2.1 The Ideal Gas Law
Problem 1 – Assume we have a quantity of 10 grams of 11.1% nitrated nitrocellulose
(C6H8N2O9) and it is heated to a temperature of 1000K and changes to gas somehow without
changing chemical composition. If the process takes place in an expulsion cup with a volume of
10 in3, assuming ideal gas behavior, what will the final pressure be in psi?
lbf
lbf
This problem is again straight-forward except for those pesky units – but we’ve done this before.
We start with equation (VW-2)
( )
RTmcbpg
=−V
(VW-2)
Rearranging, we have
cb
RTm
pg
−
=V
Here we go
( )
( ) ( ) ( ) ( )
( )
( )
( ) ( )
−
−
=
lbm
in
0.32
kg
lbm
2.2
g
kg
1000
1
g10in10
K1000
ft
in
12
kJ
lbfft
6.737
kg
kgmol
252
1
Kkgmol
kJ
314.8
g
kg
1000
1
g10
3
3
ONHC 9286
p
=2
in
lbf
2.314p
So you can see that the real gas behavior is somewhat different than ideal gas behavior at this
low pressure – it makes more of a difference at the greater pressures.
Again please note that this result is unlikely to happen. If the chemical composition were reacted
we would have to balance the reaction equation and would again have to use Dalton’s law for the
=2
OH in
lbf
257,1
2
p
( ) ( ) ( )
( )
( )
( )
( ) ( )
−
−
=
in
ft
12
1
lbfft
kJ
6.737
1
lbm
in
0.32
kg
lbm
2.2
g
kg
1000
1
g10in10
g
kg
000,1
1
g10
kg
kgmol
252
1
K1000
K–kgmol
kJ
314.8
kgmol
kgmol
5
3
3
ONHC
ONHC
ONHC
ONHC
ONHC
ONHC
CO
CO
9286
9286
9286
9286
9286
9286
p
=2
CO in
lbf
571,1p
( ) ( ) ( )
( )
( )
( )
( ) ( )
−
−
=
in
ft
12
1
lbfft
kJ
6.737
1
lbm
in
0.32
kg
lbm
2.2
g
kg
1000
1
g10in10
g
kg
000,1
1
g10
kg
kgmol
252
1
K1000
K–kgmol
kJ
314.8
kgmol
kgmol
1
3
3
ONHC
ONHC
ONHC
ONHC
ONHC
ONHC
N
N
9286
9286
9286
9286
9286
9286
2
2
p
=2
Nin
lbf
314
2
p
=
+
+
=2222 in
lbf
142,3
in
lbf
314
in
lbf
571,1
in
lbf
257,1p
Problem 3 – A hypothetical “air mortar” is to be made out of a tennis ball can using a tennis ball
as the projectile. The can has a 2-1/2” inside diameter and is 8” long. If a tennis ball of the same
diameter weighs 2 oz. and initially rests against the rear of the can, to what air pressure must one
pressurize the can to in order to achieve a 30 ft/s launch of the tennis ball? Assume that the
tennis ball can be held against this pressure until released, that it perfectly obturates and also
assume an isentropic process and ideal gas behavior with
= 1.4 for air.
4
( )
33
3
3
sphere in181.8in5.2
66
V===
d
2
V
UV sphere
0−=
( )
( )
333
0in045.2in
2
181.8
in136.6V =−=
( )
( )
in75.6in
2
5.2
in8=−=L
The formula for muzzle velocity given an isentropic expansion of air was provided in the notes
as equation (IG-28)
( )
( ) ( )
( ) ( )
−
−
−
−+
−
=1
1
1
1
2lLl
m
lRTm
V
p
ig
m
(IG-28)
If we assume ideal gad behavior we recognize that the term mgRTi is the initial pressure times the
initial chamber volume thus we can write
igi RTmp =
0
V
Making this substitution we have
( )
( ) ( )
( ) ( )
−
−
−
−+
−
=1
1
1
0
1
V2 lLl
m
lp
V
p
i
m
2
2m
c
pV
g
W
E=
lbfft
−
2.5 Combustion
Problem 6 – Calculate the A-F ratio for the combustion of the following fuels. Calculate the ratio
with both theoretical air and 10% excess air.
a.) Benzene – C6H6
b.) n-Butane – C4H10
c.) Ethyl Alcohol – C2H5OH
Solution:
A-F =
12.58
33.896
= 15.42
A-F =
12.58
)1.1)(33.896(
= 16.96
Ethyl Alchohol
07.46
69.413
With 10% Excess Air:
07.46
)1.1)(69.413(
Problem 7 – Let us examine a pressure vessel identical to the example problem in the text
containing 0.001 kg of methane (CH4) and 0.002 kg of air. The enthalpy of formation for the
methane is -74,850 kJ/kgmol and its molecular weight is 16.04 kg/kgmol. The reaction will
begin at 298 K and we shall remove enough heat from the vessel that the final temperature
becomes 1,000 K.
a.) determine much heat is given off.
b.) compare the result in a.) above with the example problem in this chapter.
( )
( )
4
4
4CH
5
CH
CH kgmol1023.6
kgmol
kg
04.16
kg001.0 −
=
=N
For the air we have
( )
( )
air
5
air
air kgmol1090.6
kgmol
kg
97.28
kg002.0 −
=
=N
Our balanced reaction is then
( ) ( )
( ) ( )( )
( ) ( ) ( )
( )
( )
2
5–5–
2
5–
2
5–
22
-5
4
-5
N1045.5sC1023.6H1056.9OH109.2
N79.0O0.21106.90CH106.23
+++
→++
44 CH298
CH −=−+ →TRhhN uTf
For the oxygen and nitrogen we have
( )
( )
( )
( ) ( )
−+=−+ −
→K298
Kkgmol
kJ
314.800kgmol1090.621.0 5
O298
0
O22 TRhhN uTf
( )
kJ50.6
OH298
0
OH 22 −=−+ →TRhhN uTf
( )
( )
( ) ( )
−+=−+ −
→K000,1
Kkgmol
kJ
314.8664,200kgmol1056.9 5
OH298
0
H22 TRhhN uTf
( )
kJ18.1
OH298
0
H22 +=−+ →TRhhN uTf
( )
( )
( )
( ) ( )
−
+=−+ −
→K000,1
Kkgmol
kJ
314.8
kgmol
kJ
468,210kgmol1090.679.0 5
N298
0
N22 TRhhN uTf
( )
kJ72.0
22 N298
0
N=−+ →TRhhN uTf
( ) ( )
( ) ( )
−
+=−+ −
→K000,1
Kkgmol
kJ
314.8
kgmol
kJ
795,110kgmol1023.6 5
C298
0
CTRhhN uTf
( )
kJ217.0
C298
0
C+=−+ →TRhhN uTf
The enthalpies of the products are then given by
( )
kJ38.4kJ217.0kJ72.0kJ18.1kJ50.6 −=+++−=−
produprod
i
iTRhN
The heat given off by the reaction is then calculated as
( )
( )
kJ606.0kJ986.4kJ38.4 +=−−−=Q
The example in the text generated a heat output of 3.37 kJ so we have actually removed
2.764 kJ more energy by reducing the temperature of the products.
Problem 8 – A really interesting person takes the tennis ball mortar we built in problem 3 and
modifies it – squirting in and igniting 0.003 oz. of acetylene gas (C2H2(g)). If we assume the
combustion kinetics are fast enough such that the energy release occurs before the ball can move
we want to determine the muzzle velocity of the tennis ball. Proceed along the following steps:
a.) Balance the stoichiometric reaction equation for acetylene.
b.) Balance the actual equation neglecting the volume the acetylene occupies in the
chamber. Assume the air initially in the chamber is at 14.7 psia and 77 ºF.
c.) Determine the increase in internal energy of the gas as we have done in class
d.) Assuming the gas is calorically perfect (U = mgcvT) and that cv = 0.33 BTU/lbm-
ºR for the mixture, determine the increase in temperature of the gas.
NOTE: you will have to do c.) and d.) by iteration, first assuming a final reaction temperature,
carrying out the calculation for u and seeing if the T you get matches – if not iterate again –
once you get an answer within say 10% that is good enough.
e.) Based on the result of d.) above, determine the initial pressure on the tennis ball
assuming the specific gas constant of the products is R = 80 ft-lbf/lbm-ºR.
f.) Use the result of e.) and possibly your results from problem 2.) to determine the
muzzle velocity of the tennis ball. Assume
= 1.4
g.) Determine the temperature of the gases at shot exit.
( )
( )
lbmol10201.7
lbmol
lbm
038.26
1
lbm0001875.0 6−
=
==
fuel
fuel
fuel n
m
N
( )
( )
lbmol10020.3
lbmol
lbm
97.28
1
lbm0000875.0 6−
=
==
air
air
air n
m
N
( ) ( ) ( )
−
→
−
lbmol
lbfin
9.337,9
ft
in
12
BTU
lbfft
16.778
lbmol
BTU xx
For the products we have
Product Enthalpy of formation (BTU/lbmol) Enthalpy of formation (in-lbf/lbmol)
H2O (g) -104,040 -971,515,116
N2 0 0
C(s) 0 0
H2 0 0
Since the reactants will be invariant here let’s deal with them first.
We calculate UR first
()()()
TRhhNTRhhNTRhhNU u
f
u
f
u
f
R−++−++−+=
2222 NOHC
()()()()
TRhhNTRhhNTRhhNTRhhNU u
f
u
f
u
f
u
f
p−++−++−++−+=
CNHOH 222
( )
( )
( )
( )
( )
( )
( )
( )
−
−++
−
−++
−
−++
−
−+−=
−
−
−
−
lbmol
lbfin
000,080,3700lbmol10440.1
lbmol
lbfin
000,080,37752,887,1000lbmol10385.2
lbmol
lbfin
000,080,37906,526,960lbmol10932.5
lbmol
lbfin
000,080,37854,102,123116,515,971lbmol10269.1
5
6
6
6
p
U
Comparing this to the expression given in the problem statement
TcmU vg =
( )
( ) ( ) ( ) ( )
R5372000
ft
in
12
BTU
lbfft
16.778
Rlbm
BTU
33.0lbm0001875.00000875.0 −
−
+=U
lbfin240,1 −=U
Based on this let’s try a final temperature of 4000 ºR.
( ) ( )
( )
−
=
−
−
=lbmol
lbfin
000,160,74
ft
in
12R000,4
Rlbmol
lbfft
1545TRu
Product Enthalpy at temperature (BTU/lbmol) Enthalpy at temperature (in-lbf/lbmol)
H2O (g) 36,251 338,508,213
N2 27,587 257,604,647
C(s) 0 0
H2 26,071 243,448,391
We calculate Up in at 4000 ºR as
( )
( )
( )
( )
( )
( )
( )
( )
−
−++
−
−++
−
−++
−
−+−=
−
−
−
−
lbmol
lbfin
000,160,7400lbmol10440.1
lbmol
lbfin
000,160,74647,604,2570lbmol10385.2
lbmol
lbfin
000,160,74391,448,2430lbmol10932.5
lbmol
lbfin
000,160,74213,508,338116,515,971lbmol10269.1
5
6
6
6
p
U
( )
( ) ( ) ( ) ( )
R5374000
ft
in
12
BTU
lbfft
16.778
Rlbm
BTU
33.0lbm0001875.00000875.0 −
−
+=U
lbfin934,2 −=U
Now let’s try a final temperature of 7000 ºR.
( ) ( )
( )
−
=
−
−
=lbmol
lbfin
000,780,129
ft
in
12R000,7
Rlbmol
lbfft
1545TRu
Product Enthalpy at temperature (BTU/lbmol) Enthalpy at temperature (in-lbf/lbmol)
H2O (g) 76,146 711,043,733
N2 54,109 491,901,896
C(s) 0 0
H2 52,678 505,264,431
We calculate Up in at 7000 ºR as
( )
( )
( )
( )
( )
( )
( )
( )
−
−++
−
−++
−
−++
−
−+−=
−
−
−
−
lbmol
lbfin
000,780,12900lbmol10440.1
lbmol
lbfin
000,780,129896,901,4910lbmol10385.2
lbmol
lbfin
000,780,129431,264,5050lbmol10932.5
lbmol
lbfin
000,780,129733,043,711116,515,971lbmol10269.1
5
6
6
6
p
U
lbfin6.679 −=
p
U
Then
PR UUU −=
( )
( )
lbfin773,5lbfin6.679lbfin453,6 −=−−−=U
Comparing this to the expression given in the problem statement
TcmU vg =
( )
( ) ( ) ( ) ( )
R5377000
ft
in
12
BTU
lbfft
16.778
Rlbm
BTU
33.0lbm0001875.00000875.0 −
−
+=U
lbfin477,5 −=U
This answer is close to 5% so we’re good to go. The temperature is now 7,000 ºR – pretty hot! –
that’s why acetylene works so good as a cutting torch fuel. Now the initial pressure on the
tennis ball comes through the ideal gas equation of state.
RTpi
=
( )
( )
( ) ( )
( )
−
−+
=ft
in
12R000,7
Rlbm
lbfft
80
in045.2
lbm0001875.00000875.0
3
i
p
=2
in
lbf
5.903
i
p
The muzzle velocity of the tennis ball is then
( )
( ) ( )
( ) ( )
−
−
−
−+
−
=1
1
1
0
1
V2 lLl
m
lp
V
p
i
m
( )( ) ( )
( )
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
−
−
−+
−
=−−
−
2
0.4–
4.114.11
0.4
14.1
3
2
slbf
ftlbm
2.32in417.075.6417.0
4.11
oz
lbm
16
1
oz2
ft
in
12
in417.0in045.2
in
lbf
5.9032
m
V
=s
ft
7.64
m
V
For the temperature of the gases when muzzle exit occurs we can use equation (IG-19)
( )
1
V
V−
=
c
i
TT
(IG-19)
between the point of obturation and the end of the exhaust and assume a “smeared” specific heat
ratio of 1.3 for the product gases. Also assume the combustion begins at 500 K and completes at
1,500 K. Assume the total enthalpy at 500 K for n-Heptane (C7H16) is -120,000 kJ/kgmol.
( )
( )
( )
( )
( )
( )
++
++
−=
−
−−
kgmol
kJ
920,50kgmol1036.41
kgmol
kJ
097,60kgmol1011
kgmol
kJ
000,120kgmol101
5
55
R
H