Question

the mean kinetic energy of molecule of hydrogen at 0°c is 5.60×10-21 J and molar gas constant is 8.31J/mol/k. Calculate the avogadro's number. ​

Answer 1

mean kinetic energy of a molecule of a hydrogen at 0° is 5.60 × 10^-21 J

we know, mean kinetic energy of an ideal gas is given by, K.E = 3/2nRT

here, n is no of moles of gas.

R is universal gas constant.

T is temperature in Kelvin.

here, R = 8.31 J/mol/k, T = 0° = 273K

so, K.E = 5.6 × 10^-21 = 3/2 × n × 8.31 × 273

or, n = 2 × 5.6 × 10^-21/(3 × 8.31 × 273)

≈ 0.0016456 × 10^-21

≈ 0.16456 × 10^-23

we know, number of mole = no of atom of substance/Avogadro's number

or, 0.16456 × 10^-23 = 1/Avogadro's number

[ as we know, mean kinetic energy means kinetic of each hydrogen atom that's why no of atom = 1 ]

or, Avogadro's number = 1/0.16456 × 10^-23

= 1/0.16456 × 10²³

≈ 6.0768 × 10²³

Answer 2

Average Kinetic energy = K.E/ mol /Av.No.

=3 RT/2×N

=3/2 KT

K=5.621×10^−14×2 / 3×273

(Since T = 273 K)

=1.372×10^−16 erg molecule^ −1 K

Now Avogadro No. = R/K=8.314×10^7 / 1.372×10^−16

=6.059×10^23

$Hope this helps you$