## Monday, December 8, 2008

### Physics 9B Practice Final Problem #2

Consider a cubic meter of air in the room. T=300K. P=105N/m3. 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.
NA = 6.02 x 1023/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 1023 = 2.41 x 1025 molecules

Part B
• U = 1⁄(γ−1) nRT
• We know that γNitrogen = 7/5
• U = 5/2 ⋅ 40.1 moles ⋅ 8.3 J/molK ⋅ 300K
• U = 2.50 x 105J
Part C
Notice that the internal energy for this problem is different than the translational energy, which is calculated like so:
• KTR = 3/2 nRT
• KTR = 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.
• KTR⁄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 mv2
• v = 514 m/s