Page 103 - Essentials of physical chemistry

The First Law of Thermodynamics 65

Example
Consider the expansion of 100 mL of N 2 gas at 100 atm in a gas bottle at 308C to 1 atm of pressure
through a nozzle so rapidly as to achieve an adiabatic process. What is the temperature of the
expanded N 2 if the final volume is 10 L?
5
is very close to a diatomic ideal gas, so we use R,
N 2 C V ¼
2
3R 2R R
translation plus rotation and raise the whole equation to the power to find the
2 2 C V

R
C V C V
" #
T 2 R ðÞ V 1
ratio of the temperatures as ¼ to solve for the ratio of temperatures as
T 1 V 2

R R 1R
2:5R
T 2 V 1 C V V 1 C V 0:100 ðÞ

,so we find that T 2 ¼ T 1 ¼ (303:15 K) ¼ 48:046 K.

¼
T 1 V 2 V 2 10:0
This rapid adiabatic expansion is sufficient to cool the nitrogen to below its boiling point of 778K,
so this is a way to make liquid nitrogen. There is a temperature for each gas called the Joule–
Thomson inversion temperature and cooling occurs if the initial temperature is below that tempera-
ture but the gas heats upon expansion if the initial temperature is above the inversion temperature.
At room temperature He is above its inversion temperature and will actually heat up upon
expansion. Although there is also a pressure effect, there are absolute temperatures for this effect.
For He the temperature is 518K, for H 2 2028K, for N 2 6218K, and for O 2 it is 7648K (see discussion
at http:==en.citizendium.org=wiki=Joule-Thomson_effect). Thus, air (N 2 þ O 2 ) can be liquefied by
adiabatic expansion starting from room temperature and 1 atm, but He and H 2 must be precooled to
below their Joule–Thomson inversion temperatures.
DIESEL ENGINE COMPRESSION
C V C V
Now back to the diesel engine. We start from V 1 T R ðÞ ¼ V 2 T R ðÞ and assume air is close to an ideal
1 2
gas, at least initially, and allow for experimental measurement of C V and C P . Although we will see
that C V and C P have to be measured experimentally since it is a mixture of gases (mostly N 2 and O 2 ),
PV
to obtain
nR
air does behave like an ideal gas up to about 100 atm. Thus, we can substitute T ¼
C V C V
P 1 V 1 R ðÞ P 2 V 2 R ðÞ
V 1 ¼ V 2 :
nR nR

R
This equation can be rearranged and raised to the power to obtain an equation useful
C V
to understand the main principle of the diesel engine. Please note that we will define a new

C P
and use the ideal gas relationship that C P ¼ C V þ R and
number as g ¼
C V
C V C V R C V þ R C P
.
þ 1 ¼
¼
R R R R R
¼
þ
Thus,
C V C V C V C V
R
R
V R ð þ1Þ ðÞ ¼ V R ð þ1Þ ðÞ ,
P
P
1 1 2 2

R

R C P C V C P C V C V
R
R
P
P
so raise the whole equation to the power. Finally we reach V 1 R ðÞ ðÞ ¼ V 2 R ðÞ ðÞ
2
1
C V

C P C P
g
g
and P 1 V C V ¼ P 2 V C V or P 1 V ¼ P 2 V .
1 2 1 2