Question

In a gas, 10% of molecules have a velocity of 2 km/sec, and 8% of molecules have a velocity of 1.5 km/sec, and 82% of molecules have a velocity of 1 km/sec. The most probable velocity of molecules is:
(A)- 1km/sec
(B)- 1.5 km/sec
(C)- 2 km/sec
(D)- 15 km/sec

Answer 1

The most probable speed as described by Maxwell Boltzmann is the speed possessed by maximum molecules of gas. The Maxwell-Boltzmann equation forms the basis of the kinetic theory of gases and describes the distribution of speeds of gaseous molecules at a certain temperature.

Complete step by step answer:
-The Kinetic Molecular theory is used to describe the motion of a molecule of an ideal gas under a controlled set of conditions. However, when ideal gas is taken into consideration, it is impossible to measure the velocity of each molecule at every instant of time.
-Therefore, the Maxwell-Boltzmann distribution is used to determine how maximum molecules are moving between velocities v and v + dv.
-Maxwell-Boltzmann distribution a classical system of distinguishable particles such as gas molecules and gave the distribution function for the magnitude of the velocity of gaseous molecules as-

$\rho (v)=4\pi {{\left( \dfrac{m}{2\pi {{K}_{B}}T} \right)}^{\frac{3}{2}{{v}^{2}}{{e}^{\dfrac{-m{{v}^{2}}}{2KT}}}}}$

where m = mass of one molecule;
${{K}_{B}}$= Boltzmann constant
T = thermodynamic temperature
-According to Maxwell-Boltzmann, the speed at which the distribution graph reaches its maximum is known as the most probable speed of gas molecules.
-Differentiating the distribution function formula and considering the derivative to be zero, will give the most probable speed of molecules.

-According to the question, in a gas,
8% of molecules have a speed of 1.5 km/sec;
10% of molecules have a speed of 2 km/sec; and
82% of molecules have a speed of 1 km/sec.
-So, the molecule having a speed of 1 km/hour will have the most probable speed because this speed is followed by 82% of molecules.
So, the correct answer is “Option A”.

Note: The magnitude of the most probable velocity of molecules of the gas as described by Maxwell-Boltzmann distribution is given by-
${{v}_{p}}=\sqrt{\dfrac{2{{K}_{B}}T}{m}}=\sqrt{\dfrac{2RT}{{{M}_{m}}}}$