In a classroom at 290 K, there are typically 3×10^27 air molecules each having f
ID: 3279941 • Letter: I
Question
In a classroom at 290 K, there are typically 3×10^27 air molecules each having five degrees of freedom and mass 4.8×10^-26 kg. The room has dimensions 3 m by 6 m by 7 m. You are going to calculate how many quantum states are accessible to this system.
In a classroom at 290 K, there are typically 3x1027 air molecules each having five degrees of freedom and mass 4.8x10-26 kg. The room has dimensions 3 m by 6 m by 7 m.You are going to calculate how many quantum states are accessible to this system (a) What is the thermal energy of this system? (b) The magnitude of the momentum of any molecule can range from 0 to po. Estimating po to be roughly twice the root mean square momentum, what is the volume in momentum space that is available to any particle? (c) We are going to calculate the number of accessible quantum states for the molecules' translational motions, ignoring the rotational states because the latter turn out to be relatively few in comparison. How many different quantum states would be accessible to any particle if it were all by itself? (d) If all the particles were identical, what would be the corrected number of states per particle, (e) How many different quantum states are accessible to the entire system? (f) Answer parts (d) and (e) but now assuming that 79% are identical nitrogen molecules and 21% are identical oxygen molecules (of about the same mass).Explanation / Answer
Average KE of a molecule is 0.5kBT per degree of freedom
each molucule has 5 degrees of freedom
average KE of each molecule = 2.5kBT
There are 3.0e+27 molecules in the room and T = 290 K
Total thermal energy of the system = 2.5*1.38e-23*290*3.0e+27 = 3.0e+7 J
KE= p2/2m
Total mass M = 4.8e-26kg
mass of each molucule m = 4.8e-26/3.0e+27 = 1.6e-53 kg
mean p2 = mean KE *2m = 2.5*1.38e-23*290*2*1.6e-53 = 2.72e-49
po = 2*RMS(p) = 2* sqrt(2.72e-49) = 1.04e-24
This the momentum in each direction x,y,z
volume in momentum space available for each molecule = (1.05e-24)3 = 1.15e-72
quantum vibrational energy is roughly 0.3ev where as the room temp the average energy of the molecule kBT = 0.025ev which is not sufficient ot excite to any higher state and all the molecules will be in the ground state and there is only one quantum state for each molecule, i.e the ground state.