What is the approximate most probable velocity of oxygen? If the kinetic energy of one mole of oxygen is 3741.3 J.

Options

A. `sqrt(43851)" J kg:^(-1)`
B. `sqrt(48321)" J kg"^(-1)`
C. `sqrt(155887)" J kg"^(-1)`
D. `sqrt(3950)" J kg"^(-1)`

Answer

C

Explanation

`:'` kinetic energy `(KE)=3/2 RT`
where R = gas constant
T = temperature
`:. RT=2/3 (KE)` ...(i)
Most probable velocity `(MP)=sqrt((2RT)/M)` ...(ii)
where M = molecular mass of oxygen = 32
Given `implies KE =3741.3 J`
Substituting the value of RT from Eq. (i), we have
`MP =sqrt((2xx2(KE))/(3xxM))`
`=sqrt((2xx2xx3741.3)/(3xx32)xx1000)`
`=sqrt(155.88xx1000)=sqrt(155887)"J kg"^(-1)`
Hence, option (c) is the correct answer.