Solution

advertisement
CHEM 241 (Dr. Ehteram A Noor)
Solutions for Chapter (I) Homework
(I-1) A bulldozer pushes 500 kg of dirt 100 m with a force of 1500 N. It
then lifts the dirt 3 m up to put it in a dump truck. How much work
did it do in each situation?
Solution
F 1 1500 N
L1  100 m
m  500 kg
L2  3 m
W1  ?
W2  ?
(1) W 1 F1  L1
W1  1500  100  150 kJ
(2) W2  F1  L2
W2  m  a  L2
Since a  9.8 m s -2
W2  500  9.8  3  14.7 kJ
____________________________________________________________________
(I-6) A quantity of air is taken from state a to b
along a path that is a straight line in the Pdiagram (As shown from Fig. (II-12)).
1
If Va  0.0700m 3 , Vb  0.1100m 3 ,
2
5
Pa  1.00 105 Pa , Pb  1.40  10 Pa ,
what is the work W done by the gas in
this process? Assume that the gas may be treated as ideal gas .
Solution
W=- [Area of triangle (1) + area of quadrilateral (2)]
V
1
( Pb  Pa )(Vb  Va )
2
Area of quadrilate ral  Pa (Vb  Va )
1
W  [ ( Pb  Pa )(Vb  Va )  Pa (Vb  Va )]
2
1
1
W  (Vb  Va )( Pb  P a  Pa )
2
2
1
W   (Vb  Va )( Pb  Pa )   Area of trapezoid
2
The negative signe of work indicates that the work is done by the system
Area of triangle 
Trapezoid (‫)شبه منحرف‬
W  1 / 2  105  (0.11  0.07)(1  1.4)
W  4.8kJ
____________________________________________________________________
Solutions for Chapter (II) Homework
(II-1)
When a quantity of monoatomic ideal gas expands at constant
pressure of 4.0  10 4 Pa , the volume of the gas increases from
2.0  10 3 m 3 to 8.0  10 1 m 3 . What is the work done by the gas?
Solution
Pext  4  10 4 Pa
V1  2  10  3 m 3
V2  8  10 1 m3
W   PV
W  4.0  10 4 (8  10 1  2.0  10 3 )
W  31.92 kJ
____________________________________________________________________
(II-2) One mole of an ideal gas at 300 K is allowed to expand
isothermally from a volume of 1.00 L to a volume of 5L. Find the
work done by the gas and the heat energy absorbed from the
surroundings.
Solution
T  cons tan t  300 K
V1  1L
V2  5L
W  nRT ln
W ?
q?
V2
V1
W  1  8.314  300  ln
5
 4.014 kJ
1
E  0
E  q  W  q  W
q  (4.014)  4.014 kJ
____________________________________________________________________
(II-8) Calculate q, W , E and H if 1.00 mole of an ideal gas with
3
C V  R undergoes a reversible adiabatic expansion from an
2
3
initial volume Vi  5.25m 3 to final volume V f  25.5m . The
initial temperature is 300 K.
Solution
Since the process is adiabatic:
Vi  5.25 m3
q0
V f  25.5 m3
H  0
Ti  300 K
E  W
W  nCV (T2  T1 )
From the adiabatic relations:
TiVi  1  T f V f 1
 1 
R
2
 R
 0.67
Cv
3R
300  5.250.67  T f  25.50.67
T f  104.05 K
3
 8.314  (104.05  300)
2
W  2.444kJ
E  2.444kJ
W  1
____________________________________________________________________
(II-9) One mole of He in state defined by Ti  300 K and Pi  23.0 atm
undergoes an isothermal reversible expansion until Pf  2.5 atm .
Calculate W assuming the gas is described by the ideal gas law.
W  nRT ln
P1
P2
W  1  8.314  300 ln
23
 55.54 kJ
2.5
____________________________________________________________________
(II-13) A-A body of air which occupies 1000 cm3 at 30 oC and 76 cmHg,
expands isothermally and reversibly to a volume 2000 cm3. What
is (a) the final pressure? (b) the work done? (c) the heat absorbed?
Solution
From Boyl , s law :
P1V1  P2V2
PV
P2  1 1
V2
76
P1 
 1 atm
V1  1L
76
1 1
P2 
 0.5 atm
2
V
V
W  nRT ln 2   P1V1 ln 2
V1
V1
W  1 1 ln 2  0.693 L atm
L atm  101.3 J
W  70.20 J
For isothermal process :
E  0
q  -W
q  70.20 J
V2  2 L
Download
Random flashcards
Radiobiology

39 Cards

Radioactivity

30 Cards

Nomads

17 Cards

Create flashcards