The molecules of a given mass of a gas have rms velocity of `200 m//s at 27^(@)C and 1.0 xx 10^(5) N//m_(2)` pressure. When the temperature and pressure of the gas are respectively `127^(@)C and 0.05 xx 10^(5) Nm^(-2)`, the rms velocity of its molecules in `ms^(-1)` is

Options

A. `(400)/(sqrt3)`
B. `(100sqrt2)/(3)`
C. `(100)/3`
D. `100sqrt2`

Answer

A

Explanation

It is given that
`v_(rms)=200ms^(1),T_(1)=300K,P_(1)=10^(5)Nh^(-2)`
`T_(2)=400K,P_(2)=0.05xx10^(5)Nm^(-2)`
As, rms velocity of gas molecules,
`therefore" "v_(rms)propsqrtT" "(thereforev_(rms)=sqrt((3RT)/(m)))`
For two different cases,
`rArr" "((v_(rms))_(1))/((v_(rms))_(2))=sqrt(T_(1)/(T_(2)))rArr(200)/((v_(rms))_(2))=sqrt((300)/(400))=sqrt((3)/(4))`
`rArr" "(v_(rms))_(2)=(2)/(sqrt3)xx200=(400)/(sqrt3)ms^(-1)`