Starting from the kinetic theory expression for pressure, derive the ideal gas equation $PV = nRT$ and show how the kinetic interpretation of temperature ($\overline{KE} \propto T$) follows from it.

Starting from the kinetic theory expression for pressure, derive the ideal gas equation PV = nRT and show how

Exercise · Q10

PV = nRT

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✓ Free question

Starting from the kinetic-theory pressure relation, $PV = \tfrac{1}{3}Nm\overline{v^2}$ , and noting that $\tfrac{1}{2}m\overline{v^2}$ is the average translational kinetic energy per molecule, this can be rewritten as

PV = \frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)

Define temperature (kinetic theory's own definition, matched to the empirical ideal gas equation) so that the average kinetic energy per molecule is $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_BT$ . Substituting,

PV = \frac{2}{3}N\cdot\frac{3}{2}k_BT = Nk_BT

For $n$ moles, $N=nN_A$ and $k_B=R/N_A$ , so $Nk_B = nN_A\cdot(R/N_A) = nR$ , giving

PV = nRT

the familiar ideal gas equation, recovered here directly from kinetic theory rather than assumed. This same substitution also shows that the average translational kinetic energy per molecule, $\tfrac{3}{2}k_BT$ , is directly PROPORTIONAL to the absolute temperature $T$ -- the kinetic interpretation of temperature (Section 9.5).

✓Final answer

PV=nRT

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 mark
$Q. Find kinetic energy of 1 litre of an ideal gas at S.T.P.$
›Reveal solutionSolution

The total kinetic energy of an ideal gas equals $\tfrac{3}{2}PV$ , which can be evaluated directly at STP without knowing the number of moles.

For an ideal gas, total translational kinetic energy $K = \tfrac{3}{2}nRT$ . Since $PV = nRT$ ,

$K = \dfrac{3}{2}PV$

At STP, $P = 1.013\times10^{5}\text{ Pa}$ and $V = 1\text{ litre} = 1\times10^{-3}\text{ m}^3$ :

$K = \dfrac{3}{2}\times 1.013\times10^{5}\times 1\times10^{-3} = \dfrac{3}{2}\times101.3 = 151.95\text{ J}$

✓Final answer
$K \approx 152\text{ J}$
CBSE 2018Set ANNUAL1 markMCQ
$Q. The kinetic energy per molecule of a gas at temperature T is ____.$
(a) $(3/2)RT$
(b) $(3/2)K_BT$
(c) $(2/3)RT$
(d) $(3/2)(RT/M)$
›Reveal solutionSolution

Each translational degree of freedom carries average energy $\tfrac12 K_BT$ ; a gas molecule has 3 translational degrees of freedom, giving $\tfrac32 K_BT$ per molecule.

From the kinetic theory of gases, the mean square speed of gas molecules is related to pressure and volume by $PV = \tfrac13 Nm\overline{c^2}$ , and using the ideal gas law $PV = NK_BT$ (for $N$ molecules), we get

$\tfrac13 m\overline{c^2} = K_BT \;\Rightarrow\; \tfrac12 m\overline{c^2} = \tfrac32 K_BT.$

The left side, $\tfrac12 m\overline{c^2}$ , is exactly the average translational kinetic energy per molecule. So the mean KE per molecule is $\tfrac32 K_BT$ — note this uses Boltzmann's constant $K_B$ (per molecule), not the universal gas constant $R$ (per mole); option (a) would be the correct molar form $\tfrac32 RT$ only per mole, not per molecule.

✓Final answer
$(3/2)K_BT$

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Q.Starting from the kinetic theory expression for pressure, derive the ideal gas equation $PV = nRT$ and show how the kinetic interpretation of temperature ( $\overline{KE} \propto T$ ) follows from it.

Concept understanding — Kinetic Interpretation of Temperature

Q.Starting from the kinetic theory expression for pressure, derive the ideal gas equation PV=nRTPV = nRTPV=nRT and show how the kinetic interpretation of temperature (KE‾∝T\overline{KE} \propto TKE∝T) follows from it.

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Q.Starting from the kinetic theory expression for pressure, derive the ideal gas equation $PV = nRT$ and show how the kinetic interpretation of temperature ( $\overline{KE} \propto T$ ) follows from it.

Unlock everything free for 14 days