Grade 11Physical Chemistry
Find rms velocity of 1)CO2 gas at 27°c2)O2 gas at 27°c Answer with proof
Find rms velocity of
1)CO2 gas at 27°c
2)O2 gas at 27°c
Answer with proof
1 Answer
Arun
Kinetic Energy (Translational, in all directions) of a gas as per Kinetic theory of gases at abs. Temperature T:
per molecule: KE = 3/2 * k_B T
per mole: KE = 3/2 * R * T
k_B = Boltzmann's constant. R= Universal gas constant.
per mole: KE = 3/2 * R * T
k_B = Boltzmann's constant. R= Universal gas constant.
Kinetic Energy of a gas (as per mechanics)
KE = 1/2 m (Vrms)²
or KE = 1/2 M (Vrms)²
KE = 1/2 m (Vrms)²
or KE = 1/2 M (Vrms)²
m = mass of a molecule, M = Molar mass of the gas.
Hence, (Vrms)² = 3 k_B T /m = 3 R T /M
R = 8.3 J/°°K/mole
M = 44 gm/mole for CO2 = 0.044 kg/mole
M = 44 gm/mole for CO2 = 0.044 kg/mole
(Vrms)² = 3 * 8.314 J/°K/mole * (273+27)° / (0.044 kg/mole)
Vrms = 412.38 m/sec
Vrms = 412.38 m/sec
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