(1/2) m (v^2)av = (3/2) kT
so m v^2 is the same for O2 and O3
m1 v1^2 = m2 v2^2
m2 = 2 and m1 = 3
3 v1^2 = 2 v2^2
so
v2^2/v1^2 = 3/2
v2/v1 = (3/2)^(1/2)
TRUE
At a given temperature O2 molecules will move (3/2)1/2 faster than O3 molecules on the average.
True or False? I'm not sure if it is true or false on this one, but I think its false.
If you mix a gas, eventually all the molecules have the same kinetic energy on the average, independent of their mass or the temperature at which they were introduced.
I think this is false, but I'm not completely sure.
In the kinetic theory of gases, pressure is due to the strength of the molecular bonds within each gas molecule.
I think this is true, but again, not sure.
3 answers
If you mix a gas, eventually all the molecules have the same kinetic energy on the average, independent of their mass or the temperature at which they were introduced.
I think this is false, but I'm not completely sure.
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I think it true
(1/2) m (v^2)av = average kinetic energy the same for all the molecules at that temp = (3/2) kT
I think this is false, but I'm not completely sure.
*************************************
I think it true
(1/2) m (v^2)av = average kinetic energy the same for all the molecules at that temp = (3/2) kT
In the kinetic theory of gases, pressure is due to the strength of the molecular bonds within each gas molecule.
I think this is true, but again, not sure.
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False false false
Think of gas banging walls of the container.
It depends on how many and how fast they are moving
remember p = n R T/V
I think this is true, but again, not sure.
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False false false
Think of gas banging walls of the container.
It depends on how many and how fast they are moving
remember p = n R T/V
1 answer