At what temperature will the total K.E of 0.3 mol of He be the same as the total K.E of 0.4 mol of Ar at 400K:

a) 533K

b) 400K

c) 346K

d) 300K

Dear student,

Solve this using this hint:

et u be the average or root-mean-square speed of a gas whose molar mass is M. Since N is the Avogadro's number, the average kinetic energy is (1/2) (M/N) u² or

          M        3R T     3
 K.E.  = --- u2   = ---- = --- k T                              
         2 N        2 N     2

Sagar Singh
B.Tech IIT Delhi

KE = 3nRT/2

(KE)_He /(KE)_Ar = (moles of He)(temp of He)/(moles of Ar)(temprature of Ar)

1 = (0.3)T/(0.4)(400)

 T = 400*4/3

    =533K

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