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

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′?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2018 · Q69Moderate

Example 1 · Heat and Thermodynamics · Thermodynamic Processes

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)

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

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

$\Delta U$ change 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?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q69Moderate

Example 2 · Heat and Thermodynamics · Thermodynamic Processes

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?

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

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?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2026 · Q55Hard

Example 3 · Heat and Thermodynamics · Thermodynamic Processes

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?

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

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

The same idea in a real exam question:

NDA · Sep 2017 · Q127Easy

Example 4 · Heat and Thermodynamics · Thermodynamic Processes

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

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

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