Question

the rotation constant of CCl4 is 0.0572/cm. Evaluate the rotational partition function explicitly and plot its value as a function of temperature. At what temperature is the value within 5% of the value calculated from the approximate formula

Answer 1

For nonlinear rotors like CCl₄, the rotational partition function is approximated by

q^R = (1/σ)(kT/hc)^3/2(π/B³)^1/2., where σ = no. of indistinguishable orientations of CCl₄, i.e. 12, B= rotational constant, i.e. 0.0572 Cm^-1, k = Boltzmann's constant, i.e. 1.38*10^-23, T = 298.15 K (room temperature), h = Planck's constant, i.e. 6.625*10^-34 J.s., c = 3*10¹⁰ Cm.s^-1 and π = 3.14.

Therefore, q^R = (1/12) * (2978.6) * (129.5)

i.e. q^R = 3.2144*10⁴.

Since, q^R (T)^3/2 ; as the temperature (T) increases, corresponding rotational partition function (q^R) increases accordingly.

5% of the above value = 1.607*10³.

Then T^3/2 = (1.607*10³) * 12 * 1.73 (K)^3/2 * 0.0077

i.e. T ~ 40 K

Therefore At T = 40 K, the corresponding q^R value is 5% of the q^R value at T = 298.15 K