10.6 Partition Function for a Gas
451
For temperatures T >> θ rot,B , b << 1 and large L contribute in the sums, so the
sums can be well approximated by integrals. Since only every other L appears in
each sum, we use
L,odd
(2L + 1) exp (−bL(L + 1)) ≈
L,even
(2L + 1) exp (−bL(L + 1))
≈
1
2
∞
0
exp (−bx)dx =
1
2b
.
(10.66)
So, for high temperatures, N o /N p → 3. At room temperature (293.15 K), we find
N o /N p = 2.985. This means that if the two isomers are in equilibrium at room
temperature, 74.9% of hydrogen gas will be in the ortho state. Compare with the
gas at 80 K, which has N o /N p = 1.00, so the ortho state is then 50.0% of the gas.
For low temperatures (T << θ rot,B = 87.6 K), only the lowest terms in the
partition function contribute, so that
N o
N p
≈
3 · 3 exp (−2b)
1
= 9e
−2B/(k B T )
→ 0 ,
(10.67)
making the gas almost all para-hydrogen.
Nitrogen Isomers
Nitrogen gas also comes in different isomer forms, because the nucleus of the atom
14 N has spin one, so the nuclei of the molecule N 2 can be in a spin state with
a spin quantum number I , with molecules in three possible isomeric states para
(I = 0), meta (I = 1), and ortho (I = 2). The spin wave functions under nucleus
exchange change sign by (−1) I Since N 2 is a boson, the overall wave function must
be symmetric. The rotation which exchanges the two nitrogen nuclei will change the
orbital wave function by (−1) L , so the orbital angular momentum quantum number
L must be even if I is even, and odd if I is odd.
The nuclear degeneracies of the nitrogen molecule is 2I + 1, and so takes values
of 1, 3, 5 for I = 0, 1, 2. The even I , even L states will be degenerate (in the absence
of any strong external magnetic field). The I = 1, L = odd states start in energy
slightly higher than the I = 0, 2, L = even states because the orbital rotational
energies are
L =
¯
h 2
2I B
L(L + 1) = BL(L + 1)
(10.68)
so the lowest energy for odd L is 2B while for even L the lowest rotational energy
is zero.
In all, then, the rotational molecular partition function will be
451
For temperatures T >> θ rot,B , b << 1 and large L contribute in the sums, so the
sums can be well approximated by integrals. Since only every other L appears in
each sum, we use
L,odd
(2L + 1) exp (−bL(L + 1)) ≈
L,even
(2L + 1) exp (−bL(L + 1))
≈
1
2
∞
0
exp (−bx)dx =
1
2b
.
(10.66)
So, for high temperatures, N o /N p → 3. At room temperature (293.15 K), we find
N o /N p = 2.985. This means that if the two isomers are in equilibrium at room
temperature, 74.9% of hydrogen gas will be in the ortho state. Compare with the
gas at 80 K, which has N o /N p = 1.00, so the ortho state is then 50.0% of the gas.
For low temperatures (T << θ rot,B = 87.6 K), only the lowest terms in the
partition function contribute, so that
N o
N p
≈
3 · 3 exp (−2b)
1
= 9e
−2B/(k B T )
→ 0 ,
(10.67)
making the gas almost all para-hydrogen.
Nitrogen Isomers
Nitrogen gas also comes in different isomer forms, because the nucleus of the atom
14 N has spin one, so the nuclei of the molecule N 2 can be in a spin state with
a spin quantum number I , with molecules in three possible isomeric states para
(I = 0), meta (I = 1), and ortho (I = 2). The spin wave functions under nucleus
exchange change sign by (−1) I Since N 2 is a boson, the overall wave function must
be symmetric. The rotation which exchanges the two nitrogen nuclei will change the
orbital wave function by (−1) L , so the orbital angular momentum quantum number
L must be even if I is even, and odd if I is odd.
The nuclear degeneracies of the nitrogen molecule is 2I + 1, and so takes values
of 1, 3, 5 for I = 0, 1, 2. The even I , even L states will be degenerate (in the absence
of any strong external magnetic field). The I = 1, L = odd states start in energy
slightly higher than the I = 0, 2, L = even states because the orbital rotational
energies are
L =
¯
h 2
2I B
L(L + 1) = BL(L + 1)
(10.68)
so the lowest energy for odd L is 2B while for even L the lowest rotational energy
is zero.
In all, then, the rotational molecular partition function will be