Question

Find the total translational kinetic energy of

2.5 L of oxygen gas held at a temperature of
6
◦C and a pressure of 2.5 atm.
Answer in units of J.

Answer 1

953.7 J

Step-by-step explanation:

The average translational kinetic energy of the molecules in a gas is given by

`KE=(3)/(2)kT`

where

`k=1.38\cdot 10^(-23) J/K` is the Boltzmann constant

T is the absolute temperature of the gas

Here we have:

`T=6^(\circ)C+273=279 K` is the absolute temperature of the gas

Therefore, the average translational kinetic energy of each molecule is:

`KE=(3)/(2)(1.38\cdot 10^(-23))(279)=5.78\cdot 10^(-21) J`

Now in order to find the total translational kinetic energy of all molecules, we have to find the number of molecules in the gas.

We can do it by using the equation of state for an ideal gas:

`pV=nRT`

where here:

p = 2.5 atm is the gas pressure

V = 2.5 L is the volume

`R=0.082J/mol K` is the gas constant

`T=279 K` is the temperature

Solving for n, we find the number of moles:

`n=(PV)/(RT)=((2.5)(2.5))/((0.082)(279))=0.273 mol`

So the number of molecules contained in this gas is:

`N=nN_A=(0.273)(6.022\cdot 10^(23))=1.65\cdot 10^(23)`

where
`N_A` is Avogadro number. Therefore, the total translational kinetic energy in the gas is:

`KE_(tot)=N\cdot KE = (1.65\cdot 10^(23))(5.78\cdot 10^(-21))=953.7 J`