Chapter
4 The First
Law
of Thermodynamics
4-246 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the other
side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal
equilibrium is established is to be determined for the cases of the piston being fixed and moving freely.
Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the
container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible.
Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is Cv =
0.743 kJ/kg. °C for N2, and R = 2.0769 kPa.m3/kg.K is Cv = 3.1156 kJ/kg. °C for He (Tables A-I and
A-2)
Analysi"
The mass of each gas in the cylinder is
mN2
/~V11
-)
-(500
N2 = (0.2968kPa
= (
kPaXl
.m3/k;
m3
.K,:
R~
,
mHe
=
= 4.77
?53
-(500
He = (2.0769
~
kPa)(I
kPa .m3/kg
kg
K )
m3
.K~ 298 K)
= 0.808
kg
R~
Taking the entire contents of the cylinder as our system, the 1st law relation can be written as
Msystem
cEin -Eou, ,
'--v--'
Netenergytransfer Change
in internal,
kinetic,
by heat,work,andmass potential,
etcenergies
O=tJ.U=(tJ.U)Nl
O=[mCv(T2
+(tJ.U)He
-T1)]Nl
+[mCv(T2
-T])]He
Substituting,
(4. 77kgXo. 743kJ/kg.o C Xr f -80 1 c + (0.808kg){1.1156kJ/kgoO
It
gives
Tr=
C Xr f -251
c = 0
57.2°C
where T f is the final equilibrium
temperature in the cylinder .
The answer would be the same if the piston were not free to move since it would
pressure, and not the specific heats.
effect only
Discussion Using the relation PV = NRu T , it can be shown that the total number of moles in the cylinder
0.170 + 0.202 = 0.372 krnol, and the final pressure is 510.6 kPa.
4-187
is