Question

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
A. `(400)/(sqrt(3))`
B. `(100sqrt(2))/3`
C. `100/3`
D. `100 sqrt(2)`

Answer 1

Correct Answer - A
Here, `upsilon_(1) = 200 m//s, T_(1) = 27^(@)C = 27+273`
`300 K`
`P_(1) = 1.0 xx 10^(5) N//m^(2), T_(2) = 127^(@)C = 127+273`
`=400K`
`P_(2) = 0.05 xx 10^(5) N//m^(2), upsilon_(2)=?`
As `upsilon_(rms) = sqrt((3RT)/(m))` , therefore, `upsilon_(rms) prop sqrt(T)`
There is no effect of change in pressure
`:. (upsilon_2)/(upsilon_1) = sqrt((T_2)/(T_1)) = sqrt(400/300) = 2/(sqrt(3))`
`upsilon_(2) = (2)/(sqrt(3)) upsilon_(1) = (2 xx 200)/(sqrt(3)) = (400)/(sqrt(3))m//s`.