Missing question (found on internet):
What is the total translational kinetic energy of all the gas molecules in the sample? Express your answer using two significant figures.
Solution:
The average translational kinetic energy of the molecules in a gas is given by:[tex]K= \frac{3}{2}k_B T [/tex]
where
kB is the Boltzmann's constant
T is the absolute temperature of the gas.
In our problem, the absolute temperature is
[tex]T=21.0^{\circ}C + 273 = 294 K[/tex]
therefore the average translational kinetic energy of the molecules is
[tex]K= \frac{3}{2}(1.38 \cdot 10^{-23} J/K)(294 K)=6.09 \cdot 10^{-21}J [/tex]
In this problem, we have 1.2 mol of this gas, and since one mole of ideal gas contains a number of molecules equal to Avogadro number, the total number of molecules in our gas is
[tex]N=n N_A = (1.2 mol)(6.022 \cdot 10^{23} mol^{-1} )=7.35 \cdot 10^{23} [/tex]
So the total translational kinetic energy of all molecules of the gas is
[tex]K_{tot} = NK = (7.35 \cdot 10^{23})(6.09 \cdot 10^{-21} J)=4476 J=4.5 \cdot 10^3 J[/tex]
