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 a gas is given by
`P=(1)/(3) rhooverline(v^(2))`
Where `rho` is the density of the gas and `overline(v^(2))` the mean square velocity of its molecules
`thereforeV_(rms)=sqrt(overline(v^(2)))=sqrt((3P)/(rho))`
here, `P=1.0xx10^(5)Nm^(-2),rho=1.98" kg "m^(-3)`
`thereforeV_(rms)=sqrt((3xx1.0xx10^(5))/(1.98))=389m//s`
Again, from the kinetic theory, the root mean square speed is directly proportional to the square root of the absolute temperature.
`V_(rms)propsqrt(T)`
`((V_(rms))_(30))/((V_(rms))_(0))=sqrt((273+30)/(273))=sqrt((303)/(273))=1.053`
`therefore(V_(rms))_(30)=(V_(rms))_(0)xx1.053=389xx1.053=410m//s`