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Question
Calculate the average molecular kinetic energy
- per kmol
- per kg
- per molecule
of oxygen at 127°C, given that the molecular weight of oxygen is 32, R is 8.31 J mol−1K−1 and Avogadro’s number NA is 6.02 × 1023 molecules mol−1.
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Solution
Data: T = 273 + 127 = 400 K,
molecular weight = 32
∴ molar mass = 32 kg/kmol,
R = 8.31 Jmol-1K-1,
NA = 6.02 × 1023 molecules mol-1
(i) The average molecular kinetic energy per kmol of oxygen = the average kinetic energy per mol of oxygen × 1000
`= 3/2 "RT" xx 1000`
= `3/2(8.31)(400)(10^3)"J"/"kmol"`
= (600)(8.31)(103)
= 4.986 × 106 J/kmol
(ii) The average molecular kinetic energy per kg of oxygen
= `3/2 "RT"/"M"_0`
= `(4.986xx10^6 "J"//"mol")/(32 "kg"//"kmol")`
= 1.558 × 105 J/kg.
(iii) The average molecular kinetic energy per molecule of oxygen
= `3/2 "RT"/"N"_"A"`
= `(4.986 xx 10^6 "J"//"mol")/(6.02 xx 10^23 "molecule"//"mol")`
= 8.282 × 10-21 J/molecule
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