction of heat flow.

Practice this conceptself-check · 5 quick reps

Try it yourself

The statement 'heat cannot flow by itself from a body at lower temperature to a body at higher temperature' is which law of thermodynamics?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

1.
A system that exchanges NO heat with its surroundings is called?
2.
Which law says heat won't flow cold-to-hot on its own?
3.
Which process has molar heat capacity C_V?
4.
Which is larger, C_P or C_V?
5.
Which law is energy conservation, ΔU = Q − W?

From the bank · past-year question

Example 4Heat and ThermodynamicsEASY

The statement that 'heat cannot flow by itself from a body at a lower temperature to a body at a higher temperature', is known as

[Q127 · Sep · 2017]

First law = energy; second law = direction

The first law (ΔU = Q − W) is conservation of energy and is direction-blind. The second law adds the arrow: heat flows hot → cold spontaneously, never the reverse without work. Statements about 'cannot flow by itself' point to the SECOND law.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The ideal gas law
Ideal gas law and the combined gas law
$PV = nRT \qquad \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
First law of thermodynamics
First law of thermodynamics
$\Delta U = Q - W$
Named processes — isothermal, adiabatic, isochoric, isobaric
Identify a process by substituting PV = nRT
$P = kT \;\Rightarrow\; V = \frac{nR}{k} = \text{const} \;\Rightarrow\; \text{isochoric},\; C = C_V$

Reference tables (1)

The second law and a process summary table5 rows

Process / law	What is held / stated	Key consequence
Isothermal	Temperature constant	$\Delta U = 0$ ; $PV = \text{const}$ ; all heat becomes work
Adiabatic	No heat exchanged (Q = 0)	Insulated; temperature still changes (compression heats the gas)
Isochoric	Volume constant (W = 0)	$\Delta U = Q$ ; molar heat capacity $C_V$ ; $P \propto T$
Isobaric	Pressure constant	Molar heat capacity $C_P$ (and $C_P > C_V$ ); $V \propto T$
Second law	Heat won't flow cold → hot unaided	External work needed to move heat uphill (refrigerator); sets the direction of natural processes NDA 2017 — 'heat cannot flow by itself from a lower to a higher temperature' is the SECOND law of thermodynamics.

The first law is energy bookkeeping (ΔU = Q − W); the second law sets the one-way direction of heat flow.

Watch out for (5)

Temperature in the gas law is ALWAYS in kelvin
→ The ideal gas law
Internal energy of an ideal gas depends only on temperature
→ First law of thermodynamics
Don't guess the process — substitute PV = nRT
→ Named processes — isothermal, adiabatic, isochoric, isobaric
Adiabatic means no HEAT exchange, not no temperature change
→ Named processes — isothermal, adiabatic, isochoric, isobaric
First law = energy; second law = direction
→ The second law and a process summary table

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Heat and ThermodynamicsEASY

A system that does NOT allow exchange of heat with its surrounding is called

[Q140 · Apr · 2025]

Example 2Heat and ThermodynamicsHARD

In a certain process,

PV^2 = \text{constant}

for an ideal gas. If initial temperature is

T_1

, final temperature

T_2

, initial volume

V_1

, final volume

V_2

, then which one is correct?

[Q54 · Apr · 2026]

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