Physics · Kinetic Theory · NEET
Temperature is not heat and not speed. In kinetic theory, the absolute temperature T is a measure of the average translational kinetic energy of the molecules. NCERT gets this by combining PV = (2/3)E with PV = N kB T, giving E = (3/2) N kB T. Dividing by the number of molecules N, the average KE per molecule = (3/2) kB T. So a higher T simply means the molecules, on average, move with more kinetic energy.
A monatomic molecule moves freely in 3 directions (x, y, z), so it has 3 translational degrees of freedom. Each degree of freedom carries an average energy of (1/2) kB T (law of equipartition). Adding the three gives 3 x (1/2) kB T = (3/2) kB T. This (3/2) kB T is the average TRANSLATIONAL kinetic energy per molecule and holds for every ideal gas.
No. At the same temperature, one molecule of hydrogen and one molecule of oxygen have the SAME average kinetic energy, (3/2) kB T, because it depends only on T. What differs is speed: since (1/2) m (v-bar squared) = (3/2) kB T, lighter molecules (small m) must move faster to carry the same energy. So equal KE, but different speeds.
It is linked to the AVERAGE kinetic energy per molecule, (3/2) kB T. The total translational kinetic energy of the whole gas is E = (3/2) N kB T = (3/2) n R T, which depends on how many molecules N (or moles n) you have. Two containers at the same T have the same average KE per molecule but different totals if they hold different amounts of gas.
Because average KE is proportional to T, heating raises the average kinetic energy. If you double the absolute temperature (in kelvin), the average kinetic energy doubles. Note the direct proportionality is with kinetic energy, not with speed: since KE goes as v-bar squared, doubling T multiplies the rms speed only by root 2, not by 2.
The relation average KE = (3/2) kB T is a direct proportionality that must pass through zero. Only the absolute (kelvin) scale has its zero at the point of zero molecular kinetic energy. If you used Celsius, 0 C would wrongly imply zero KE. Always convert: T(K) = t(C) + 273. For example 27 C = 300 K.
Increase in temperature of a gas filled in a container would lead to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules: average KE per molecule = (3/2) kB T, where kB is the Boltzmann constant.
Average KE per molecule = (1/2) m (v-bar squared) = (3/2) kB T. The total translational KE of N molecules is E = (3/2) N kB T = (3/2) n R T.
Yes. It depends only on temperature, so at a given T every ideal gas has average KE = (3/2) kB T per molecule. Only the molecular speeds differ, because lighter molecules move faster.
According to this ideal-gas relation, at T = 0 K the average translational kinetic energy would be zero, since KE = (3/2) kB T. This defines absolute zero in kinetic theory.
It doubles, because average KE is directly proportional to absolute temperature T. But the rms speed increases only by a factor of root 2, since KE depends on the square of speed.
