Question

What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.

Answer 1

i

`J_(m) = 20`

ii

`J_(m) = 22.5`

Step-by-step explanation:

From the question we are told that

The first temperatures is
`T_1 =  25^oC =  25 +273 =298 \ K`

The second temperature is
`T_2 =  100^oC =  100 +273 = 373 \ K`

Generally the equation for the most highly populated rotational energy level is mathematically represented as

`J_(m) = [ (RT)/(2B)]  ^{(1)/(2) } - (1)/(2)`

Here R is the gas constant with value
`R =8.314 \ J\cdot K^(-1) \cdot mol^(-1)`

Also

B is given as
`B=\ 0.244 \ cm^(-1)`

Generally the energy require per mole to move 1 cm is 12 J /mole

So
`0.244 \ cm^(-1)` will require x J/mole

`x =  0.244 *  12`

=>
`x =  2.928 \ J/mol`

So at the first temperature

`J_(m) = [ (8.314 * 298  )/(2*  2.928 )]  ^{(1)/(2) } - 0.5`

=>
`J_(m) = 20`

So at the second temperature

`J_(m) = [ (8.314 * 373  )/(2*  2.928 )]  ^{(1)/(2) } - 0.5`

=>
`J_(m) = 22.5`