A sealed container of volume V =0 .10m^3 holds a sample of N =3 .0×10^24 atoms of helium gas in equilibrium. The distribution of speeds of the helium atoms shows a peak at 1.1×10^3 ms^−1.

Question

A sealed container of volume V =0 .10m^3 holds a sample of N =3 .0×10^24 atoms of helium gas in equilibrium. The distribution of speeds of the helium atoms shows a peak at 1.1×10^3 ms^−1.
(i) Calculate the temperature and pressure of the helium gas. 
(ii) What is the average kinetic energy of the helium atoms? 
(iii) What is the energy corresponding to the maximum in the energy distribution

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I have the temperature as 194K. for the Pressure I need to use P=nKT I assume but im a little lost on the value of n. I have the number of helium atoms as 3.0×10^24 so do I times this by the amu and then by Avocado(joke) constant to get the n value of moles?

Any help would be greatly accepted! Thank you.

Damon · Answer

number of mols, n = 3*10^24/6*10^23 = 5

Anon · Answer

So if I know the average kinetic energy of one mole of helium atoms. Can I just times this by five to obtain the answer to ii) ? this seems deceptively simple.

bobpursley · Answer

yes, if you know the avg for one mole, multiply it by 5.