CHE 351 HW#5
a) Since the tank is nonconducting, Q=0; and since it is rigid, W=0. Thus,
U =Q+W=0
But,
b)
U =C v T
P1 V 1 P 2 V 2
=
T1
T2
. Thus, there is no change in temperature
6
or
( 43 ) = P ( 4 )
2
373
373
P2=2
A reversible process to bring the sy
CHE 351 Thermodynamics Sample Test Problems
1. And ideal gas at 298 K and 2 bars is compressed in such a way that its final pressure and
temperature are 20 bar and 350 K.
a) Calculate the change in internal energy if C P=5/2R.
b) Design a reversible traje
CHE 435 F 2016
Quiz 1
NAME:_
1. Convert the differential equation below to Laplace domain and write the resulting equation in
partial fraction expansion format. Assume that all initial conditions are 0 and u(t) is a unit step
change at time 0. One of the
CHE 351 HW#3
U =Q+W
(since Q=0
(
U =W )
H=mCp T =( 20 kg ) 4.18
kJ
( 3020 C )=836 kJ
kg C
)
U = H(PV )
From appendix E, Table F.1 we get V of water at 20 and 30 celsius.
V = 1.002 cm3/g (1.002 x 10-3 Kg/m3) at 20 Celsius
V = 1.004 cm3/g (1.004 x 10-3
CHE 351 HW#4
=
Molar volume and molar density are relates as
1 V
=
V P T
( )
1
P
( )
( 1 )
P
=
T
(1 )( P )
2
T
( )
=
=
T
1 V
V T
( ) =
1
T
d=
P
( )
( )
( 1 )
T
=
P
(
( ) T )
1
2
P
P
( T ) dT +( P ) dP=dT + dP
P
T
d
=dT +dP
2
ln
=(P2P 1)
For constant T