13-72
13-96 A mixture of gases is placed in a spring-loaded piston-cylinder device. The device is now heated until the pressure
rises to a specified value. The total work and heat transfer for this process are to be determined.
Properties The molar masses of Ne, O2, and N2 are 20.18, 32.0, 28.0 kg/kmol, respectively and the gas constants are
0.4119, 0.2598, and 0.2968 kJ/kgK, respectively (Table A-1). The constant-volume specific heats are 0.6179, 0.658, and
0.743 kJ/kgK, respectively (Table A-2a).
Analysis The total pressure is 200 kPa and the partial pressures are
PNe  y Ne Pm  (0.25)(200 kPa)  50 kPa
PO2  y O2 Pm  (0.50)(200 kPa)  100 kPa
PN2  y N2 Pm  (0.25)(200 kPa)  50 kPa
25% Ne
50% O2
25% N2
(by pressure)
0.1 m3
10C, 200 kPa
The mass of each constituent for a volume of 0.1 m3 and a temperature of 10C are
m Ne 
P NeV m
(50 kPa)(0.1 m 3 )

 0.04289 kg
R NeT
(0.4119 kPa  m 3 /kg  K)(283 K)
mO2 
P O2V m
(100 kPa)(0.1 m 3 )

 0.1360 kg
R O2T
(0.2598 kPa  m 3 /kg  K)(283 K)
P N2V m
(50 kPa)(0.1 m 3 )

 0.05953 kg
R N2T
(0.2968 kPa  m 3 /kg  K)(283 K)
 0.04289  0.1360  0.05953  0.2384 kg
Q
m N2 
m total
P
(kPa)
The mass fractions are
m
0.04289 kg
 0.1799
mf Ne  Ne 
mm
0.2384 kg
mf O2
m
0.1360 kg
 O2 
 0.5705
m m 0.2384 kg
mf N2
m
0.05953 kg
 N2 
 0.2497
mm
0.2384 kg
2
500
200
1
0.1
V (m3)
The constant-volume specific heat of the mixture is determined from
cv  mf Ne cv , Ne  mf O2 cv ,O2  mf N2 cv , N2
 0.1799  0.6179  0.5705  0.658  0.2497  0.743  0.672 kJ/kg  K
The moles are
N Ne 
m Ne
0.04289 kg

 0.002126 kmol
M Ne 20.18 kg/kmol
N O2 
m O2
0.1360 kg

 0.00425 kmol
M O2 32 kg/kmol
N N2 
m N2
0.05953 kg

 0.002126 kmol
M N2 28 kg/kmol
N m  N Ne  N O2  N N2  0.008502 kmol
Then the apparent molecular weight of the mixture becomes
Mm 
mm
0.2384 kg

 28.04 kg/kmol
N m 0.008502 kmol
The apparent gas constant of the mixture is
R
8.314 kJ/kmol  K
R u 
 0.2964 kJ/kg  K
Mm
28.05 kg/kmol
The mass contained in the system is
m
P 1V1
(200 kPa)(0.1 m 3 )

 0.2384 kg
RT1
(0.2964 kPa  m 3 /kg  K)(283 K)
Noting that the pressure changes linearly with volume, the final volume is determined by linear interpolation to be
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
13-73
500  200 V 2  0.1


V 2  0.4375 m 3
1000  200 1.0  0.1
The final temperature is
T2 
P 2V 2
(500 kPa)(0.4375 m 3 )

 3096 K
mR
(0.2384 kg)(0.2964 kPa  m 3 /kg  K)
The work done during this process is
P P
(500  200) kPa
(0.4375  0.1) m 3  118 kJ
Wout  1 2 (V 2 V1 ) 
2
2
An energy balance on the system gives
Qin  Wout  mcv (T2  T1 )  118  (0.2384 kg)(0.672 kJ/kg  K )(3096  283) K  569 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.