Answer to Question #127897 in Physical Chemistry for Goldi

Question

Plot the rotational partition function of N2 as a function of temperature from 10 to 300 K. At what temperature (in K) does the approximation of qr=TΘr (Eq. 5.49) result in less than 1% error?

Answer 1

The rotational partition function of bimolecular gas is:

The value of  the rotational constant B for N₂ is 1.998 cm^-1. So called symmetry number σ=2 for a homonuclear molecules like N₂.  The value of k_BT in cm^-1 can be obtained by dividing it by hc, i.e., (k_BT/hc) = 209.7 cm^-1 at 300 K. Therefore q_rot at 300K can be calculated as: 209.7 / (1,998 * 2) = 52.477. At 10K q_rot = 7 / (1,998 * 2) = 1.75. The intermediate values are plotted on the graph:

To find the temperature when temperature the approximation of qr=TΘr (Eq. 5.49) result in less than 1% error we need to the following calculations. Eq. 5.49 for rotational partition function q_rot is:

This 5.49 expression is simplified. The full expression is Eq. 5.51:

Lets take into account first two members of this range only, because the others are negligible. Thus we need to solve the equation:

T = 1.998 / 0.03 = 66.6 K

Answer: At temperatures lower than 66.6K the error is higher then 1%.