Two gaes A and B are contained in two separate, but otherwise identical containers. Gas A consists of monatomic molecules, each with atomic mass of 4u whereas Gas B consists of rigid diatomic molecules, each with atomic mass of 20 u, If gas A is kept at `27^(@)C`, at what temperature should gas B be kept so that both have the same rms speed ?

Options

A. `27^(@)C`
B. `54^(@)C`
C. `270^(@)C`
D. `62^(@)C`

Answer

C

Explanation

For monatomic gas A,
atomic mass of molecules of gas A, `m_(A)=4 u`
and temperature of gas, `T_(A)=27^(@)C`
For diatomic gas B,
`m_(B)=2xx20u=40 u`
Temperature of gas, `B=T_(B)`
Since, rms speed of the molecules of gas, A = rms speed to molecules of gas, B
`(v_(rms))_(A)=(v_(rms))_(B)`
`sqrt((3RT_(A))/m_(A))=sqrt((3RT_(B))/m_(B))`
`implies T_(A)/m_(A)=T_(B)/m_(B) implies 27/(4u)=T_(B)/(40 u)`
`implies T_(B)=270^(@)C`