Five samples of argon gas are described in the table below. Rank t he samples in
ID: 1031431 • Letter: F
Question
Five samples of argon gas are described in the table below. Rank t he samples in order of increasing average kinetic energy of the atoms in them That is, select "1" next to the sample in which the argon atoms have the lowest average kinetic energy. Select "2" next to the sample in which have the next lowest average kinetic energy, and so on. sample average kinetic energy of atoms in sample amount pressure temperature 2.1 mol 19 atm 2.5 mol1.7 atm 2.8 mol 2.1 atm 1.7 mol 1.0 atm 67, °C 91. C (Choose one) 77, °C (Choose one) (Choose one) I3 atm 8 °C(Choose one)Explanation / Answer
Average molecular Kinetic energy=3/2kT ,
where k=boltzmann's constant=1.381*10^-23 J/K,T=273-67=206K
1)n=2.1 mol=2.1mol*Na Na=avogadro's number=6.022*10^23 molecules/mol
N=number of atoms=2.1mol*(6.022*10^23 atoms/mol)=1.265*10^24 atom
K.E average=(1.265*10^24 atom)*3/2kT =(1.265*10^24 atom)*3/2*(1.381*10^-23 J/K)*206K=5398.12185J
2) N=2.5mol*(6.022*10^23 atoms/mol)=1.505*10^24 atom
T=273-91=182K
K.E average =(1.505*10^24 atom)*3/2*(1.381*10^-23 J/K)*182K=5674.0456J
3)N=2.8mol*(6.022*10^23 atoms/mol)=1.686*10^24 atom
T=273-77=196K
K.E average=(1.686*10^24 atom)*3/2*(1.381*10^-23 J/K)*196K=6845.3960J (1)
4) N=1.7mol*(6.022*10^23 atoms/mol)=1.024*10^24 atom
T=273-80=193K
K.E average=(1.024*10^24 atom)*3/2kT =(1.024*10^24 atom)*3/2*(1.381*10^-23 J/K)*193K=4093.9469J (5)
5) N=2.3mol*(6.022*10^23 atoms/mol)=1.385*10^24 atom
T=273-58=215K
K.E average=(1.385*10^24 atom)*3/2*(1.381*10^-23 J/K)*215K=6168.4091J
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