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3 years ago

Calculate the rms speed of an oxygen gas molecule, o2, at 17.0 ∘c .

2 answers:

6 0

The RMS of O2 at 17 degrees is calculated as follows

RMs= ( 3RT/m)^1/2 where

R= ideal gas constant = 8.314
T= temperature= 17+273= 290 K
M= molar mass in KG = 32/1000= 0.032 Kg

Rms is therefore= sqrt (3x 8.314 x290/0.032 ) = sqrt( 226036.875

RMs=475.43

NeTakaya3 years ago

6 0

The μ rms speed = <u>475,433 m / s </u>

<h3> Further explanation </h3>

Gas particles move randomly (both speed and direction, as vector)

Average velocities of gases can be expressed as root-mean-square velocity. (μ rms)

$\rm \mu=\sqrt{\dfrac{3RT}{M_m} }$

R = gas constant, 8,314 J / mol K

T = temperature, K

Mm = molar mass of the gas particles , Kg

μ rms = root mean square velocity of Gas Particles.,m/s

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 = 17 °C = 17 + 273 = 290 K

then:

$\rm \mu=\sqrt{\dfrac{3\times8.314\times 290}{0.032} }\\\\\mu_{rms}=475,433\:\frac{m}{s}$

<h3> Learn more </h3>

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Step 1:

Based on equation(1), at the first eq point:

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For the second equivalence point setup an ICE table:

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