Question

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 43.4°C and (b) 89.1°C. What is the translational kinetic energy per mole of an ideal gas at (c) 43.4°C and (d) 89.1°C?

Answer #1

Average kinetic energy of translatioal kinetic energy of a
molecule is

k avg = ( 3/ 2) kT

here k = 1.38*10-23 J / K

a) at T = 43.4 o C = 43.4 + 273 K =316.4 K

k avg = ( 3 /2 ) ( 1.38*10^-23 ) ( 316.4 )

=6.54948e-21 J

b) at T = 89.1 o C = 89.1 +273 K

= 362.1K

k avg = ( 3 /2 ) ( 1.38 *10-23 ) ( 362.1 )

=7.49547e-21 J

c ) translational kinetic energy per mole of an ideal gas
at43.4°C

k mol = ( Av ) ( k avg )

here Av = avagadros number ( 6.02*10^23 )

= ( 6.02*10^23 ) ( 6.54948e-21 )

= 3942.78696 J/ mol

d) at T = 89.1 o C

k mol = ( 6.02*10^23 ) * 7.49547e-21

=4512.27294 J / mol

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