What is the stuffing in a closed box doing?

It is converting the adiabatic pressurization of the box into isothermal (partially). The positive pressure of a compression transmits energy to the stuffing and the air reduces temperature, therefore reduces pressure. The opposite part of the cycle is when the negative pressure of a rarefaction takes the energy of the stuffing, increasing its temperature and increasing its pressure. The stuffing is acting as a moderator. As we will see, if the stuffing was perfect we can get an apparent increase in box size of y=1.4 times.

Solution:

Let "d" be the differential operator
Let "~" be "approximately equal to"

Every gas has two molar heat capacities. One is when the gas is at constant volume (Cv), and the other when the gas is at constant pressure (Cp). We can relate the energy (Q) we add to the gas to the change in its temperature (T):

     dQ = n Cv dT                                 (1)
     dQ = n Cp dT                                 (2)

We can also measure the amount of energy bestowed upon the gas by relating the change in volume, with constant pressure, and relating change in pressure with constant volume:

     dQ = P dV                                    (3)
     dQ = -V dP                                   (4)

If we assume there is no loss of energy to outside the system (adiabatic), then all these dQ are equal. These four equations allow us to cancel out dQ and dT.

     P dV = n Cv dT                               (5)
    -V dP = n Cp dT                               (6)

     P dV   -V dP
=>   ---- = -----                                 (7)
      Cv      Cp

Now we rearrange and integrate:

     Cp    1        1
=>   -- * --- dV + --- dP = 0                     (8)
     Cv    V        P

     Cp
=>   -- * ln(V) + ln(P) = Constant                (9)
     Cv

Let y=Cp/Cv (which is always > 1) and we get:

     y
=> V P = Constant                                (10)

Let the adiabatic volume be Va. Let the isothermal volume be Vi. Lets analyze the differences in pressure between adiabatic and isothermal pressurization of the gas, with original volume Vo and pressure Po.

           1/y
    1     P
    -- = -------- (adiabatic)                    (11)
    Va        1/y
         Vo Po

     1     P
    -- = -----   (isothermal)                    (12)
    Vi   Vo Po

We must take a second order Taylor series of (11) if we want to get an approximate linear relationship between these two types of compression. The Taylor series I mention is:

     1       1    /   1/y      (1/y-1)       \
    -- ~ -------- | Po   - y Po       (P-Po) |   (13)
    Va        1/y |                          |
         Vo Po    \                          /

We now differentiate (12) and (13) wrt P:

    d (1/Vi)     1
    -------- = -----                             (14)
       dP      Vo Po

    d (1/Va)   1     1
    -------- = - * -----                         (15)
       dP      y   Vo Po

Now cancel 1/(Vo*Po) in (14) and (15) to conclude:

    d (1/Va)   1   d (1/Vi)
    -------- = - * --------                      (16)
       dP      y      dP

=>      y Vi = Va                                (17)

Therefore if we convert from isothermal to adiabatic pressurization we will notice an increase in volume of y times. An adiabatic enclosure (speaker box) is y times larger than an equivalent isothermal box.

Lets look at some application. The values of Cv and Cp have been measured to be:

These values take into account the distribution if 'heat' energy into its various kinetic forms. We can see that since our atmosphere is made of mostly Nitrogen and Oxygen our value for y is 1.40.