How do you work out the total translational kinetic energy of N2 gas?

Question

What is the total translational kinetic energy in a test chamber filled with nitrogen (#N_2#) at #2.23×10^5# Pa and #20.7°C#?

The dimensions of the chamber are #1.00 m × 6.00 m × 6.50 m. #

The atomic weight of nitrogen is #28.0# g/mol, Avogadro's number is #6.022 × 10^23# molecules/mol and the Boltzmann constant is #1.38 × 10^-23# J/K.

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Answer 1 · 2018-03-11T12:27:32

75 J

Volume of the chamber(#V#)=39 #m^3#
Pressure=#(2.23*10^5)/(1.01*10^5)#=2.207 atm

Temp=293.7 K

BY Equation of state;

n=#p*v/(RT)#=3.5696 moles

total molecules=3.5696*6.022#*10^23#=21.496#*10^23#

now energy for Each diatomic molecule

=#(DOF)*1/2*k*t#

For a diatomic gas degree of freedom =5

Therefore energy= (no of molecule) *(energy of each molecule)

Energy=#5*21.496*10^23*0.5* 1.38 *10^-23#
=74.168 J