Solutions to Problems for Examination I
1. Calculate the potential energy of a 50.00 kg object 100.0 meters above the ground (assuming
the ground is the zero potential energy level) [49.03]
Potential Energy = m g h
= (50.00 kg) (9.080665 m/sec 2 ) (100.0 m) = 49033. kg-m 2 / sec 2
= 49.033 X 10 3 Joules = 49.03 kJ
2. During hurricane Irene, the pressure in the eye of the storm fell to 968. millibar. Convert this to
Torr, Pascals and Atmospheres. [0.955, 9.68 X 104 , 726.]
968. millibar = 0.968 bar
( 0.968 bar ) (100 kPa/bar) = 96.8 kPa = 9.68 X 10 4
Pascals
1 atm
!
$
#"
& = 0.95534 = 0.955 atm
101.325 kPa %
( 0.95534 atm ) ( 760 Torr/atm ) 726.06 = 726. Torr
( 96.8 kPa )
3. Convert J2/N3 into base units [m2]
Joule = kg-m 2 /sec 2
Newton = kg-m/sec 2
J 2 = kg 2 -m 4 /sec 4
N 2 = kg 2 -m 2 /sec 4
J2
kg 2 -m 4 /sec 4
=
= m2
2
2
2
4
N
kg -m /sec
4. 10.00 grams of oxygen is confined to a 2.500 L container at a pressure of 1200. Torr. What is the
temperature of the gas. [-119.2]
n=
10.00 grams
10.00 grams
=
= 0.312512 moles
2 (15.9994 g/mol) 31.9988 g/mol
P = 1200. Torr (1 atm/760 Torr ) = 1.57895 atm &
(1.57895 atm ) (101.325 kPa/1 atm ) = 159.9868 kPa
V = 2.500 L
T =
PV
399.967
(159.9868 kPa ) ( 2.500 L )
=
=
n R ( 0.312512 moles ) ( 8.31447 kPa-L/mol-K ) 2.59837
T = 153.93 K
(153.93 - 273.15) = -119.22° C
T = 153.9 K = -119.2° C
5. A collection O2, CO2 and H2 gases are in a container. Assuming all the gases behave ideally,
which gas has the highest kinetic energy and which gas has the highest velocity? (explain your
answers) [??]
Since kinetic energy depends only on the Temperature (KE = 3/2 RT), all three gases have the
SAME kinetic energy. Since KE = ½ mv2 , the heaviest gas (CO) moves the slowest and the
lightest gas (H2) moves the fastest
6. Name a observable phenomena that confirms the non-ideality of a gas. (explain your answer).
[class notes?]
Ideal gases have no attractive forces, attractive forces allow gases to form liquids (forming a liquid
is observable)
7. Sketch the variation of the compressibility factor as a function of pressure and fully explain the
shape of the graph. [see the book]
8. 2.500 moles of an ideal gas at a temperature of 300.0 K is initially at a pressure of 100.0 kPa
undergoes the following isothermal transformations:
Process I. State 1  State 2 where P2 is 500.0 kPa while the external pressure is held
constant at 10.00 atm.
Process II. State 2  State 3 where the external pressure is always equal to the internal
pressure and V3 is four times V2
Process III. State 3  State 4 an expansion by a pressure equal to the final pressure in
which 3.125 X 103 kJ of work are done by the gas.
a) Identify P, V, and T for all four states
b) Draw three P versus V graphs to illustrate the three processes
b) Calculate the work in each of the three processes
[The volumes are 12.47, 49.89, 62.36 and 100.0]
Pressure (kPa)
Volume (L)
Temperature (K)
State 1
100.0
62.36
300.0
State 2
500.0
12.42
300.0
State 3
125.0
49.89
300.0
State 4
62.36
100.0
300.0