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Calculate the average molecular kinetic energy (i) per kmol (ii) per kg (iii) 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

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Question

Calculate the average molecular kinetic energy 

  1. per kmol 
  2. per kg 
  3. per molecule 

of oxygen at 127°C, given that the molecular weight of oxygen is 32, R is 8.31 J mol−1K1 and Avogadro’s number NA is 6.02 × 1023 molecules mol1.

Numerical
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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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Chapter 3: Kinetic Theory of Gases and Radiation - Exercises [Page 74]

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Balbharati Physics [English] Standard 12 Maharashtra State Board
Chapter 3 Kinetic Theory of Gases and Radiation
Exercises | Q 18 | Page 74

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