Question

Calculate the average molecular kinetic energy (i) per kmol (ii) per kg (iii) per molecule of oxygen at 127 ºC, given that molecular weight of oxygen is 32, R is 8.31 J mol^-1 K^-1 and Avogadro’s number NA is 6.02 x 10²³ molecules mol^-1.

Answer 1

Given : T = 273+127 = 400 K,  molecular weight=32 \ molar mass = 32 kg/kmol,

R = 8.31 Jmol^-1 K^-1, N_A = 6.02 x 10²³ molecules mol^-1

(i) The average molecular kinetic energy per kmol of oxygen = the average kinetic energy per mol of oxygen x 1000

= \(\frac32\) RT x 1000 = \(\frac32\) (8.31) (400) (10³)

= (600)(8.31)(10³) = 4.986 × 10⁶ J/kmol

(ii) The average molecular kinetic energy per kg of oxygen

= \(\frac{3}{2}\frac{RT}{M_o}\)

= \(\frac{4.986×10^6}{32}\)

=  1.558 x 10⁵ J/kg.

(iii) The average molecular kinetic energy per molecule of oxygen

= \(\frac{3}{2}\frac{RT}{N_A}\)

= \(\frac{4.986×10^6}{6.02×10^{23}}\)

= 8.282 x 10^-21 J/molecule