SOLUTION TO EXERCISE 14

According to the Ideal Gas Law, P = nkT.  Since T = 0 C = 273 K is the temperature
of S.T.P. conditions, the number density of the gas in the vacuum chamber is
directly related to the number density of gas at S.T.P. by its pressure.  Since,
at S.T.P., one mole of particles occupies 22.4 liters,

      0.000000001 * 6 * 10^23 
  n = ----------------------- = 2.69 * 10^10 per cm^3 in the vacuum chamber.
       22.4 l * 1000 cm^3/l 


(a) Again, since P = nkT, and the comet tail has T = 273 K, the same as the
vacuum chamber, we can compare pressures by comparing number densities.

  P in vacuum chamber     2.69 * 10^10 per cm^3
 --------------------- = ----------------------- = 2.69 * 10^6
    P in comet tail           10,000 per cm^3

So the pressure in the chamber is about 2.69 million times greater 
than in the comet tail!  Clearly, vacuum experiments in Earth 
laboratories still have a long way to go before they can mimic the 
vacuum of space.


(b) If the comet tail's temperature is actually 3 K rather than 273 K, then
the Ideal Gas Law gives us guidance again: since P = nkT, the pressure in this
comet tail would be lower than that of the tail at 273 K by a factor of the
temperature ratio.  Thus,

 P in comet tail at 273 K     273 K
-------------------------- = ------- = 91
 P in comet tail at 3 K         3 K

So the pressure in the vacuum chamber is greater that that of the 3 K comet tail
by a factor of

   2.69 * 10^6 * 91 = 2.45 * 10^8  (that is, a factor of 245 million).  Very low!