## Archive for May, 2009

### Physics Friday 74

May 29, 2009

Previously, I discussed the heat capacity of an ideal gas, and the relationship to the molecular degrees of freedom f. Now, another important property for a gas is the ratio of the specific heat for constant pressure to that for constant volume, known as the heat capacity ratio, or adiabatic index:
. Recalling the derivation of the specific heats from the internal energy and enthalpy, we see that for an ideal gas, we can also express the adiabatic index as .

Recalling that in terms of f, our ideal gas has heat capacities  and , so . Thus, a monatomic ideal gas has f=3, and so , while a diatomic gas with five degrees of freedom (such as nitrogen or oxygen at room temperature) has . For comparison, room temperature air has an adiabatic index measured at approximately 1.403, very close to the 7/5=1.4 given above. Note that as f increases, as occurs for real gases with increasing temperature, γ decreases toward unity.

To see why γ is called the adiabatic index, we recall that an adiabatic process is one in which no heat is transfered to or from the working fluid. We recall that the increment in internal energy dU is the sum of the heat added  and the work done ; as the heat transfered is zero (, dS=0), we have
. Now, as , we get

Now, putting the ideal gas law  in differential form, . Solving for dT and using the difference relation , we see

Equating with the above condition for dT in an adiabatic process:
.
Integrating the last, we obtain
, which we can combine using the properties of logarithms to get:
 for an adiabatic process; thus the use of the term adiabatic index. The above condition for pressure and volume is known as the adiabatic condition.

May 25, 2009

Find .

Solution:

### Physics Friday 73

May 22, 2009

For an ideal gas of N molecules, each with f total available degrees of freedom for each molecule (f≥3), we have equation of state  and internal energy , as discussed here and here. Now, we can use this to find the heat capacity for a given quantity of an ideal gas. The heat capacity is defined as the ratio of a small amount of heat (energy) added to the body to the resulting small temperature increase, with the system under some constraint (to define a particular path through the phase space of the system):

Now, recalling that we can define temperature via , we can define heat capacity as
.
For gases, the two conditions we usually consider are constant volume, CV, and constant pressure, CP.

For a closed system, the internal energy has increment  (see here). Thus, we see that for constant volume (dV=0), , and so .
Now, as our internal energy is , we see that 
With enthalpy defined as H=U+PV, it has increment , which can combine with the internal energy increment equation above to get , and so our heat capacity at constant pressure is .
Using our ideal gas law, , so , and thus .

The specific heat capacity is the heat capacity for a fixed quantity of substance, usually either a unit mass or unit mole. Recalling that for n moles of gas, Nk=nR, we have , so for the molar specific heats, we divide the above by n to get .

Next, we can also note that the heat capacity has the same units as Boltzmann’s constant k, and as it scales with quantity, it scales with the number of molecules. Thus, we can define a dimensionless specific heat capacity:
. The above difference relation between heat capacities for constant pressure and constant volume then becomes ĉPĉV=1, and in terms of f, we see  and .

Recalling the previous discussion of the internal degrees of freedom, we note that for a real gas, f is dependent on temperature: as T increases past various threshold temperatures, more degrees of freedom “unfreeze” and f increases. Thus actual specific heat capacities also depend on temperature, usually increasing with rising temperatures.

### Monday Math 72

May 18, 2009

What are the conditions on real numbers x, y, z such that ?

Solution

### Physics Friday 72

May 15, 2009

When we previously derived the ideal gas law PV=NkT=nRT from a particle bouncing in a box, we used the Maxwell speed distribution, which was in turn derived using the Boltzmann factor with regards to the kinetic energy of the translational motion of the molecules.
Now, if we want to consider the internal energy U of the gas, and the properties deriving from it, such as specific heat capacity, we must consider not only the translatation of the molecules, but also other possible motions of the molecule; in particular, the number of degrees of freedom.
For a molecule of n atoms, there are 3n degrees of freedom; 3 of these are translation of the center of mass of the molecule (as space is three-dimensional). The remaining 3n-3 are internal degrees of freedom, such as rotations, bond length vibrations, and bond angle changes. (See here.)
We note that the internal degrees of freedom are in fact quantized, and thus, if the energy levels for a mode are spaced wide enough, that mode may not be accessisible at energies, and thus temperatures, below certain levels. So, we must consider the available degrees of freedom; the equipartition theorem tells us that in thermal equilibrium, all the available degrees of freedom should have the same average kinetic energy.
To give examples, we first consider a monatomic gas, such as the inert gases: there are only 3 degrees of freedom, and the internal energy is simply the total kinetic energy . For nitrogen gas, a diatomic molecule, there are six degrees of freedom: three translational modes, two rotational modes, and vibration of the bond length. At room temperature, all of these except for the vibration are accessable, and each of the available modes has an average energy per molecule of , for internal energy , five-thirds that of a monatomic gas at the same temperature.

### Monday Math 71

May 11, 2009

Let x, y, and z be positive real numbers. Then prove that .

Due to (permutation) symmetry, we may assume without loss of generality, that . Then , and so by Chebyshev’s inequality:


And by Nesbitt’s inequality,
,
so


### Physics Friday 71: The Isothermal Atmosphere

May 8, 2009

Consider an ideal gas in a vertical column with horizontal cross-sectional area A, under uniform gravity with gravitational accelleration g, at a uniform temperature T (the “isothermal atmosphere” model). What, then, is the probability of finding a particle at a given height z above the “floor” of our column?

The ideal gas law tells us that for an ideal gas of N particles, the themperature T, pressure P, and volume V obey the relation PV=NkT. Solving for the pressure, . We can note, though, that  is the number density of molecules. Thus, we have . For our case, we have our properties varying with height z: .

Now, consider the volume of gas in a thin layer of height dz at an altitude z: The volume is A dz, so the number of molecules is n(z)A dz, and if the mass of a gas molecule is m, this gives the layer a mass of mn(z)A dz. This layer then has a weight of mgn(z)A dz. This force must be balanced by a pressure gradient:
P(z+dzAP(zA=A dP=-mgn(z)A dz
and so we can cancel A, and we have differential equation
. Solving  for n(z), we get , and plugging this in, we get:
. This is a simple seperable differential equation; our solution is
, and our pressure should decline exponentially with height. Further, , so the density declines exponentially as well.

Returning to our original problem, the probability of a particular molecule being found at a particular height should be proportional to the number density at that height. Thus, . However, mgz is the gravitational potential energy of a molecule at a height z. Thus, our exponential proportionality we just found is, in fact, simply the Boltzmann factor.

(I vaguely recall reading somewhere, though I don’t recall where, that reasoning similar to the above, along with observations of how the atmosphere changes with altitude, was what originally motivated the development of the Boltzmann factor).

### Monday Math 70

May 4, 2009

The harmonic mean of a set of positive real numbers  is given by . (So , and the reciprocal of the harmonic mean is the arithmetic mean of the reciprocals). This mean finds uses in problems involving averages of rates, and also in some electronic circuit problems (resistors or inductors in parallel, capacitors in series).

Now, let us apply the AM-GM inequality to the reciprocal of the harmonic mean:
,
where  is the geometric mean. Thus,
. Therefore, for any set of positive numbers, we see the arithmetic, geometric, and harmonic means obey , with equality if and only if the numbers in the set are all equal.

For an example, one can use this result to prove a classic inequality called Nesbitt’s inequality: for positive real numbers x, y, z
.

Let . Then the AM-HM inequality above tells us that:

so

expanding the product on the left:
.
Q.E.D.

### Physics Friday 70

May 1, 2009

We noted previously that for an ideal gas, the molecule speeds should follow the Maxwell speed distribution; from this, we predict a rms velocity . We can rewrite this in terms of the molar mass M and gas constant R: . Let us consider a “room” temperature of 290 K. The gas constant has value 8.3145 J K-1 mol-1, so this tells us that 3RT for our situation is 7234 J/mol. Let us then compute the vrms for several gasses:

Gas Molar Mass (kg/mol) vrms (m/s)
H2 0.00202 1.89×103
He 0.00400 1.34×103
CH4 0.0160 672
Ne 0.0202 599
N2 0.0280 508
C2H6 0.0301 490
O2 0.0320 475
Ar 0.0399 425
CO2 0.0440 405
C3H8 0.0441 405
Kr 0.0838 293
Xe 0.131 235