Physics · Thermal Properties of Matter

Ideal Gas Equation & Absolute Temperature

A liquid-in-glass thermometer reads slightly differently depending on the liquid inside it, because every liquid expands by its own law. A gas does not have this defect: all gases at low density expand the same way, so a gas makes a far better thermometric standard. NCERT §10.4 builds the ideal gas equation $PV = \mu RT$ from Boyle's and Charles's laws, uses it to define the constant-volume gas thermometer, and extrapolates the straight pressure–temperature line back to the one temperature where the gas would lose all its pressure: $-273.15\,^\circ\text{C}$, the absolute zero that anchors the Kelvin scale.

Why a gas makes the best thermometer

The variables that describe a fixed mass of gas are its pressure $P$, its volume $V$ and its temperature $T$. Experiment shows that all gases at low densities behave the same way as these three are varied — the chemical identity of the gas drops out. That universality is exactly what a temperature standard needs: a liquid-in-glass thermometer disagrees with another between the fixed points because mercury and alcohol expand by different laws, but two gas thermometers agree because all dilute gases obey one common law.

That one common law is assembled from three experimental relationships, each found by holding one of the three variables fixed and watching how the other two respond. Learn them as a set, because the ideal gas equation is just their product.

The three gas laws

Each gas law fixes one quantity and links the other two. The grid below is the whole experimental foundation in one view; the conditions matter as much as the formulae, because NEET distractors swap them.

LawHeld constantRelationshipShape
Boyle's law (R. Boyle, 1662)Temperature $T$, amount of gas$PV = \text{constant}$, i.e. $P \propto \dfrac{1}{V}$$P$–$V$ curve is a rectangular hyperbola
Charles's law (J. Charles)Pressure $P$, amount of gas$\dfrac{V}{T} = \text{constant}$, i.e. $V \propto T$$V$–$T$ is a straight line through the origin (in K)
Gay-Lussac's lawVolume $V$, amount of gas$\dfrac{P}{T} = \text{constant}$, i.e. $P \propto T$$P$–$T$ is a straight line through the origin (in K)

NCERT names Boyle's and Charles's laws explicitly and notes that low-density gases obey them; the third member, Gay-Lussac's (the pressure law), is the one that runs the gas thermometer. All three straight-line statements only pass through the origin when $T$ is measured on the absolute (Kelvin) scale — that is the deep hint that there exists a natural zero of temperature.

Combining into PV = nRT

Since $PV$ is constant at fixed temperature (Boyle) and $V/T$ is constant at fixed pressure (Charles), the combined quantity $PV/T$ must itself be constant for a given quantity of gas. Writing this constant in a form that scales with the amount of gas gives the ideal gas equation:

$$PV = \mu R T$$

where $\mu$ is the number of moles of gas, $T$ is the absolute temperature in kelvin, and $R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$ is the universal gas constant — the same value for every gas.

The word ideal signals that this law holds exactly only in the limit of low density, where molecules are far apart and intermolecular forces are negligible. Real gases follow it closely at ordinary pressures and high temperatures, and deviate at low temperature and high pressure — which is precisely when they are about to liquefy. Any gas obeying $PV = \mu RT$ at all states is called an ideal gas.

Two forms — moles and molecules

The same law appears in two NEET-relevant forms depending on how you count the gas. Counting in moles uses $R$; counting individual molecules uses the Boltzmann constant $k$. They are algebraically identical because a mole contains Avogadro's number $N_A$ of molecules and $R = N_A k$.

FormCounts the gas asConstant usedValue
$PV = \mu R T$$\mu$ molesUniversal gas constant $R$$8.31\ \text{J mol}^{-1}\text{K}^{-1}$
$PV = N k T$$N$ moleculesBoltzmann constant $k$$1.38 \times 10^{-23}\ \text{J K}^{-1}$
Bridge: $N = \mu N_A$ and $R = N_A k$, with $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.

The detailed molecular justification of $PV = NkT$ — pressure as the rate of momentum transfer by colliding molecules, and the link between $T$ and mean kinetic energy — belongs to the kinetic theory of gases, a separate chapter. In this chapter the ideal gas law is taken as the experimental statement summarised by the three gas laws.

The constant-volume gas thermometer

Holding the volume of a fixed quantity of gas constant collapses the ideal gas law to Gay-Lussac's form, $P \propto T$. Pressure then becomes a clean, linear reading of absolute temperature. This is the working principle of the constant-volume gas thermometer, the instrument that calibrates all other thermometers.

gas fixed V bulb at temperature T fixed mark h P = P₀ + ρgh raise / lower to restore mark
Figure 1. Constant-volume gas thermometer. The gas bulb sits in the bath whose temperature $T$ is wanted. The right limb of the mercury manometer is raised or lowered until the mercury in the left limb returns to the fixed mark, restoring the gas to its reference volume. The gas pressure is then $P = P_0 + \rho g h$. Because $V$ is fixed, $P \propto T$, and the temperature follows from the pressure ratio against the triple point.

To read a temperature, the thermometer is first placed at the triple point of water, where its pressure $P_{tp}$ is recorded at the assigned temperature $273.16\ \text{K}$. At any unknown temperature $T$ the pressure $P$ is measured at the same fixed volume, and since $P \propto T$,

$$T = 273.16\ \text{K} \times \frac{P}{P_{tp}}.$$

Extrapolation to absolute zero

Plotting the measured pressure of a low-density gas against temperature at constant volume gives a straight line. Real gases deviate from the ideal prediction at low temperature — they liquefy and then solidify long before the line reaches the axis — but over a wide range the relationship is linear, and the line behaves as though the pressure would fall to zero if the substance kept behaving as a gas. The temperature at which the extrapolated line meets the temperature axis (zero pressure) is the absolute minimum temperature for an ideal gas. Measured this way it comes out to $-273.15\,^\circ\text{C}$, designated absolute zero.

P t (°C) 0 −273.15 °C (0 K) gas A gas B gas C solid = measured · dashed = extrapolated
Figure 2. Constant-volume $P$–$t$ lines for three different low-density gases. The measured (solid) portions have different slopes and intercepts, yet every extrapolated (dashed) line strikes the temperature axis — zero pressure — at the same point, $-273.15\,^\circ\text{C}$. That common intercept is absolute zero; its independence from the gas used is what makes the ideal gas a universal thermometric standard.

The remarkable result, and the reason the gas thermometer is fundamental, is that every gas extrapolates to the same intercept. Different gases give lines of different slope, but all of them point back to one temperature where the predicted pressure vanishes. This happens because at low density all gases obey the identical law $P \propto T$, so the zero of pressure must fall at one shared temperature regardless of which gas drew the line.

The Kelvin scale and the triple point

Absolute zero is the foundation of the Kelvin (absolute) temperature scale, named after Lord Kelvin. On this scale $-273.15\,^\circ\text{C}$ is taken as the zero point, $0\ \text{K}$, and the size of one kelvin is made identical to the size of one degree Celsius. The two scales therefore differ only by an offset:

$$T(\text{K}) = t_C\,(^\circ\text{C}) + 273.15$$

The modern Kelvin scale is defined by a single fixed point, not two. The original Celsius scale used two fixed points — the ice point and the steam point — but these depend on pressure and are hard to reproduce precisely. The triple point of water is far better: it is the unique temperature and pressure at which ice, liquid water and water vapour coexist in equilibrium, occurring at exactly one combination — $273.16\ \text{K}$ and $6.11 \times 10^{-3}\ \text{Pa}$. Because this state is perfectly reproducible, the Kelvin scale assigns it the exact value $273.16\ \text{K}$, and absolute zero supplies the natural zero. One defining point plus a fixed zero is all the scale needs.

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Related drill

For how Celsius, Fahrenheit and Kelvin convert into one another and why the steam point is no longer exactly $100\,^\circ\text{C}$, see temperature scales.

Worked examples

NCERT Exercise 10.4(b)

On the Kelvin absolute scale one fixed point is the triple point of water, assigned $273.16\ \text{K}$. What is the other fixed point on this scale?

Reasoning. The Kelvin scale needs only one experimental fixed point because its zero is supplied by nature — absolute zero. So the "other" fixed point is the zero of the scale itself, the temperature of zero molecular activity.

Answer: the other fixed point is absolute zero, $0\ \text{K}\,(=-273.15\,^\circ\text{C})$. Modern thermometry fixes the scale with the triple point ($273.16\ \text{K}$) and the natural zero, rather than with two reproducible-but-pressure-dependent points like ice and steam.

NCERT Exercise 10.5

Two constant-volume gas thermometers A (oxygen) and B (hydrogen) read these pressures: at the triple point of water, $P_A = 1.250 \times 10^5\ \text{Pa}$ and $P_B = 0.200 \times 10^5\ \text{Pa}$; at the normal melting point of sulphur, $P_A = 1.797 \times 10^5\ \text{Pa}$ and $P_B = 0.287 \times 10^5\ \text{Pa}$. What absolute temperature does each thermometer give for the melting point of sulphur, and why do they differ slightly?

Use $T = 273.16 \times P/P_{tp}$ for each.

Thermometer A: $T_A = 273.16 \times \dfrac{1.797}{1.250} = 392.7\ \text{K}$.

Thermometer B: $T_B = 273.16 \times \dfrac{0.287}{0.200} = 391.9\ \text{K}$.

Why the small difference. Oxygen and hydrogen are real gases, so at the working densities each deviates a little from the ideal $P \propto T$ law, and by different amounts. The remedy is to repeat the readings at successively lower gas densities and extrapolate the result to the limit of zero pressure; in that limit both gases behave ideally and the two thermometers agree.

Quick calculation

A fixed mass of an ideal gas at $27\,^\circ\text{C}$ and pressure $P$ is heated at constant volume until its pressure doubles. Find the new temperature.

Convert first: $T_1 = 27 + 273 = 300\ \text{K}$. At constant volume Gay-Lussac's law gives $P/T = \text{constant}$, so $\dfrac{P}{300} = \dfrac{2P}{T_2}$.

Solve: $T_2 = 600\ \text{K} = 327\,^\circ\text{C}$. Note that doubling the absolute temperature, not the Celsius reading, doubles the pressure — the trap is to write $T_2 = 54\,^\circ\text{C}$ by doubling $27\,^\circ\text{C}$.

Quick recap

Ideal gas & absolute temperature in one breath

  • Boyle ($PV=$ const at fixed $T$), Charles ($V/T=$ const at fixed $P$) and Gay-Lussac ($P/T=$ const at fixed $V$) combine into $PV/T=$ const, i.e. $PV = \mu R T$ with $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$.
  • Same law per molecule: $PV = NkT$, with $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ and $R = N_A k$.
  • $T$ in the gas law is always in kelvin: $T(\text{K}) = t(^\circ\text{C}) + 273.15$.
  • Constant-volume gas thermometer: $P \propto T$, so $T = 273.16\ \text{K} \times P/P_{tp}$.
  • Extrapolating the $P$–$T$ line to zero pressure gives absolute zero, $-273.15\,^\circ\text{C} = 0\ \text{K}$; all gases share this intercept.
  • The Kelvin scale uses one defining fixed point — the triple point of water, $273.16\ \text{K}$ (not $273.15$).
  • Kinetic-theory derivation of $PV=NkT$ is a separate chapter.

NEET PYQ Snapshot — Ideal Gas & Absolute Temperature

NEET tests this subtopic mostly through the kelvin-conversion trap and the absolute-zero / triple-point numbers. Convert to kelvin first, every time.

Concept check

A given mass of an ideal gas at $127\,^\circ\text{C}$ is heated at constant pressure until its volume doubles. The final temperature is:

  1. $254\,^\circ\text{C}$
  2. $527\,^\circ\text{C}$
  3. $800\,^\circ\text{C}$
  4. $400\,^\circ\text{C}$
Answer: (2) 527 °C

Kelvin first. $T_1 = 127 + 273 = 400\ \text{K}$. Charles's law at constant $P$: $V/T=$ const, so doubling $V$ doubles $T$ → $T_2 = 800\ \text{K} = 527\,^\circ\text{C}$. Option (1) is the trap of doubling the Celsius value.

Concept check

Which statement about the constant-volume gas thermometer and absolute zero is incorrect?

  1. At constant volume the pressure of a low-density gas is proportional to its absolute temperature.
  2. The $P$–$T$ lines of different gases extrapolate to the same temperature where pressure is zero.
  3. That common intercept, $-273.15\,^\circ\text{C}$, is absolute zero and can be physically reached by cooling a real gas.
  4. The triple point of water, $273.16\ \text{K}$, is the single defining fixed point of the Kelvin scale.
Answer: (3)

Why (3) is wrong. Absolute zero is inferred by extrapolating the straight line to zero pressure; every real gas liquefies and solidifies long before reaching it, so it is never physically attained. Statements (1), (2) and (4) are all correct NCERT facts.

FAQs — Ideal Gas & Absolute Temperature

Short answers to the questions NEET aspirants get wrong most often on this subtopic.

Why must temperature be in Kelvin in PV = nRT?
Because the ideal gas law states that PV is proportional to the absolute temperature, and proportionality requires a scale whose zero is the true zero of molecular activity. The Celsius scale puts zero at the ice point, so a Celsius temperature can be negative while pressure and volume can never be. Substituting a Celsius value into PV = nRT gives nonsense — for example 0 °C would predict zero PV. Always convert: T(K) = t(°C) + 273.15 before using the equation.
What is the difference between 273.15 K and 273.16 K?
273.15 K is the normal melting point of ice (0 °C) at one atmosphere, and it is the offset in the conversion t(°C) = T − 273.15. 273.16 K is the triple point of water — the single temperature and pressure (6.11 × 10⁻³ Pa) at which ice, water and vapour coexist — and it is the one fixed point that defines the Kelvin scale. The two differ by 0.01 K and NEET deliberately swaps them in distractors.
Why does the constant-volume gas thermometer use pressure to measure temperature?
If the volume of a fixed quantity of gas is held constant, the ideal gas law reduces to P proportional to T. Pressure then becomes a direct, linear reading of absolute temperature. The instrument measures the pressure of a low-density gas trapped at constant volume; multiplying the triple-point pressure ratio by 273.16 K gives the unknown temperature.
Why do all gases extrapolate to the same absolute zero?
At low densities every gas obeys the same ideal gas law independent of its chemical identity, so a plot of pressure versus temperature at constant volume is a straight line for every gas. Each line, extended backward, crosses the temperature axis at the same point, −273.15 °C, where the predicted pressure falls to zero. That common intercept is absolute zero, and its universality is what makes the gas a true thermometric standard.
What is the difference between PV = nRT and PV = NkT?
They are the same law written with different counts of particles. PV = nRT counts the gas in moles n with the universal gas constant R = 8.31 J mol⁻¹ K⁻¹. PV = NkT counts the gas in individual molecules N with the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹. Since N = n·Nₐ and R = Nₐ·k, the two forms are identical; choose whichever count the problem gives you.
Can a real gas actually reach absolute zero?
No. Long before −273.15 °C every real gas liquefies and then solidifies, so the straight pressure–temperature line is an extrapolation, not a measured value — the gas stops being a gas. Absolute zero is the limiting temperature inferred from the trend, the foundation of the Kelvin scale, and the third law of thermodynamics states it cannot be reached in a finite number of steps.