Physics · Kinetic Theory · NEET
Yes, at constant pressure lambda increases with temperature. Write n using PV = NkT, so n = P/(kT). Put this in lambda = 1/(root2 pi d^2 n) to get lambda = kT / (root2 pi d^2 P). At fixed P, lambda is directly proportional to T. Heating the gas at constant pressure spreads molecules out (lower n), so each one travels farther between hits.
At constant temperature lambda is inversely proportional to P. From lambda = kT / (root2 pi d^2 P), doubling the pressure halves the mean free path. Higher pressure packs molecules closer (higher number density n), so collisions happen sooner and the free path is shorter.
Only one molecule is moving through the crowd of others. The chance of hitting something in a given distance depends on how many targets sit per unit volume, which is just n to the first power. The collision cross-section pi d^2 handles the size part. So lambda has n and d^2 in the denominator, but n appears only once.
Shorter. lambda is inversely proportional to d^2. A molecule with twice the diameter has four times the collision cross-section pi d^2, so it collides four times as often and its mean free path becomes one fourth (at the same n). Bigger targets are easier to hit.
No. The formula lambda = 1/(root2 pi d^2 n) contains only diameter and number density, not mass or speed. Speed and mass affect the collision frequency (collisions per second) and the time between collisions, but the average distance between collisions depends only on how crowded and how big the molecules are.
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.
lambda = 1 / (root2 x pi x d^2 x n), where d is the diameter of a molecule and n is the number of molecules per unit volume (number density). The root2 factor comes from accounting for the relative motion of all molecules.
Three things: molecular diameter d (lambda goes as 1/d^2), number density n (lambda goes as 1/n), and, when expressed through the gas law, temperature and pressure (lambda = kT / root2 pi d^2 P). It does not depend on molecular mass or speed.
Inversely, at constant temperature. lambda = kT / (root2 pi d^2 P), so raising pressure lowers the mean free path because molecules get packed closer.
At high altitude the air is thinner, so number density n is much lower. Since lambda is proportional to 1/n, the mean free path becomes much longer high up in the atmosphere.
For air at ordinary temperature and pressure it is of the order of about 10^-7 m (roughly 1000 angstroms), which is far larger than the molecular size of about a few angstroms.
