A room contains oxygen and hydrogen molecule in the ratio 3:1 . The temperature of the room is `27^(@)C` .The molar mass of ` O_(2)` is 32 g `mol^(-1)` and for `H_(2)` 3 g `mol^(-1)` . The value of gas constant R is 8.32 J `mol^(-1)K^(-1)`
calculate:
(a) rms speed of oxygen and hydrogen molecule.
(b) Average kinetic energy per oxygen molecule and per hydrogen molecule.
(c) Ratio of average kinetic energy of oxygen molecules and hydrogen molecules.

Explanation

(a) Absolute Temperaure `T=27^(@)C+27+273=300K`
Gas constant `R=8.32J mol^(-1)K^(-1)`
For Oxygen molecule : Molar mass `M=32g=32xx10^(-3) kg mol^(-)`
rms speed `nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx300)/(32xx10^(-3)))=483.73ms^(-1)~~484ms^(-1)`
For Hydrogen molecule: Molar mass `M=2xx10^(-3) kg mol^(-1)`
rms speed `nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx300)/(2xx10^(-3)))=1934ms^(-1)=1.93ms^(-1)`
Note that the rms speed is inversely proportional to `sqrt3` and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature.`1934/484cong`
(b) The average kinetic energy per molecule is`3/2kT`. It depends only on absolute temperature of the gas and is independent of the nature of molecules. Since both the gas molecules are at the same temperature, they have the same average kinetic energy per molecule. k is Boltzmann constant.
`3/2kT=3/2xx.138xx300xx6.21xx10^(-21)J`
(c) Average kinetic energy of total oxygen molecules `=3/2N_(o)kT` where `N_(O)`- number of oxygen molecules in the room.
Average kinetic energy of total hydrogen molecules `3/2= N_(H)kT` where `N_(H)-` number of hydrogen molecules in the room. It is given that the number of oxygen molecules is 3:1