Question

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

Answer 1

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)m
where 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 Kg

so, 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

Answer 2

Given :
T=300K
m = molar mass /Avogadro's number =2 × 10⁻³/6.023 × 10²³ Kg
We know that average kinetic energy of a molecule of gas at temperature T is:
K.E=3/2 k where k is boltzmann constant =1.38x10⁻²³
T=temperature

Relation between momentum and Kinetic energy :
P²=2(K.E)m
where P is momentum, m is mass
Average momentum of hydrogen molecule is :
Pav=√(2xK.Exm)=√(2x3/2kTxm)

=√[2x3/2x1.38x10⁻²³x300x2 × 10⁻³/6.023 × 10²³ ]
=√3x1.38x10⁻²³x300x2 × 10⁻³/6.023 × 10²³
=√[2484/6.023]x10⁻⁴⁹=√412.41x10⁻⁴⁹
=√41.241x10⁻⁴⁸
=6.42x10⁻²⁴ kg-m/s

∴Average Kinetic energy of hydrogen molecule is : 6.42x10⁻²⁴ kg-m/s