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

Answer:

i

$J_{m} = 20$

ii

$J_{m} = 22.5$

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

     $J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}$

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

     $J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5$

=>  $J_{m} = 20$

So at the second temperature

           $J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5$

=>  $J_{m} = 22.5$