phy1020.DVI

Now Vis the total mass of gas, which we’ll callm,sowehave

vsndD

r



NkBT

m

: (11.9)

The mass per atom (or molecule) ismaDm=N,sowehave

vsndD

s



kBT

ma

D

r



RT

M

; (11.10)

whereRDkBNAD8:3144621J mol^1 K^1 is the molar gas constant andNAis Avogadro’s number, and
MDm=.N=NA/is the molar mass of the gas (kilograms per mole). Since the molecular (or atomic) weight
is in grams per mole, this mean thatMis just the molecular (or atomic) weight divided by 1000.
Using Eq. (11.10), we can see where the empirical relation for the speed of sound in air, Eq. (11.2), comes
from. Air consists of about 78% nitrogen (N 2 ), 21% oxygen (O 2 ), and 1% argon (Ar). Since the gases are
mostly diatomic, we will take D1:40. To find the mass per molecule, we’ll compute a weighted average
based on composition. Since N 2 has a molecular weight of 28, O 2 has a molecular weight of 32, and Ar has
an atomic weight of 40, we compute the weighted average molecular weight of air to be

maD.0:7828/C.0:2132/C.0:0140/D28:96: (11.11)

To convert this to mass in kilograms, we multiply this by the atomic mass unituD1:660538921 10 ^27 kg
to getmaD4:8089 10 ^26 kg. Substituting these results into Eq. (11.10), we get

vsndD

s



kBT

ma

(11.12)

D

s

.1:40/.1:3806488 10 ^23 J=kg/T

4:8089 10 ^26 kg

(11.13)

D20:0472

p

T (11.14)

in SI units. NowTis the absolute temperature, and let’s letTcbe the temperature in degrees Celsius. Since
the two are related byTDTcC273:15, we have

vsndD20:0472

p

TcC273:15 (11.15)

D.20:0472/

p

273:15

r

Tc

273:15

C 1 (11.16)

D331:32

r

1 C

Tc

273:15

(11.17)

We now use the series expansion (valid forjxj<1; see Appendix D)

.1Cx/1=2D 1 C

1

2

x

1

8

x^2 C

1

16

x^3 

5

128

x^4 C

7

256

x^5  (11.18)

 1 C

1

2

x (11.19)

and we have

vsnd331:32



1 C

1

2

Tc

273:15



(11.20)

D331:32C0:6065Tc (11.21)

 331 C0:60Tc (11.22)

and we have just derived the empirical relation, Eq. (11.2).