Answer to Question #89048 in Molecular Physics | Thermodynamics for Promise

Question

(a) What is the average translational kinetic energy of an ideal-gas molecule at 27°C?
(b) What is the total random translational kinetic energy of the molecules in 1 mole of this gas?
(c) What is the root-mean-square speed of oxygen molecules at this temperature?

Answer 1

(a) The average translational kinetic energy of an ideal-gas molecule is calculated as

$$E=(3/2)kT$$

where T is a temperature in Kelvins

$$Tk \approx Tc+273$$

, and

$$k=1.38×10^{−23} J/K$$

is a Boltzmann constant

So we can calculate the energy as

$$E=(3/2)kT=(3/2)*1.38×10^(−23)*(273.3+27)$$

$$E=6.21*10^{e-21}$$

(b) The total random translational kinetic energy of the molecules in 1 mole of this gas is equal to the sum of the energies of all molecules,

$$E_{total}=\sum{E}=E*N_{Avogadro}$$

$$E=6,023*(10^23)*6.21*10^(-21)=3740283J$$

where Navogadro is the number of molecules in one mole (Navogadro= 6,023*10^(23))

(c) The root-mean-square speed of oxygen molecules at this temperature can be calculated from the relation

$$E_{trans}=mv^2/2=(3/2)kT$$

where m is the mass of the oxygen molecule (m(O-molecule) = 2*m(O-atom) = 2.66*10^(-26))

$$v= \sqrt{3kT/m}$$

$$v= \sqrt{3*1.38*10^{-23}*300/{(2.66*10^{-26})}}=683.31m/s$$