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The average translational kinetic energy...

The average translational kinetic energy of nitrogen gas molecules is 0.02 eV(1 eV = `1.6xx10^(-19)J`). Calculate the temperature of the gas. Boltzmann constant `k=1.38xx10^(-28)J//K`.

A

193.1 K

B

1011 K

C

212 K

D

154.5 K

Text Solution

Verified by Experts

The correct Answer is:
D
Translational K.E. of nitrogen molecule = `(3)/(2)kT`
`(3)/(2)kT=0.02eV=0.02xx1.6xx10^(-19)J`
`T=(0.02xx1.6xx10^(-19))/(1.38xx10^(-23))xx(2)/(3)orT=154.5K`
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The average translational kinetic energy of nitrogen gas molecule is 0.02eV (1eV - 1.6 xx 10^(-19)J) . Calculate the temperatuire of the gas. Boltzmann constant k = 1.38 xx 10^(-23) J//K .

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Knowledge Check

  • The translational kinetic energy of 10^(20) molecules of nitrogen at a certain temperature is 0.629J. What is the temperature in .^(@)C ?

    A
    `43.3^(@)C`
    B
    `23^(@)C`
    C
    `30.^(@)C`
    D
    `15.8^(@)C`

  • The average kinetic energy of gas molecule at 27^(@)C is 6.21xx10^(-21) J. Its average kinetic energy at 127^(@)C will be

    A
    `12.2xx10^(-21)J`
    B
    `8.28xx10^(-21)J`
    C
    `10.35xx10^(-21)J`
    D
    `11.35xx10^(-21)J`

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