Home/ Questions/Q 452696
Next
In Process
Jamshed Emran Bhattacharyya
Asked: 2022-11-10T07:40:32+05:30 In:

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u=((2E)/(m))^(1//2)`
B. `u=((3E)/(2m))^(1//2)`
C. `u=((E)/(2m))^(1//2)`
D. `u=((E)/(3m))^(1//2)`

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u=((2E)/(m))^(1//2)`
B. `u=((3E)/(2m))^(1//2)`
C. `u=((E)/(2m))^(1//2)`
D. `u=((E)/(3m))^(1//2)`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. `v_(rms)=sqrt((3RT)/(M))`
    `KE(E) “per mole”=(3)/(2)RT`
    `RT=(2)/(3)KE`
    `v_(rms)=sqrt((3)/(M)xx(2)/(3)KE)`
    `=sqrt((2KE)/(M))=((2KE)/(M))^(1//2)`

    • 0
Amrita Sarma
Asked: 2022-11-09T00:10:12+05:30 In:

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u=((2E)/m)^(1//2)`
B. `u=((3E)/(2m))^(1//2)`
C. `u=(E/(2m))^(1//2)`
D. `u=(E/(3m))^(1//2)`

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u=((2E)/m)^(1//2)`
B. `u=((3E)/(2m))^(1//2)`
C. `u=(E/(2m))^(1//2)`
D. `u=(E/(3m))^(1//2)`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Correct Answer – A

    • 0
Gayatri Malhotra
Asked: 2022-11-04T11:17:32+05:30 In:

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u = ((2E)/(m))^(1//2)`
B. `u = ((E)/(2m))^(1//2)`
C. `u = ((3E)/(2m))^(1//2)`
D. `u = ((E)/(3m))^(1//2)`

For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is
A. `u = ((2E)/(m))^(1//2)`
B. `u = ((E)/(2m))^(1//2)`
C. `u = ((3E)/(2m))^(1//2)`
D. `u = ((E)/(3m))^(1//2)`
Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Correct Answer – A
    According to the kinetic gas eqauation ,
    `pV = (1)/(3) mnc^(2)`
    For a single molecule of mass `m , n = 1` .
    Thus ,
    `u_(rms) = sqrt((3 pV)/(m))` (1)
    According to gas equation , for one mole of gas , we have
    `pV = RT`
    For a single molecule , we have
    `pV = (R )/(N_(0))T = kT` (2) lt brgt where `k` is called the Boltzmann constant.
    Average kinetic energy per molecule `(E)` is given as
    `E = (3)/(2) kT`
    or `kT = (2)/(3) E` (3)
    Combining Eqs. (1),(2), and (3) , we have
    ` u_(rms) = sqrt((3 xx (2)/(3)E)/(m)) = sqrt((2E)/(m))`

    • 0