The ratio of kinetic energy of equal no of moles of hydrogen and helium at same temperature is
Answers
Answer:
Irrespective of their molar masses,
He
ATOM has approximately
60
%
the average molar kinetic energy that
H
2
MOLECULE does at
25
∘
C
and
1 atm
.Well, I assume you mean only
He
is an atom, and not a molecule...
He
is an atom, and
H
2
is a molecule. Only then is one's molar mass twice that of the other.
As an atom,
He
can only translate, i.e. shoot forward and backwards via cartesian degrees of freedom, having only translational average kinetic energy
K
trans
.
H
2
can vibrate, at vibrational frequency
ω
≈
4394.49 cm
−
1
.
H
2
can also rotate, with
2
degrees of freedom.This theorem applies at the so-called high temperature limit, where the energy level spacings of each kind of motion (translational, rotational, vibrational) are much smaller than
k
B
T
where
k
B
is the Boltzmann constant.
This (pretty much) always applies for translational DOFs.
This usually applies for rotational DOFs, except for molecules with particularly low moments of inertia. This applies for
H
2
with no problem (for
H
2
, it would be a problem below approximately
88 K
).
For vibrational DOFs, given the high vibrational frequency of
H
2
, we can safely say that ignoring the vibrational DOFs of
H
2
is more accurate than including them.
We can break up the kinetic energy into translational and rotational parts, then, to a good approximation:
⟨
κ
⟩
≈
⟨
κ
⟩
trans
+
⟨
κ
⟩
rot
, for
H
2
For
He
,
⟨
κ
⟩
rot
≈
0
, because atoms cannot rotate and turn out to look any different under the hard sphere approximation.
So, the ratio of the average kinetic energies is:
⟨
κ
⟩
H
2
⟨
κ
⟩
He
=
⟨
κ
⟩
H
2
,
trans
+
⟨
κ
⟩
H
2
,
rot
⟨
κ
⟩
He
,
trans
=
3
2
R
T
+
2
2
R
T
3
2
R
T
=
5
2
×
2
3
=
5
3
So, on average, according to the equipartition theorem at high enough temperatures, an ensemble of
H
2
molecules will have
1.67
times the kinetic energy of an analogous ensemble of
He
atoms.