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Homework answers / question archive / For a CO2 gas molecule, calculate (a) the average translational energy ?KE? at 300 K and at 1000 K, (b) the root-mean-square speed Urms at 300 K and 1000 K, and (c) the average speed ?U? at 300 K and 1000 K
For a CO2 gas molecule, calculate (a) the average translational energy ?KE? at 300 K and at 1000 K, (b) the root-mean-square speed Urms at 300 K and 1000 K, and (c) the average speed ?U? at 300 K and 1000 K. Show all work and include units.
a) 6.21 X 10-21 J , 2.07 X 10-20 J
b) 412.39 m/sec, 752.924 m/sec
c) 379.93 m/s, 693.66 m/s
Step-by-step explanation
a) Average translational energy = 3/2 KbT
Kb = Boltzmanns constant = 1.38 x 10-23 J/K
At 300 K
Average translational energy = 3/2 KbT
= 3/2 x 1.38 x 10-23 J/K x 300 K
= 6.21 X 10-21 J
At 1000 K
Average translational energy = 3/2 KbT
= 3/2 x 1.38 x 10-23 J/K x 1000 K
= 2.07 X 10-20 J
b)
Root mean square velocity μrms = (3RT/M)½
R = ideal gas constant = 8.3145 (kg·m2/sec2)/K·mol
M = mass of a mole of the gas in kilograms. = 0.0044 kg/mol
At 300 K
Root mean square velocity μrms = (3RT/M)½
= (3 x 8.3145 (kg·m2/sec2)/K·mol x 300K x/0.044 kg/mol)1/2
= 412.39 m/sec
At 1000k
Root mean square velocity μrms = (3RT/M)½
= (3 x 8.3145 (kg·m2/sec2)/K·mol x 1000K x/0.044 kg/mol)1/2
= 752.924 m/sec
c)
Average molecular speed = (8RT/pi M)1/2
v = molecular speed
R =Ideal gas constant. (8.314 8.314 J/K/mole)
T: Absolute Temperature in Kelvin.
m: molar mass
At 300 K
Average molecular speed = (8RT/pi M)1/2
= (8 X 8.314 8.314 J/K/mole) X 300K/ pi x 0.044)1/2
= 379.93 m/s
At 1000 K
Average molecular speed = (8RT/pi M)1/2
= (8 X 8.314 8.314 J/K/mole) X 1000K/ pi x 0.044)1/2
= 693.66 m/s
