Home
Class 11
PHYSICS
At what temperature will the average vel...

At what temperature will the average velocity of oxygen molecules be sufficient to escape from the earth. Given mass of oxygen molecule `= 5.34 xx 10^(-26) kg`. Boltzmann constant, `k = 1.38 xx 10^(-23) J "molecule"^(-1) K^(-1)`. Escape velocity of earth `= 11.0 km s^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
`(1452pi)/(25)xx10^(3)K`

At every temperature there will be some molecules with the required escape speed, as theoretically
`0 le |vecV| le prop` at all temperatures.
`V_(arm) = sqrt((8RT)/(piM)) :. sqrt((8RT)/(piM)) = 11 xx 10^(3) m//s`
put `R = 25//3J-mol^(-1)K^(-1)`
`M = 0.032` is `SI` units `(kg//mol)`
`(11^(2) xx 10^(3xx2)) xx (pi)/(8) xx (3)/(25) xx 0.032 = T`
`T = (1452pi)/(25) xx 10^(3)K`
Promotional Banner

Topper's Solved these Questions

  • KTG & THERMODYNAMICS

    RESONANCE|Exercise SECTION (E)|3 Videos
  • KTG & THERMODYNAMICS

    RESONANCE|Exercise SECTION (F)|3 Videos
  • KTG & THERMODYNAMICS

    RESONANCE|Exercise SECTION (B)|3 Videos
  • KINETIC THEORY OF GASES AND THERMODYNAMICS

    RESONANCE|Exercise Exercise|64 Videos
  • MAGNETIC FIELD AND FORCES

    RESONANCE|Exercise Exercise|65 Videos

Similar Questions

Explore conceptually related problems

At what temperature the kinetic energy of a molecule will be equal to 2.8 xx 10^(-20) J ? Boltzmann constant (k_(B)) = 1.4 xx 10^(-23) J "molecule"^(-1)K^(-1)

At what temperature , will the rms speed of oxygen molecules be sufficient for escaping from the earth ? Taken m = 2.76 xx 10^(-26) kg, k = 1.38 xx 10^(-23) J//K and v_(e) = 11.2 km//s .

At what temperature the root mean square velocity is equal to escape velocity from the surface of earth for hydrogen and for oxygen ? Given radius of earth = 6.4 xx 10^(6) m , g = 9.8 ms^(-2) . Boltzmann constant = 1.38 xx 10^(-23) J "molecule"^(-1) .

The diameter of a gas molecule is 2.4 xx 10^(-10) m . Calculate the mean free path at NTP. Given Boltzmann constant k = 1.38 xx 10^(-23) J molecule^(-1) K^(-1) .

At what temperature, hydrogen molecules will escape from the earth's surface ? (Take, radius of earth R_(e)=6.4xx10^(6)m , mass of hydrogen molecule m=0.34 xx 10^(-26) kg, Boltzmann constant k=1.38xx10^(-23)JK^(-1) and acceleration due to gravity = 9.8 xx ms^(-2) ) also take rms speed of gas as v_(rms) = sqrt((3kT)/(m)) .

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0^(@)C. Mass of a helium molecule =6.64xx10^(-27)kg and Blotzmann constant =1.38xx10^(-23) J K^(-1).

Calculate the temperature at which the average K.E. of a molecule of a gas will be the same as that of an electron accelerated through 1 volt. Boltzmann constant = 1.4 xx 10^(-23) J "molecule"^(-1) K^(-1) , charge of an electron = 1.6 xx 10^(-19)C .

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to: [Boltzmann constant k _ B = 1.38 xx 10 ^( -23) J//K Avogadro Number N _ A = 6.02 xx 10 ^( 26) // Kg Radius of Earth : 6.4 xx 10 ^( 6 ) m Gravitational acceleration on Earth = 10 ms ^( -2) ]