a) At a given temperature T, all gases molecules have the same average translational kinetic energy, namely K = (3/2)kBT, where T is in Kelvin, kB = 1.38*10-23 J/K
T(K) = T(oC) + 273.15 = 10.0C + 273.15 = 283.15 K
KA = (3/2)*1.38*10-23 J/K *283.15 K = 5.86*10-21 J
KB = (3/2)*1.38*10-23 J/K *283.15 K = 5.86*10-21 J
Average translational kinetic energy of molecules A and B are the same: KA = KB = 5.86*10-21 J
υrms = sqrt(3kBT/m) (for 1 molecule)
υrms(A) = sqrt(3*1.38*10-23 J/K *283.15 K/3.34*10-27 kg) = 1873.42 m/s
υrms(B) = sqrt(3*1.38*10-23 J/K *283.15 K/5.34*10-26 kg) = 468.53 m/s
As mA < mB molecules A move faster than molecules B: υrms(A)>υrms(B).
b) υrms(A) = sqrt(3kTA/mA)
υrms (B)= sqrt(3kTB/mB)
υrms(A) = υrms (B);
sqrt(3kTA/mA) = sqrt(3kTB/mB)
3kTA/mA = 3kTB/mB
TA/mA = TB/mB
As mA < mB , then TB should be higher than TA in order υrms(A) to be equal to υrms(B). We should raise the temperature of the B container in order to have both gases with the same rms speed.
c) TA/mA = TB/mB
283.15/ 3.34*10-27 = TB / 5.34*10-26
TB = 4527 K
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