Page 104 - Essentials of physical chemistry

66 Essentials of Physical Chemistry

This formula is the preferred formula to calculate=estimate the temperature of a gas under
adiabatic compression and we will become familiar with it after working out some problems.
g g
Suffice it to say that when you hear the word ‘‘diesel’’ you should think of P 1 V ¼ P 2 V .
1 2
Yes, we have made some approximations and will need to make a few more, mainly in the
assumption that air remains an ideal gas up to about 100 atm, but this method is sufficiently accurate
to show that the diesel compression can raise the temperature above the flash point (FP) of most
fuels. The FP of a liquid is the minimum temperature at which the vapor of the liquid will form an
ignitable mixture with air and for cetane=hexadecane, the CRC Handbook [8] reports this as 1368C,
well below a temperature easily reached in diesel engines.

Example
Calculate the temperature of air compressed adiabatically in a one-cylinder diesel engine from 1040
cc at 258C to 40 cc. Given that C V ¼ (5=2)R, compute Q. W, DH, and DU for this compression.

5
R þ R
C V þ R 2 7
g g C P
1
2
‘‘Adiabatic þ diesel’’ means P 1 V ¼ P 2 V . g ¼ ¼ ¼ ¼ ¼ 1:40.
C V C V 5 5
R
2
Assume air is an ideal gas.
P 1 ¼ 1 atm P 2 ¼ ?
V 1 ¼ 1040 cc Q ¼ 0 ! V 2 ¼ 40 cc
T 1 ¼ 298:15 K T 2 ¼ ?

g 1:4

V 1 1040 cc P 1 V 1 P 2 V 2 P 2 V 2
P 2 ¼ P 1 ¼ (1 atm) ¼ 95:71 atm, then ¼ and T 2 ¼ T 1 ¼
V 2 40 cc T 1 T 2 P 1 V 1
(95:71 atm)(40 cc)
(298:15 K) ¼ 1097:54 K, which is very hot, DT ¼ 799:39 K.

(1 atm)(1040 cc)
Then, even though we assumed an imaginary ‘‘reversible’’ path, the
DU ¼ U after U before ¼ nC V (DT), so we need to calculate n, the number of moles.
1040 cc
ffi 0:04251 mol, so then

n ¼
298:15 K

(22, 414 cc=mol)
273:15 K

DU ¼ nC V DT ¼ (0:04251 mol)(2:5)(1:987 cal= K mol)(799:39 K) ¼þ168:81 cal:

Similar reasoning can be used for DH ¼ nC P DT, which would be the same as for DU except

C P
for ¼ g so we use the same reasoning with (after–before) and this implies that
C V
DH ¼ g(DU) ¼ (1:4)(168:81)cal ¼þ236:33 cal. This calculation of DH is mysterious because
we have not used any detailed information about the process, but we get the answer from DU using
the powerful (after–before) principle since H is a state variable. Now what about W?We assumed
Q ¼ 0, so by the first law we have DU ¼ 0 þ W,sothatmeans that W ¼þ168:81 cal.Notethat
W is positive since we certainly did work ON the air. An added comment is that the effective

1040 cc
compression ratio of this engine is ¼ 26. This is a bit higher than most diesel engines,
40 cc
which tend to have compression ratios near 22:1, but the temperature will certainly be high enough
to ignite most fuels for any compression ratio greater than about 16:1.