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.

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=nRTPV = 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 nn moles of an ideal gas: PV=nRTPV = nRT, with TT the absolute (Kelvin) temperature. Special cases (combined gas law):

  • Constant TT (Boyle's law): PV=constPV = \text{const}.
  • Constant PP (Charles's law): VTV \propto T.
  • Constant VV (Gay-Lussac's law): PTP \propto T.
  • Constant TT and VV: PnP \propto n — pressure scales with the number of molecules.

Ideal gas law and the combined gas law

PV=nRTP1V1T1=P2V2T2PV = 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

From the bank · past-year question

Example 1Heat and ThermodynamicsMODERATE
A stainless steel chamber contains Ar gas at a temperature TT and pressure PP. The total number of Ar atoms in the chamber is nn. Now Ar gas in the chamber is replaced by CO2\text{CO}_2 gas and the total number of CO2\text{CO}_2 molecules in the chamber is n/2n/2 at the same temperature TT. The pressure in the chamber now is PP'. 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=nRTPV = nRT or in PTP \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: ΔU=QW\Delta U = Q - W. If no work is done, all the heat shows up as internal energy.

Definition

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

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

(Sign convention: QQ positive when heat enters, WW positive when the gas does work by expanding.)

First law of thermodynamics

ΔU=QW\Delta U = Q - W
  • ΔU\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?
Practice this conceptself-check · 4 quick reps

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 (ΔU=0\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=nRTPV = nRT and see which variable ends up constant.

Definition

Four standard processes:

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

For a process given as an unusual equation, substitute PV=nRTPV = 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    V=nRk=const    isochoric,  C=CVP = 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
V (volume)Pstartisobaric (P fixed)isochoric (V fixed)isothermal (T fixed)adiabatic (Q = 0)

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

From the bank · past-year question

Example 3Heat and ThermodynamicsHARD
For an ideal gas, a process is described by P=kTP = kT, where kk is a constant. If the molar heat capacity for this process is CC, then which one of the following is correct?

[Q55 · Apr · 2026]

Don't guess the process — substitute PV = nRT

A process given as P=kTP = kT or PV2=constPV^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=0Q = 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.

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 / lawWhat is held / statedKey consequence
IsothermalTemperature constantΔU=0\Delta U = 0; PV=constPV = \text{const}; all heat becomes work
AdiabaticNo heat exchanged (Q = 0)Insulated; temperature still changes (compression heats the gas)
IsochoricVolume constant (W = 0)ΔU=Q\Delta U = Q; molar heat capacity CVC_V; PTP \propto T
IsobaricPressure constantMolar heat capacity CPC_P (and CP>CVC_P > C_V); VTV \propto T
Second lawHeat won't flow cold → hot unaidedExternal 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

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=nRTP1V1T1=P2V2T2PV = nRT \qquad \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
  • First law of thermodynamics

    First law of thermodynamics

    ΔU=QW\Delta U = Q - W
  • Named processes — isothermal, adiabatic, isochoric, isobaric

    Identify a process by substituting PV = nRT

    P=kT    V=nRk=const    isochoric,  C=CVP = 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 / lawWhat is held / statedKey consequence
IsothermalTemperature constantΔU=0\Delta U = 0; PV=constPV = \text{const}; all heat becomes work
AdiabaticNo heat exchanged (Q = 0)Insulated; temperature still changes (compression heats the gas)
IsochoricVolume constant (W = 0)ΔU=Q\Delta U = Q; molar heat capacity CVC_V; PTP \propto T
IsobaricPressure constantMolar heat capacity CPC_P (and CP>CVC_P > C_V); VTV \propto T
Second lawHeat won't flow cold → hot unaidedExternal 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)

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