Concise Physical Chemistry

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.