Question

Consider the isotopes of oxygen ¹⁶O and ¹⁸O. These form three diatomic molecules: ¹⁶O¹⁶O, ¹⁶O¹⁸O, and ¹⁸O¹⁸O.

(a) What are the relative average translational kinetic energies for these three diatomic molecules at 25◦ ?

(b) Describe and compare their Maxwell-Boltzmann speed distributions, and write the f(v) equation for each.

(c) ¹⁶O with isotopic mass 15.994 g/mol is much more common than ¹⁸O with isotopic mass 17.999 g/mol, so it is safe to neglect ¹⁸O¹⁸O. You have a mixture of ¹⁶O¹⁶O and ¹⁶O¹⁸O that is 99.80% ¹⁶O¹⁶O and would like to enrich it by gaseous diffusion to be 99.98% ¹⁶O¹⁶O. How many gaseous diffusion stages are required?

Answer 1

Answer:-

(a). The average traslational kinetic energy of a gas is independent of molar mass . So the the relative traslational kinetic energy for these three diatomic molecules will be same at 25^o C.

THe average translational kinetic energy = 3/2kT

Temperature, T= 25^oc= 273+25= 298 K

thus average traslational kinetic energy= 3/2*1.3806*10^-23*298

= 6.1712*10^-21 J

For all these three diatomic molecules the average traslational kinetic energy will be equal.

(b).the Maxwell-Boltzmann speed distributions for these three diatomic molecules will be different because all the molecules have different molar mass . In the  Maxwell-Boltzmann speed distributions the molar mass is in the negative power of exponential so if the molar mass increases the speed of the molecule will decrease so the distribution curve. Thus for 16O16O the velocity will be highest whereas for 18O18O, the velocity wiil be least so the distribution curve.

f(v) equation for 16O16O, 16O18O and 18O18O

f(v)=(m/2kT)^3/2e^-mu2/2kT

(c).