Physics · Kinetic Theory · NEET
Number density n is the count of molecules per unit volume, n = N/V, with SI unit per cubic metre. Moles count particles in groups of Avogadro number (6.02 x 10^23). They are linked by n = (N_A) x (number of moles / V). From the ideal gas law PV = N kB T you get the handy form n = P / (kB T). So higher pressure packs more molecules per cubic metre, and higher temperature spreads them out.
A moving molecule sweeps out a cylinder. Any other molecule whose centre lies within a distance d (the diameter, since both have radius d/2) will be hit. The cross-section area of that cylinder is pi d^2, which uses d squared. So mean free path is lambda = 1 / (sqrt(2) pi d^2 n). Double the diameter and the target area becomes 4 times bigger, so collisions rise 4 times and the mean free path drops to one fourth.
NCERT gives the order of magnitude directly: an atom is about 1 angstrom (10^-10 m) and a molecule about 2 angstrom (2 x 10^-10 m). To estimate it from data you use the mean free path formula backwards: measure lambda and number density n, then d = sqrt( 1 / (sqrt(2) pi n lambda) ). This is how kinetic theory yields molecular sizes.
Yes. From n = P / (kB T), at fixed pressure a higher temperature gives a smaller n, because the gas expands and the same molecules occupy more volume. At fixed temperature, higher pressure gives a larger n. Do not confuse this with mass density; both change with P and T in the same way here since number density and mass density differ only by the constant molecular mass m (mass density = n x m).
The collision cross-section is the effective target area a molecule presents, equal to pi d^2 (some books use sigma = pi d^2). It is not the physical face area of one molecule; it is built from the sum of the two radii, which equals the full diameter d. A larger cross-section means the molecule is easier to hit, so collisions per second go up and the mean free path goes down.
The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If the number densities of gases A and B are n_A and n_B respectively, the correct option is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
n = P / (kB T), where P is pressure, T is absolute temperature and kB = 1.38 x 10^-23 J/K is the Boltzmann constant. Equivalently n = N/V. Its SI unit is per cubic metre (m^-3).
About 2 angstrom, that is 2 x 10^-10 m, as stated in NCERT. A single atom is about 1 angstrom (10^-10 m). The average gap between molecules in a gas is about 10 or more times this size.
Both increase collisions and shorten the path. Mean free path lambda = 1 / (sqrt(2) pi d^2 n). Larger diameter d (as d^2) or larger number density n both make lambda smaller, so molecules collide more often.
It is the effective target area for a collision, equal to pi d^2, where d is the molecular diameter. A molecule sweeps a cylinder of this cross-section, and any molecule whose centre lies inside it gets hit.
No. Number density n counts molecules per unit volume (m^-3). Mass density is mass per unit volume (kg/m^3). They are related by mass density = n x m, where m is the mass of one molecule. Using n = P/(kBT), mass density = P m / (kB T).
