Calculate the root mean square (rms) average speed of the atoms in a sample of xenon gas at 0.17atm and −52.°c .

round your answer to 3 significant digits.

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Explanation:

NB: In this question, we need the mass which is not given,so we have to make assumptions for the mass of xenon.

V^2(rms)= 3KT/m(1).

Where k= Boltzmann's constant =1.381×10^−23 m^2.kg/s^2.K.

T= temperature of the gas (in Kelvin), m= mass of one molecule of the gas (in kg).

Temperature, T= 273+(-52) =221K and V= the root mean square (rms).

Assuming that one mole of the gas has ;

m=0.084 kg. Hence, the mass of one molecule using Avagadro's number is;

0.084 kg ÷ 6.02×10^23 molecule.

Therefore, mass,m= 1.395× 10^ -25.

Therefore, from equation one above, we have;

V^2(rms)= 3 × 1.381×10^-23×221k/1.395×10^25.

V^2(rms) = 9.15603×10^-21/ 1.395×10^-25.

V(rms)= √ 6.56346236559139× 10^−46.

V(rms)= 2.56×10^-23 m/s.

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