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 = 16.6^{0}C = (273+16.6) = 289.6 K

KE avg = ( 3 /2 ) ( 1.38*10^-23 ) ( 289.6 )

= 5.99*10^-21 J

b) at T = 90.8^{0}C = (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 ^{0}C

k mol = ( 6.02*10^23 ) ( 7.53*10-20 )

=4.53*10^3 J / mol

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