Calculate the rms speed of an oxygen gas molecule, o2, at 29.0 ∘c . express your answer numerically in meters per second.

Question

Calculate the root mean square (rms) speed of an oxygen gas molecule (O₂) at a temperature of 29.0 °C. Please express your answer numerically in meters per second.

Anonymous · Accepted Answer

The rms speed of a gas can be calculated using the following rule:Vrms = sqrt ( 3RT / M) where:R is the gas constant = 8.314T is the temperature = 29 + 273 = 302 degrees kelvinM is the molar mass of the gas (oxygen) = 2*16 = 32 grams = 0.032 kgSubstitute in the equation to get Vrms as follows:Vrms = sqrt [(3*8.314*302) / (0.032)] = 485.05 meters/sec

Anonymous · Answer

The rms speed of an oxygen gas molecule : 485.17 m/sFurther explanationKinetic Molecular Theory (KMT) states that a gas consists of molecules that move at a constant and random speed. The collisions between molecules are perfectly elastic so that no energy is wasted.The molecules move in straight lines until they collideEnergy because this motion is expressed as Kinetic energy (KE) which can be formulated as:  The average kinetic energy value is only affected by temperature changes. The higher the temperature, the average kinetic energy of the molecule increasesThis molecule is very small when compared to the distance between molecules, so the volume of gas contains mostly empty spaceGas particles move randomly (both speed and direction , as vector)Average velocities of gases can be expressed as root-mean-square averages. (V rms)R = gas constant, T = temperature, Mm = molar mass of the gas particlesFrom the questionR= 8.314 J/mol KT = 29 + 273 = 302 KMm = molar mass O₂ = 2.16 = 32 g/mol = 32.10-3 kg/molLearn moreinertia affect during a collisionHenry's Lawideal gasKeywords: Kinetic Molecular Theory, Collisions, The average kinetic energy, the rms speed, gas molecule

Prof. Tony Greenholt V · Answer

The rms speed of an oxygen gas molecule is 486.77 m/s.Explanation: The expression for the rms speed of the molecule is given by:where, R = Gas constant = 8.314 J/K mol  = T = Temperature of the gas = 31°C = (273 + 31)K = 304KM = Molar mass of the gas =    (Gas is Oxygen) Putting values in above equation, we get:

Anonymous · Answer

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 KelvinThe molar mass of oxygen = 0.032 kg/moler = 8.314 j/(mol-k)u = √(3 × 8.314 × 308 / 0.032)u = 489.966 m/s

Billy Hettinger · Answer

The μ rms speed = 475,433 m / sFurther explanationGas particles move randomly (both speed and direction, as vector)Average velocities of gases can be expressed as root-mean-square velocity. (μ rms)R = gas constant, 8,314 J / mol KT = temperature,KMm = molar mass of the gas particles, Kgμ rms = root mean square velocity of Gas Particles.,m/sFrom this equation shows that the velocity of the gas is inversely proportional to the molar mass of the gas particlesμ ≅ 1 / MmSo 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 / molT = 17 °C = 17 + 273 = 290 Kthen:Learn morekinetic molecular theoryThe gas particle that travels the slowest

Anonymous · Answer

The μ rms speed = 489.9661 m / sFurther explanationGas particles move randomly (both speed and direction, as vector)Average velocities of gases can be expressed as root-mean-square velocity. (V rms)R = gas constant, 8.314 J / mol KT = temperature,Mm = molar mass of the gas particlesμ 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 particlesμ ≅ 1 / mmSo 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 / molT = 35 C + 35 + 273 = 308 Kthen :Learn morekinetic molecular theoryThe gas particle that travels the slowest

Anonymous · Answer

The  RMS  of O2  at  17  degrees   is  calculated  as  followsRMs= ( 3RT/m)^1/2   whereR= ideal  gas   constant  =  8.314T=   temperature=   17+273=  290 KM=  molar  mass   in  KG =   32/1000=  0.032  KgRms  is  therefore= sqrt (3x   8.314  x290/0.032 )  =  sqrt( 226036.875RMs=475.43

Melyna Murphy

Calculate the rms speed of an oxygen gas molecule, o2, at 29.0 ∘c . express your answer numerically in meters per second.

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