Question

If he(g) has an average kinetic energy of 8750 j/mol under certain conditions, what is the root mean square speed of cl2(g) molecules under the same conditions?

Answer 1

The root mean square speed is given by V_rms = âšRT/M where r, t, and m are the rate constant, temperature and molar mass the gas Average molar kinetic energy of the gas E = 1/2 M * (V_rms)^2 = 8750 ms/1 So (V_rms)^2 = (2 * 8750) / M Molar mass of 2 chlorine atoms in kg is 2 * 35 * 10^(-3) Hence we have (V_rms)^2 = (2 * 8750)/ (2 * 35 * 10^(-3)) (V_rms)^2 = 8750/0.035 = 250000 So V_rms = âš 250000 = 500

Answer 2

Answer:

18.74 m/s is the root mean square speed of chlorine gas molecules under the same conditions.

Explanation:

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. It is determined by the equation:

$K.E=\frac{3RT}{2}$

where,

K.E = Average kinetic energy

$R$ =Universal gas constant =8.314 J /mol K

T = Temperature of the system

He has an average kinetic energy of 8750 J/mol

$8750 J/mol =\frac{3\times 8.314 J/mol K\times T}{2}$

T = $\frac{8750 J/mol \times 2}{3\times 8.314 J/mol K}$

T = 701.63 K

The formula used for root mean square speed is:

$\nu_{rms}=\sqrt{\frac{3kN_AT}{M}}$

where,

$\nu_{rms}$ = root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature =701.63 K

M = atomic mass = 0.071 kg/mole

$N_A$ = Avogadro’s number = $6.02\times 10^{23}mol^{-1}$

$\nu_{rms}=\sqrt{\frac{3\times 1.38\times 10^{-23}J/K\times 6.022\times 10^{23} mol^{-1}\times 701.63 K}{0.071 kg/mol}}$

$\nu_{rms}=18.74 m/s$

18.74 m/s is the root mean square speed of chlorine gas molecules under the same conditions.