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Chemistry - Gaseous state

Chemistry - Gaseous state

by Harish J Mohan -
Number of replies: 4

Calculate the average kinetic energy, in joules, of the molecules in 8.0g of Methane ar 27oC.

Please Help??

Thank You.

In reply to Harish J Mohan

Re: Chemistry - Gaseous state

by Varsha Venkat -

Here, you can plug in the formula, average K.E =

3/2 nRT = 3/2 x 1/2 x R x 300

(since 8 grams of methane correspond to half a mole, and 27oC correspond to 300 K)

= 225 x R = 225 x 8.314 = 1870.65 joules.

 

In reply to Varsha Venkat

Re: Chemistry - Gaseous state

by Harish J Mohan -

Yeah, exactly thats what I did but the answer provided is not this,

The working is:

Av K.E = 3/2 RT

(It says that no of moles does'nt matter),

so the answer is

3/2 x 8.314 J/K-Mol x 300.15 K

= 3743 J/Mol.

In reply to Harish J Mohan

Re: Chemistry - Gaseous state

by Srinath Murthy -

Hopefully the following clears the doubt:

Total translational KE=(3/2)nRT

Av. translational KE/mole=(3/2)RT

Av. translational KE/molecule=(3/2)kT

In reply to Srinath Murthy

Re: Chemistry - Gaseous state

by Harish J Mohan -

Yes, ok

Av translational K.E / mole is - 3/2 RT

but goven here is 1/2 mole so do we have to neglect the data regarding given moles and simply use av Translational K.E / mole?