A room contains oxygen and hydrogen moleculein the ratio 3:1 . Thetemperature of the room is 27^(@)C .The molar mass ofO_(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 ofoxygen molecules and hydrogen molecules.

A room contains oxygen and hydrogen moleculein the ratio 3:1 . Thetemp

1.	A room contains oxygen and hydrogen moleculein the ratio 3:1 . Thetemperature of the room is 27^(@)C .The molar mass ofO_(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 ofoxygen molecules and hydrogen molecules.
Answer» SOLUTION :Absolute temperature `T= 27^@C = 27 + 273 = 300K` Gas CONSTANT `R = 8.32 J "mol"^(-1) K^(-1)` For oxygen MOLECULE: molar mass ` M =32 gm//"mol" = 32 = xx 10^(-3) kg "mol"^(-1)` rms speed ` v_(rms) =sqrt( (3RT)/(M) ) = sqrt((3 xx 8.32 xx 300)/(32 xx 10^(-3) ))= 483.73 ms^(-1) ~~ 484 ms^(-1)` For hydrogenmolecule: Molar mass M ` = 2 xx 10^(-3) kg "mol"^(-1)` rms speed ` v_(rms) = sqrt( (3RT)/(M) ) = sqrt( ( 3 xx 8.32 xx 300)/(2 xx 10^(-3) )) = 1934 ms^(-1) = 1.93 K ms^(-1)` Note that the rms speed is inversely proportional to `sqrtM`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)/(484) = 4`