#### Question

Calculate the average molecular kinetic energy :

(a) per kilomole, (b) per kilogram, of oxygen at 27°C.

(R = 8320 J/k mole K, Avogadro's number = 6*03 x 10^{26} molecules/K mole)

#### Solution

Given that

Temperature = 27°C = (273 + 27) K = 300 K

(a) `"Average kinetic energy per kilo mole" = 3/2RT`

`=3/2xx 8320 J/K "mole" xx300 K`

`=3.744 xx 10^6J`

(b) `"Kinetic energy per kilogram"="Kinetic energy per kilo mole"/"Molecular weight"`

`"Molecular weight of oxygen"= 32`

`"Kinetic energy per kilogram"=(3.744xx10^6 J)/32=0.117xx10^6 J`

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#### APPEARS IN

Solution Calculate the Average Molecular Kinetic Energy : (A) per Kilomole, (B) per Kilogram, of Oxygen at 27°C. Concept: K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M..