c01 JWBS043-Rogers September 13, 2010 11:20 Printer Name: Yet to Come
PROBLEMS AND EXERCISES 13
Since thepV/Tquotients are equal to each other for any arbitrary variations inp,V,
andT, they must be equal to the same constantk:
pV
T
=const=k
which is the combined gas law.
If we accept the combined gas law, there is a gas constant for any specified quantity
of each individual gas, subject only to the restriction of ideal behavior. If we demand
thatVbe themolar volume Vm
pVm
T
=
p
(V
n
)
T
=k
wherenis the number of moles in the gas sample, then this equation becomes
pV
T
=nR
where the symbolRis used to denote auniversalgas constant applicable to one
mole of any gas in the approximation of ideal behavior. One can obtain a value for
Rby arbitrarily assigning a pressure of 1 bar atT=298.15 K to precisely 1.0 mole
of an ideal gas. We know the molar volume to be 24.790 dm^3 at this pressure and
temperature, so
R=
p
(V
n
)
T
=
1 .000(24.790)
298. 15
= 0. 083146
with units of bar dm^3 K−^1 mol−^1. If the volume is expressed in m^3 , thenRis in
bar m^3 K−^1 mol−^1 =100(0.0831)= 8 .310 J K−^1 mol−^1. Remember that the unit
JK−^1 is for a molar gas constant. Notice that the numerator is an energy. The
tabulated value is 8.3144725 J K−^1 mol−^1 (CRC Handbook of Chemistry and Physics,
2008–2009, 89th ed.)
Exercise 1.2 The Maxwell–Boltzmann Distribution
Derive
pV=^13 NAmv^2 x
whereNA,m,andvxare Avogadro’s number, the mass, and the averagex-component
of the velocity of a collection of ideal gas particles confined to a cubic boxldm on
an edge.