^{5}N/m

^{3}. To make it a little easier, assume the air is 100% (rather than the actual 80%) nitrogen molecules each with a mass of 47 x 10

^{-27}kg and an effective γ of 7/5.

N

_{A}= 6.02 x 10

^{23}/mol.

R = 8.31 J/(mol K)

a) How many molecules are there?

b) What is the internal energy for the cubic meter of gas?

c) What is the average translational kinetic energy per molecule?

d) To what speed does that correspond?

Question so kindly provided by Joe Kiskis

**Part A**

This is a simple application of PV=nRT, just solving in molecules instead of moles.

n = PV⁄RT

n = 40.1 moles

40.1 moles ⋅ 6.02 x 10

^{23}= 2.41 x 10

^{25}molecules

**Part B**

- U = 1⁄(γ−1) nRT
- We know that γ
_{Nitrogen}= 7/5 - U = 5/2 ⋅ 40.1 moles ⋅ 8.3 J/molK ⋅ 300K
- U = 2.50 x 10
^{5}J

**Part C**

Notice that the internal energy for this problem is different than the translational energy, which is calculated like so:

- K
_{TR}= 3/2 nRT - K
_{TR}= 3/2 ⋅ 40.1 mole ⋅ 8.31 J/molK ⋅ 300K - Note that we want translational energy per molecule, not total, so divide by our answer from part A.
- K
_{TR}⁄molecule = 6.21 x 10^{-21}J/molecule

**Part D**

Simply apply the energy from the last step to kinetic motion for 1 molecule:

- 6.21 x 10
^{-21}J = 1/2 mv^{2} - v = 514 m/s

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