Question

Q. 16) Calculale the average molecular kinetic energy
D per kmol
It) per kg
II) per molecule of oxygen at 127 C
Given that molecular weight of oxygen is 32 R 15 831 J'molk and avagadros number
is 6:02x10 molecule/mole.​

Answer 1

Answer:

(a) KE=

2N

3RT

, where R=8320J/Kmole, T=27+273=300K and N= Avogadro number =6.03×10

26

molecules/Kmole

or, KE=

2(6.03×10

26

)

3(8320)300

=

(0.67×10

24

)

(4160)

=62×10

−22

J/molecule

(b) K.E.=(

2

3

)kT

or, K.E.=(

2

3

)(1.38×10

−23

J/K)(27+273)K=6.21×10

−21

J/molecule.

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