Question

A 1.2-mol sample of hydrogen gas has a temperature of 21 ∘C.

Answer 1

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:
`K= (3)/(2)k_B T`
where
kB is the Boltzmann's constant
T is the absolute temperature of the gas.
In our problem, the absolute temperature is

`T=21.0^(\circ)C + 273 = 294 K`
therefore the average translational kinetic energy of the molecules is

`K= (3)/(2)(1.38 \cdot 10^(-23) J/K)(294 K)=6.09 \cdot 10^(-21)J`

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

`N=n N_A = (1.2 mol)(6.022 \cdot 10^(23) mol^(-1) )=7.35 \cdot 10^(23)`

So the total translational kinetic energy of all molecules of the gas is

`K_(tot) = NK = (7.35 \cdot 10^(23))(6.09 \cdot 10^(-21) J)=4476 J=4.5 \cdot 10^3 J`