Question

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 16.6°C and (b) 90.8°C. What is the translational kinetic energy per mole of an ideal gas at (c) 16.6°C and (d) 90.8°C?

Answer 1

Average kinetic energy of translatioal kinetic energy of a molecule is
KEavg = ( 3/ 2) kT
here k = 1.38*10^-23 J / K
a) at T = 16.6⁰C = (273+16.6) = 289.6 K
KE avg = ( 3 /2 ) ( 1.38*10^-23 ) ( 289.6 )
= 5.99*10^-21 J
b) at T = 90.8⁰C = (90.8 +273) = 363.8 K
  
KE avg = ( 3 /2 ) ( 1.38 *10^-23 ) ( 363.8 )

= 7.53*10^-21 J
c ) translational kinetic energy per mole of an ideal gas at 16.6°C
KE mol = ( Av ) ( KEavg )
here Av = avagadros number ( 6.02*10^23 )
= ( 6.02*10^23 ) ( 5.99*10-21 )
= 3.60*10^3 J/ mol
d) at T = 90.8 ⁰C
k mol = ( 6.02*10^23 ) ( 7.53*10-20 )
=4.53*10^3 J / mol