Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
 The first temperatures is [tex]T_1 =  25^oC =  25 +273 =298 \ K[/tex]
  The second temperature is  [tex]T_2 =  100^oC =  100 +273 = 373 \ K[/tex]
Generally the equation for  the most highly populated rotational energy level is mathematically represented as
   [tex]J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also Â
   B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
  Generally the energy require per mole to move 1 cm is  12 J /mole
So  [tex]0.244 \ cm^{-1}[/tex]  will require x J/mole
      [tex]x =  0.244 *  12[/tex]
=> Â Â Â Â Â [tex]x = Â 2.928 \ J/mol [/tex]
So at the first temperature
   [tex]J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]
=> Â [tex]J_{m} = 20 [/tex]
So at the second temperature
      [tex]J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]
=> Â [tex]J_{m} = 22.5 [/tex]
