###### Advertisements

###### Advertisements

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)

###### Advertisements

#### 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`

#### APPEARS IN

#### RELATED QUESTIONS

Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy

- at mean position and
- at extreme position.

State an expression for K. E. (kinetic energy) and P. E. (potential energy) at displacement ‘x’ for a particle performing linear S.H. M. Represent them graphically. Find the displacement at which K. E. is equal to P. E.

Prove the law of conservation of energy for a particle performing simple harmonic motion.Hence graphically show the variation of kinetic energy and potential energy w. r. t. instantaneous displacement.

The kinetic energy of nitrogen per unit mass at 300 K is 2.5 × 106 J/kg. Find the kinetic energy of 4 kg oxygen at 600 K. (Molecular weight of nitrogen = 28, Molecular weight of oxygen = 32)

Obtan an expression for potential energy of a particle performing S.H.M. What is the value of potential energy at (i) Mean position, and (ii) Extreme position