Question

The molecules of a given mass of 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?

Answer 1

Here, `C_(1) = 100 m//s, T_(1) = 27^(@)C = (27+273)K = 300K`
`P_(1) = 1.00 atm. C_(2) = ?, T_(2) = 127^(@)C = (127+273)K = 400K , P_(2)= 2.0 atm`.
from `(P_(1)V_(1))/(T_1) = (P_(2)V_(2))/(T_2) , (V_1)/(V_2) = (P_2)/(P_1)* (T_1)/(T_2) = 2 xx 300/400 = 3/2`
Again `P_(1)=1/3 M/(V_1) C_(1)^(2)` and `P_(2) = 1/3 M/(V_2) C_(2)^(2) :. (C_(2)^(2))/(C_(1)^(2))*(V_1)/(V_2) = (P_2)/(P_1)`
`C_(2)^(2) = C_(1)^(2) xx (P_2)/(P_1) xx(V_2)/(V_1) = (100)^(2) xx 2 xx 2/3`
`C_(2) = (100 xx 2)/(sqrt(3)) m//s`.