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

## 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.

Mar 11, 2018

75 J

#### Explanation:

Volume of the chamber($V$)=39 ${m}^{3}$
Pressure=$\frac{2.23 \cdot {10}^{5}}{1.01 \cdot {10}^{5}}$=2.207 atm

Temp=293.7 K

BY Equation of state;

n=$p \cdot \frac{v}{R T}$=3.5696 moles

total molecules=3.5696*6.022$\cdot {10}^{23}$=21.496$\cdot {10}^{23}$

now energy for Each diatomic molecule

=$\left(D O F\right) \cdot \frac{1}{2} \cdot k \cdot t$

For a diatomic gas degree of freedom =5

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

Energy=$5 \cdot 21.496 \cdot {10}^{23} \cdot 0.5 \cdot 1.38 \cdot {10}^{-} 23$
=74.168 J