Place the following gases in order.

Chemistry
Tutor: None Selected Time limit: 1 Day

Place the following gases in order of increasing average velocity at 300 K: Ar, CH4, N2, and N20.

I am very confused with this question. I thought they would be all equal because in the textbook it said that regardless of their molecular mass they will have the same average kinetic energy at the same temperature.

Aug 6th, 2015

Thank you for the opportunity to help you with your question!

you are right that at same temperature these all gases have same average kinetic energy, but this question asks about average velocity, means the speed at which each gas travels at 300K. for this you need to apply Graham's law formula which relate velocity of a gas with it's molar mass.

Argon:

Velocity= 1/√mAr

velocity = 1/√39.94

velocity= 0.158

Methane:

velocity= 1/√mMethane

             = 1/√16

             = 0.25

Nitrogen:

velocity= 1/√mNitrogen

           = 1/√14

           = 0.267

Nitrogen dioxide:

velocity= 1/√mNitrogen dioxide

            = 1/√30

            = 0.182

so, according to the above calculations the order will be;

Ar<nitrogen dioxide<methane<nitrogen

Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 6th, 2015

Thank you.

Although that did help me, but within my homework question it only give me theses answers:

a. Ar=CH4=N2=N2O  b. N2O<Ar<N2<CH4  c. N2O<N2<CH4<Ar  d. Ar<N2<N2O<CH4  e. CH4<N2<N2O<Ar

So which one and why? If you do not mind. This will help when I take the Chapter test.

Aug 6th, 2015

actually i messed up in the molar mass of nitrogen dioxide. it'll be 43.99 instead of 30. sorry for that.

Nitrogen dioxide:

velocity= 1/√mNitrogen dioxide

            = 1/√43.99

  = 0.150

so, now you can opt up option B as the right answer.



Aug 6th, 2015

actually i messed up in the molar mass of nitrogen dioxide. it'll be 43.99 instead of 30. sorry for that.

Nitrogen dioxide:

velocity= 1/√mNitrogen dioxide

            = 1/√43.99

  = 0.150

so, now you can opt up option B as the right answer.



Aug 6th, 2015

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