OnRamps UT Austin

Question

Boltzmann’s constant is 1.38066×10−23 J/K, and the universal gas constant is 8.31451 J/K · mol. If 3.1 mol of a gas is confined to a 2.8 L vessel at a pressure of 10.4 atm, what is the average kinetic energy of a gas molecule? Answer in units of J.

Answer 1

Using the expression that pv=NkT and (1/2) mv^2 = 3kT/2 one arrives at pv=(Nmv^2)/3 where p = pressure,

v= volume , N= number of molecules. N= number of moles * avogado's number( NA) where NA= 6.02x10^23/ mol. Using this information 10.4x 1000x2.8=29120 = 3.1x 6.02x10^23xKEx2/3 where KE is the average kinetic energy in J. Therefore KE= 29120/12.44x10^23 J Hence avg KE = 2340x10^-23 J or = 23.4x10^-21 J.