The density of `CO_(2)` gas at `0^(@)C` and at a pressure of `1.0 xx 10^(5) Nm^(-2)` is 1.98 `kgm^(-3)`. Calculate the root mean square velocity of its molecules at `0^(@)C` and `30^(@)C`. Assume the pressure to remain constant.

Explanation

According to the kinetic theory, the pressure P of gas is given by
`P = (1)/(3) pv^(bar2)`
where `rho` is the density of the gas and `v^(bar2)` the mean square velocity of its molecules.
`therefore v_("rms") = sqrt(v^(bar2))=sqrt((3P)/(rho))`
Here, `P = 1.0 xx 10^(5) Nm^(-2)`
`rho = 1.98 kg m^(-3)`
`therefore v_("rms") = sqrt((3xx1.0 xx 10^(5))/(1.98)`
`= 389 m//s`
Again, from the kinetic theory, the root mean square is directly proportional to the square root of the absolute temperature.
`v_("rms") prop sqrt(T)`
`((V_("rms"))30)/((V_("rms"))0)= sqrt((273+30)/(273))`
`= sqrt((303)/(273))`
`therefore (v_("rms"))_(30)= (v_("rms"))_(0) x 1.053`
`=389 xx 1.053`
`= 410 m//s`.