The relation between the mass, average speed, and temperature is given by:
u = √(3RT/M)
It is given that, temperature is 35 degree C = 35 + 273 = 308 Kelvin
The molar mass of oxygen = 0.032 kg/mole
r = 8.314 j/(mol-k)
u = √(3 × 8.314 × 308 / 0.032)
u = 489.966 m/s
The μ rms speed = 489.9661 m / s
Gas particles move randomly (both speed and direction, as vector)
Average velocities of gases can be expressed as root-mean-square velocity. (V rms)
[tex]\large {\boxed {\bold {\mu_ {rms} = \sqrt {\frac {3RT} {M_m}}}}[/tex]
R = gas constant, 8.314 J / mol K
T = temperature,
Mm = molar mass of the gas particles
μ rms = root mean square velocity of Gas Particles.
From this equation shows that the velocity of the gas is inversely proportional to the molar mass of the gas particles
μ ≅ 1 / mm
So that the greater the molar mass of the gas particles, the smaller the speed (the slowest)
Oxygen gas molecule, O₂ has a molar mass: 0.032 kg / mol
T = 35 C + 35 + 273 = 308 K
then :
[tex]\rm \mu_{rms}=\sqrt{\dfrac{3\times 8.314\times 308}{0.032} }\\\\\mu_{rms}=489.9661\frac{m}{s}[/tex]
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