A)What is the rms speed (in m/s) of oxygen molecules at 24.4 ° C? Since oxygen i

Question

A)What is the rms speed (in m/s) of oxygen molecules at 24.4 ° C? Since oxygen is a diatomic molecule, the mass of the oxygen molecule (O2) is 5.313 x 10-26 kg. B)If one wished to reduce the speed of an oxygen molecule to 57 % of the value that was obtained in the problem above, what temperature (in ° C) would be required? C) At what temperature (in K) would the rms speed of a molecule of hydrogen gas be 2230 m/s? The mass of a hydrogen molecule (H2) is 3.34 x 10-27 kg. D) If you wished to reduce the speed of the hydrogen molecules to 1200 m/s, what temperature (in K) would be required?

Explanation / Answer

A) Given that,

T = 24.4 Deg C = 297.4 K ; mass of O2 = 5.313 x 10-26 kg

We need to find the rms speed v(rms). We know that,

V(rms) = sqrt (3 RT/M) , where R is the gas constant , T is temperature and M is the molecular mass.

M = 5.313 x 10-26 kg x 6.02 x 1023 = 0.032 kg / mol

V(rms) = sqrt ( 3 x 8.3145 kg m2 s-2 k-1 mol-1 x 297.4 K / 0.032 kg mol-1 ) = 481.5 m/s

Hence, V(rms) at 24.4 deg C = 297.4 K is V(rms) = 481.5 m/s.

B)57 % of 481.5 m/s = 277.455 m/s

Again using the same relation:

V(rms) = sqrt (3 RT / M) solving for T we get,

T = V2 x M / 3 R = (277.455)2 x 0.032 kg mol-1 / 3 x 8.3145 kg m2 s-2 k-1 mol-1

T = 92.11 K = 92.11 - 273 = -181 Deg C

C) V(rms) = 2230 m/s M = 3.34 x 10-27 kg x 6.02 x 1023 = 0.00201 K mol-1

V(rms) = sqrt (3 RT / M)

T =  V2 x M / 3 R = (2230)2 x 0.00201 kg mol-1 / 3 x 8.3145 kg m2 s-2 k-1 mol-1

T = 400.7 K

D) V(rms) = 1200 m/s

T =  V2 x M / 3 R = (1200)2 x 0.00201 kg mol-1 / 3 x 8.3145 kg m2 s-2 k-1 mol-1

T = 116.04 K

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