Answer to Question #83936 in Molecular Physics | Thermodynamics for Tran Thi Tuyet Mai

Question

Two identical gas containers A and B contains the masses of molecules mA = 3.34×10-27 kg and mB = 5.34×10-26 kg,
respectively. Both gases are under the same pressure and are at 10.0°C.
a) Compare the translational kinetic energy per molecule of molecules A, B and root-mean-square (rms) speeds of
them?
b) In order have both gases with the same rms speed, we need to raise the temperature of only one of these
containers. Which gas should be raised the temperature?
c) At what temperature will you accomplish your goal?

Answer 1

a) At a given temperature T, all gases molecules have the same average translational kinetic energy, namely K = (3/2)k_BT, where T is in Kelvin, k_B = 1.38*10^-23 J/K

T(K) = T(^oC) + 273.15 = 10.0C + 273.15 = 283.15 K

K_A = (3/2)*1.38*10^-23 J/K *283.15 K = 5.86*10^-21 J

K_B = (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: K_A = K_B = 5.86*10^-21 J

υ_rms = sqrt(3k_BT/m) (for 1 molecule)

υ_rms(A) = sqrt(3*1.38*10^-23 J/K *283.15 K/3.34*10^-27 kg) = 1873.42 m/s

υ_rms(B) = sqrt(3*1.38*10^-23 J/K *283.15 K/5.34*10^-26 kg) = 468.53 m/s

As m_A < m_B molecules A move faster than molecules B: υ_rms(A)>υ_rms(B).

b) υ_rms(A) = sqrt(3kT_A/m_A)

υ_rms (B)= sqrt(3kT_B/m_B)

υ_rms(A) = υ_rms (B);

sqrt(3kT_A/m_A) = sqrt(3kT_B/m_B)

3kT_A/m_A = 3kT_B/m_B

T_A/m_A = T_B/m_B

As m_A < m_B , then T_B should be higher than T_A in order υ_rms(A) to be equal to υ_rms(B). We should raise the temperature of the B container in order to have both gases with the same rms speed.

c) T_A/m_A = T_B/m_B

283.15/ 3.34*10^-27 = T_B / 5.34*10^-26

T_B = 4527 K