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Question Number 180902 by nadovic last updated on 19/Nov/22

Calculate the root mean square   speed of the molecules of a Helium  gas kept in a gas cylinder at 400K.        [Take R = 8.3 Jmol^(−1) K^(−1) ]  The answer provided is 1.58 kms^(−1)   Please I need help with the solution

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{root}\:\mathrm{mean}\:\mathrm{square}\: \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{molecules}\:\mathrm{of}\:\mathrm{a}\:{Helium} \\ $$$$\mathrm{gas}\:\mathrm{kept}\:\mathrm{in}\:\mathrm{a}\:\mathrm{gas}\:\mathrm{cylinder}\:\mathrm{at}\:\mathrm{400K}. \\ $$$$\:\:\:\:\:\:\left[{Take}\:\mathrm{R}\:=\:\mathrm{8}.\mathrm{3}\:{Jmol}^{−\mathrm{1}} {K}^{−\mathrm{1}} \right] \\ $$$${The}\:{answer}\:{provided}\:{is}\:\mathrm{1}.\mathrm{58}\:{kms}^{−\mathrm{1}} \\ $$$$\mathrm{Please}\:\mathrm{I}\:\mathrm{need}\:\mathrm{help}\:\mathrm{with}\:\mathrm{the}\:\mathrm{solution} \\ $$

Answered by Sam09 last updated on 19/Nov/22

M = 4 gmol^(−1)  = 4×10^(−3)  kgmol^(−1)   rms = (√((3RT)/M))  = (√((3×8.3×400)/(4×10^(−3) ))) ms^(−1)   = 1578 ms^(−1)   = 1578×10^(−3)  kms^(−1)   = 1.578 kms^(−1)

$${M}\:=\:\mathrm{4}\:{gmol}^{−\mathrm{1}} \:=\:\mathrm{4}×\mathrm{10}^{−\mathrm{3}} \:{kgmol}^{−\mathrm{1}} \\ $$$${rms}\:=\:\sqrt{\frac{\mathrm{3}{RT}}{{M}}} \\ $$$$=\:\sqrt{\frac{\mathrm{3}×\mathrm{8}.\mathrm{3}×\mathrm{400}}{\mathrm{4}×\mathrm{10}^{−\mathrm{3}} }}\:{ms}^{−\mathrm{1}} \\ $$$$=\:\mathrm{1578}\:{ms}^{−\mathrm{1}} \\ $$$$=\:\mathrm{1578}×\mathrm{10}^{−\mathrm{3}} \:{kms}^{−\mathrm{1}} \\ $$$$=\:\mathrm{1}.\mathrm{578}\:{kms}^{−\mathrm{1}} \\ $$

Commented by nadovic last updated on 11/Dec/22

Thanks Sir

$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$

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