The molecules of a given mass of a gas have root mean square speeds of `100 ms^(-1) "at " 27^(@)C` and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at `127^(@)C` and 2.0 atmospheric pressure?

Options

A. `(200)/(sqrt3)`
B. `(100)/(sqrt3)`
C. `(400)/(3)`
D. `(200)/(3)`

Answer

A

Explanation

Here, `v_(rms)=100ms^(-1),T_(1)=27^(@)C=(27+273)K=300K`
`P_(1)=1 atm,v_(rms2)=?,T_(2)=127^(@)C=(127+273)K=400 K`
`P_(2) =2atm`
From `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(V_(1))/(V_(2))=(P_(2))/(P_(1))2xx(300)/(400)=3/2`
Again `P_(1)=1/3 (M)/(V_(1))v_(rms1)^(2)and P_(2)=1/3(M)/(V_(2))v_(rms_(2))^(2)`
`therefore(v_(rms_(2))^(2))/(v_(rms_(1))^(2))xx(V_(1))/(V_(2))=(P_(2))/(P_(1))`
`v_(rms_(2))^(2)=v_(rms_(1))^(2)xx(P_(2))/(P_(1))xx(V_(2))/(V_(1))=(100)^(2)xx2xx2/3`
`v_(rms_(2))=(200)/(sqrt3)ms^(-1)`