7) Calculate the average molecular kinetic energy
1.per kmol 2. per k gram of oxygen at
27°C (R= 8.320 J/kmol avagodoos no. -
6.03 x 10^23 molecules /k mol
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Answer:
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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