. Find the average momentum of molecules of hydrogen gas in a container at temperature 300 K.

							We know, formula of average Kinetic energy of a molecule of gas at Temperature T is K.E = 3/2kT where , k is boltzmann constant { k = universal gas constant / Avogadro`s constant = 1.38 × 10^-23 } T is the temperature .also we know, relation between momentum and kinetic energy is P² = 2(K.E)mwhere P is the momentum , m is the mass now, for above explanation we can get, average momentum of Hydrogen molecule is P_{av} = \sqrt{2\times\:KE\times\:m}P_{av} = \sqrt{2\times\:KE\times\:m} \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  =  \sqrt{2 \times  \frac{3}{2}kT \times m} \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:     =  \sqrt{3kTm} here , K = 1.38 × 10^-23 T = 300K and , m = molecular mass/Avogadro`s number = 2 × 10^-3/6.023 × 10^23 Kgso, Pav = √{3 × 1.38 × 10^-23 × 300 × 2 × 10^-3/6.023 × 10^23 }= √{ 412.41 × 10^-49}= √{41.241} × 10^-24 Kg.m/s = 6.42 × 10^-24 kg.m/s hence, average momentum of Hydrogen molecule is 6.42 × 10^-24 kg.m/s
						

							
P av = M*V av
And we know that average velocity of gas molecules in any container is always zero as particles are moving randomly in all direction.
So av momentum is zero