9.A – Calorimetry: Basics
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. In a closed system what basic fact must always hold true for the energy content of the overall system? Energy Is neither lost nore gained, only transferred between
components of the system
2. What is the difference between heat and temperature, including units and symbol?
Heat (q) measured in J = total KE of particles (meaning dependent on amount of material)
Temperature (T) measured in K = reflects the average KE of particcles (meaning independent of the amount of material as it is averaged out)
3. What property and two conditions determine the phase of a sample of matter is? Property = Equilbirum
Conditions = Temperature and Pressure
9.B – Calorimetry: Temperature Conversions and Calculations
Remember:
°C = K – 273.15
K = °C + 273.15
AND
Sig Figs are by position of least precise decimal as it is +/- order of operation
K, NOT °K !!!!!!!!!!!
Instructions: ON A SEPARATE SHEET, convert °C to K or vice versa. . Remember, Sig Figs, NW = NC, Boxed Answers and N3 (this includes labeling and in your set-up!!!)
3. 30.02 K – 273.15 = -243.13 °C
5. 240 K – 273.15 = -30°C
1. 153.2°C + 273.15 = 426.4 K
7. 1234°C + 273.15 = 1507 K
2. -33.0°C + 273.15 = 240.1 K
4. -95.9400 °C + 273.15 = 177.21 K
6. 235.23 K – 273.15 = -37.92°C
8. 924.9 K – 273.15 = -651.8 °C
Instructions: ON A SEPARATE SHEET, calculate the change in temperature ( ΔT ) from the following data. Remember, Sig Figs, NW = NC, Boxed Answers and N3.
Remember:
ΔT = Tfinal - Tinitial
9. 153.2°C to 12.4°C
11. -95.9400°C to -104.2°C
-104.2 °C – 95.9400°C = - 8.3 °C
12.4 °C – 153.2°C = -140.8 °C
10. -33.0°C to 49.0°C
49.0 °C – ( – 33.0°C) = 82.0 °C
12. 1234°C to 49.04°C
49.04 °C – 1234°C = - 1185 °C
9.C – Calorimetry: Phase Diagrams
Instructions: Answer the question 1 – 9 in relation to the following phase diagram. Remember N3
1. Which letter denotes the solid phase?
A_
2. Which letter denotes the liquid phase?
C_
3. Which letter denotes the gas phase?
B_
4. Which letter denotes the triple point?
d_
In your own words, what is the definition of a triple point? The
condition of equlibrium between solid, liquid and gas due to
pressure and temperature conditions
5. What is the melting point at 1 atm of pressure?
60°C _
6. What is the boiling point at 1 atm of pressure?
100°C _
7. Above what temperature is it impossible to liquefy this substance, no
matter what the pressure?
110°C _
8. At what temperature and pressure do all three phases coexist?
@ Triple Point =
45°C
0.5 atm _
9. At a constant temperature, what would you do to cause this
substance to change from the liquid phase to the solid phase? ?
INCREASE in pressure
Instructions: Refer to the phase diagram below when answering the questions 10 – 25 NOTE: “Normal” refers to 1.0 atm and remember N3
Error = +/- 10° / OR +/- 0.02 atm
10. What are the values for temperature and pressure at STP? T= 0°C _P= 1.00 atm _
11. What is the normal freezing point of this substance?
100°C _
normal boiling point of this substance?
340°C _
normal melting point of this substance?
100°C _
12. What is the phase (s, l, g) of a substance at 2.0 atm and 100 °C?
s_
13. What is the phase (s, l, g) of a substance at 0.75 atm and 100 °C?
l_
14. What is the phase (s, l, g) of a substance at 0.5 atm and 100 °C?
g_
15. What is the phase (s, l, g) of a substance at 1.5 atm and 50 °C?
s_
16. What is the phase (s, l, g) of a substance at 1.5 atm and 200 °C?
l_
17. What is the phase (s, l, g) of a substance at 1.5 atm and 800 °C?
g_
18. What is the condition of the triple point of this substance? T= 180°C _P= 0.70 atm _
19. A sample was at an initial pressure of 1.25 atm and a temperature of 300 0 C and was lowered
to a pressure of 0.25 atm, what phase transition(s) would occur? boiling _
l→g
20. A sample was at an initial pressure of 0.5 atm and a temperature of 2000 C was lowered to a
temperature of -2000 C, what phase transition(s) would occur? desublimation _
g→s
21. If this substance was at a pressure of 2.0 atm, at what temperature would it melt?
180°C _
boil?
Never _
22. If this substance was at a pressure of 0.75 atm, at what temperature would it melt?
90°C _
boil?
180°C _
23. At what temperature do the gas and liquid phases become indistinguishable from each other? 80°C _
24. At what pressure would it be possible to find this substance in the gas, liquid, and solid phase? 0.70 atm_
25. If I had a quantity of this substance at a pressure of 1.00 atm and a temperature of -1000 C, what phase change(s) would occur if I increased the temperature to 6000 C?
At what temperature(s) would they occur? (NOTE: multiple answers needed for this question)
Melt @ 100°C
Boil @ 340°C _
s→l
l→g
9.D – Calorimetry: Calometric Calculations
1. A sample of mercury is heated from 25.5 ˚C to 52.5 ˚C. In the process, 3050J of
heat are absorbed. What mass of mercury was in the sample? The specific heat
of mercury is 0. 140 J/g ˚C.
3. A cube of gold with a mass of 192.4 g is heated from 30.0 ˚C to some higher
temperature with the absorption of 921 J of heat. The specific heat of gold is
0.129 J/g ˚C. What is the final temperature of the gold?
Ti = 25.5˚C
Tf = 52.5˚C
ΔT = 27.0˚C
Ti = 30.0˚C
Tf = X ˚C
ΔT = Y ˚C
q
m= Xg
Cp =
q
m = 192.4 g
Cp =
= 3050 J
0.140 J/g•˚C
2. A block of aluminum with a mass of 140g is cooled from 98.4 ˚C to 62.2 ˚C with a
release of 1137J of heat. From this data, calculate the specific heat of aluminum.
= 921 J
0.129 J/g•˚C
4. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0 ˚C to
42.0 ˚C. From this data, what is the specific heat of lead?
Ti = 98.4˚C
Tf = 62.2˚C
ΔT = -36.2˚C
Ti = 12.0˚C
Tf = 42.0 ˚C
ΔT = 30.0 ˚C
q
m = 140 g
Cp =
q
m = 58.3 g
Cp =
= -1137 J
X J/g•˚C
= 226 J
X J/g•˚C
Instructions: ON A SEPARATE SHEET OF PAPER, preform the following calorimetric calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N3 (this includes
labeling and in your set-up!!!)
NOTE: I HAVE PULLED OUT VARIABLES WITH MY WORK TO TRY AND HELP YOU FIND THE ERRORS IN YOUR CALCUATIONS
5. A sample of mercury is heated from 25.5 ˚C to 52.5 ˚C. In the process, 3050J of
heat are absorbed. What mass of mercury was in the sample? The specific heat
of mercury is 0. 140 J/g ˚C.
Ti
= 25.5˚C
Tf = 52.5˚C
ΔT = 27.0˚C
q
= 3050 J
m = Xg
Cp =
Base Equation:
q = m Cp ΔT
Required Work to Show:
Answer: 807 J
Algebraic Rearrangement:
m=
3050 J
0.140 J/g•˚C x 27.0 ˚C
0.140 J/g•˚C
m= _ q _
Cp ΔT
_
6. A block of aluminum with a mass of 140g is cooled from 98.4 ˚C to 62.2 ˚C with a
release of 1137J of heat. From this data, calculate the specific heat of aluminum.
Ti
= 98.4˚C
Tf = 62.2˚C
ΔT = -36.2˚C
q
= -1137 J
m = 140 g
Cp =
Base Equation:
q = m Cp ΔT
Required Work to Show:
Answer: 0.224 J/g•˚C
Algebraic Rearrangement:
Cp =
-1137 J
140 g x -36.2 ˚C
X J/g•˚C
Cp = _ q _
m ΔT
_
7. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0 ˚C to
42.0 ˚C. From this data, what is the specific heat of lead?
Ti
= 12.0˚C
Tf = 42.0 ˚C
ΔT = 30.0 ˚C
q
= 226 J
m = 192.4 g
Cp =
Base Equation:
q = m Cp ΔT
Required Work to Show:
Algebraic Rearrangement:
Cp =
226 J
58.3 g x 30.0 ˚C
Answer: 0.129 J/g•˚C
X J/g•˚C
Cp = _ q _
m ΔT
_
Required Work to Show:
Ti
= 32.2˚C
Tf = 45.0 ˚C
ΔT = 12.8 ˚C
q
= XJ
m = 132 g
Cp =
Base Equation: q = m Cp ΔT
Required Work to Show:
Answer: 650. J
Algebraic Rearrangement:
q = 132 g x 0.385 J/g•˚C x
0.385 J/g•˚C
_
14. If a 38 g sample of water releases 621 J of heat energy and cools to 4.0˚C. What
was the initial temperature of the water?
Ti
= X ˚C
Tf = 4.0 ˚C
ΔT = Y ˚C
q
= 621 J
m = 38 g
Cp = 4.18 J/g•˚C
q = m Cp ΔT
Algebraic Rearrangement:
ΔT = Tf - Ti
Required Work to Show:
ΔT =
Ti
Answer: 7.9˚C
621 J
38 g x 4.18 J/g•˚C
= 4.0˚C + 3.9 ˚C
ΔT = _ q _
m Cp
Ti = Tf + ΔT
_ = 3.9 ˚C
NA
12.8 ˚C
15. If 41 grams of water at 24 ˚C absorbs 2208 J of heat energy, what will be the
final temperature of the water?
9. What is the change in heat energy when 75.0 g of water at 9.8 ˚C is raised to 22.4
˚C and the specific heat of liquid water is 4.18 J/g °C?
Ti
= 9.8 ˚C
Tf = 22.4 ˚C
ΔT = 12.6 ˚C
q
= XJ
m = 75.0 g
Cp =
Base Equation: q = m Cp ΔT
Required Work to Show:
Answer: 3950 J
418.6 J
26.86 g x 39.9 ˚C
Answer: 0.391 J/g•˚C
Base Equation:
8. What is the change in heat energy when 132 g of copper at 32.2 ˚C is raised to
45.0 ˚C? The specific heat of copper is 0.385 J/g °C.
Cp =
Algebraic Rearrangement:
q = 75.0 g x 4.18 J/g•˚C x
Ti
= 24 ˚C
Tf = X ˚C
ΔT = Y ˚C
q
= 2208 J
m = 41 g
Cp =
Base Equation:
q = m Cp ΔT
4.18 J/g•˚C
NA
12.6 ˚C
Algebraic Rearrangement:
ΔT = Tf - Ti
Required Work to Show:
ΔT = _ q _
m Cp
Tf = ΔT + Ti
ΔT =
Tf
Answer: 37 ˚C
4.18 J/g•˚C
2208 J
41 g x 4.18 J/g•˚C
= 24˚C + 13˚C
_ = 13 ˚C
10. What is the change in heat energy when 125 g of water at 10.0 ˚C is raised to
90.0 ˚C
Ti
= 10.0 ˚C
Tf = 90.0 ˚C
ΔT = 80.0 ˚C
q
= XJ
m = 125 g
Cp =
Base Equation: q = m Cp ΔT
Required Work to Show:
Answer: 41800 J
Algebraic Rearrangement:
q = 125 g x 4.18 J/g•˚C x
4.18 J/g•˚C
NA
80.0 ˚C
11. What is the change in heat energy when 64.82 g of aluminum metal at 100.0 ˚C
is cooled to 82.0 ˚C? The specific heat of aluminum is 0.897 J /g ˚C.
Ti
= 100.0 ˚C
Tf = 82.0 ˚C
ΔT = ―18.0 ˚C
q
= XJ
m = 64.82 g
Cp =
Base Equation: q = m Cp ΔT
Required Work to Show:
Answer: ―1050 J
0.897 J/g•˚C
Algebraic Rearrangement:
NA
q = 64.82 g x 0.897 J/g•˚C x ―18.0 ˚C
16. How much heat energy is absorbed by 10 g of silver if it increases in temperature
from 10 ˚C to 310 ˚C? The specific heat of silver is 0.235 J/g °C.
Ti
= 10 ˚C
Tf = 310 ˚C
ΔT = 3.0 x 10 2 ˚C
q
= XJ
m = 10 g
Cp =
0.235 J/g•˚C
Algebraic Rearrangement:
q = 10 g x 0.235 J/g•˚C x
NA
3.0 x 10 2 ˚C
Base Equation: q = m Cp ΔT
Required Work to Show:
Answer: 700 J
17. What is the change in temperature in a 128 g sample of titanium if it absorbs
2808 J of heat energy at a temperature of 2.2 ˚C? The specific heat of titanium is
0.523 J/ g ˚C.
Ti
= 2.2 ˚C
Tf = Not Needed
ΔT = Y ˚C
q
= 2808 J
m = 128 g
Cp = 0.523 J/g•˚C
Base Equation:
12. What is the mass of a sample of iron if that sample lost 2300J of heat energy
when it cooled from 80 ˚C to 30 ˚C? The specific heat of iron is 0.449 J /g ˚C.
Ti
= 80˚C
Tf = 30˚C
ΔT = - 50˚C
q
= - 2300 J
m = Xg
Cp =
Base Equation:
q = m Cp ΔT
Required Work to Show:
Answer: 100 g
Algebraic Rearrangement:
m=
- 2300 J
0.449 J/g•˚C x - 50 ˚C
q = m Cp ΔT
Required Work to Show:
m= _ q _
Cp ΔT
_
without sig figs = 102.44 FYI
13. What is the specific heat of metal if its mass is 26.86 g and it requires 418.6 J of
heat energy to raise its temperature from 27.4 ˚C to 67.3 ˚C?
Ti
= 27.4˚C
Tf = 67.3 ˚C
ΔT = 39.9 ˚C
q
= 418.6 J
m = 26.86 g
Cp =
q = m Cp ΔT
Algebraic Rearrangement:
ΔT =
Answer: 41.9 ˚C
ΔT = _ q _
m Cp
2808 J
128 g x 0.523 J/g•˚C
_ =
0.449 J/g•˚C
20. What is the change in temperature in a 128 g sample of water if it absorbs 2808J
of heat energy at a temperature of 3.21 ˚C?
Ti
= 3.21 ˚C
Tf = Not Needed
ΔT = Y ˚C
q
= 2808 J
m = 128 g
Cp = 4.18 J/g•˚C
Base Equation:
Base Equation:
Algebraic Rearrangement:
X J/g•˚C
Cp = _ q _
m ΔT
q = m Cp ΔT
Required Work to Show:
Answer: 5.25˚C
Algebraic Rearrangement:
ΔT =
2808 J
128 g x 4.18 J/g•˚C
ΔT = _ q _
m Cp
_ =
9.E – Calorimetry: Heating and Cooling Curves
Instructions: Answer the question 1 – 2 using this heating curve for water.
1. Identify the sections where the following phases are found:
a. D E _gas b. A B _solid
c. B D E _liquid
d. B _solid and liquid
e. D _liquid and gas
2. Identify by letter (A-E) in which section the following are found:
a.
B _ Freezing (if cooling)
b E _ Particles farthest apart
c.
D _ Boiling
d. A _ Particle motion is most restricted
e.
B _ Heat of fusion
f. B D _ All areas where energy change is potential only
g.
D _ Heat of vaporization
h. B C D E _ All areas where particles move past each other.
i.
A _ Least kinetic energy
j.
A C E _ All areas where kinetic energy is changing
k
A _ most potential energy
l.
B D _ All areas where phase changes occur
m.
A C E _ All areas in which the heat is making the particles move faster
n.
B D _ All areas in which the heat is breaking the attractions or bonds between particles
o.
B D _ All areas in which the particles are not changing their speed
Heat Removed
Instructions: Given the following BP and MP information, sketch and label heating and cooling curves for the substances under the following changes in temperature AND
phase. BPAg = 2162˚C MPAg = 962˚C
BPPb = 1749˚C MPPb = 328˚C
BPCu = 2927˚C MPCu = 1085˚C
BPN = -196˚C MPN = -210.˚C
3.
Water going from 24˚C to 123 ˚C.
6.
Water going from 10˚C to steam at100˚C
9.
Water going from 130˚C to -30˚C.
4.
Water going from -24˚C to 110˚C.
7.
Pb going from 900˚C to -90˚C.
10. Ice going from 0˚C to water at 10˚C.
5.
Silver going from 1300˚C to 50˚C.
8.
Cu going from 150˚C to 1200˚C.
11. Water at 0˚C going to -50˚C.
9.F – Calorimetry: The Full Picture
Instructions: ON A SEPARATE SHEET OF PAPER, (1) Sketch the heating or cooling curves (BE SURE TO LABEL!!!) associated with each of the following temperature
changes and: (2) Calculate the amount of heat energy lost or gained during each temperature change given the information below right. Remember, Sig Figs, NW = NC,
Boxed Answers and N3 (this includes labeling and in your set-up!!!)
1.
45.0 g of H2O changing in temperature from 110.0C to 85.0C
q1 =
q2 =
m Cl ΔTl
m HVAP
=
=
45.0 g • 4.18 J/g•˚C • -15.0 C
45.0 g • -2260 J/g
= -2820 J
= -102000 J
q3 =
m Cg ΔTg
=
45.0 g • 2.08 J/g•˚C • -10.0 C
qT
= + -936 J
= -106000 J
2.
3.
4.
65.2 g of H2O changing in temperature from -35.0C to 105.0C.
q1 =
q2 =
m Cs ΔTs
m HFUS
=
=
65.2 g • 2.11 J/g•˚C • 35.0 C
65.2 g • 334 J/g
= 4820 J
= 21800 J
q3 =
q4 =
m Cl ΔTl
m HVAP
=
=
65.2 g • 4.18 J/g•˚C • 100.0 C
65.2 g • 2260 J/g
= 27300 J
= 147000 J
q5 =
m Cg ΔTg
=
65.2 g • 2.08 J/g•˚C • 5.0 C
qT
= + 680 J
= 202000 J
15 g of H2O(g) changing in temperature from 100.0C to -75.0C
q1 =
m HVAP
=
15 g • - 2260 J/g
m Cl ΔTl
m HFUS
=
=
15 g • 4.18 J/g•˚C • -100.0 C
15 g • - 334 J/g
=
-6300 J
= -5.0 x 102 J
q4 =
m Cs ΔTs
=
15 g • 2.11 J/g•˚C • -75.0 C
qT
= + -2400 J
= -48,000 J
340.5g of Au changing in temperature from 100.C to 1234C
q1 =
m HFUS
=
340.5 g • 64.4 J/g
7.
m C ΔT
=
340.5 g • 0.129 J/g•˚C • 1134 C
qT
57.2 g of O2 changing in temperature from -400.0C to 150.0C
q1 =
m HVAP
=
57.2 g • 13.8 J/g
q2 =
m HFUS
=
57.2 g • 213 J/g
q3 =
6.
-34000 J
q2 =
q3 =
q2 =
5.
=
m C ΔT
=
57.2 g • 0.918 J/g•˚C • 550.0 C
qT
=
21900 J
= + 49810 J
=
26200 J
=
=
789 J
12200 J
= + 28900 J
=
41900 J
8934 g of H2O changing in temperature from -90.0C to 140.0C
q1 =
q2 =
m Cs ΔTs
m HFUS
=
=
8934 g • 2.11 J/g•˚C • 90.0 C
8934 g • 334 J/g
= 1.70 x 106 J
= 2980000 J
q3 =
q4 =
m Cl ΔTl
m HVAP
=
=
8934 g • 4.18 J/g•˚C • 100.0 C
8934 g • 2260 J/g
= 3730000 J
= 20200000 J
q5 =
m Cg ΔTg
=
8934 g • 2.08 J/g•˚C • 40.0 C
qT
= + 743000 J
= 29400000 J
41.5 g of H2O changing in temperature from 130.0C to H2O(s) at 0.00C
q1 =
q2 =
m Cg ΔTg
m HVAP
=
=
41.5 g • 2.08 J/g•˚C • -30.0 C
41.5 g • - 2260 J/g•˚C
=
=
q3 =
q4 =
m Cl ΔTl
m HFUS
=
=
41.5 g • 4.18 J/g•˚C • -100.0 C
41.5 g • -334 J/g
qT
=
-17300 J
= + -13900 J
= -127600 J
-2590 J
-93800 J
(1700000)
8.
57.2 g of He changing in temperature from -295.0C to 269.5C
q1 =
m HVAP
=
57.2 g • 5.23 J/g
q2 =
m HFUS
=
57.2 g • 20.9 J/g
q3 =
m C ΔT
=
57.2 g • 5.19 J/g•˚C • 564.5 C
qT
=
299 J
= 1.20 x 103 J
(1200)
= + 16800 J
=
18300 J
9.G – Gases: KMT
Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. What theory explains the behavior of gases with respect to conditions such as temperature and pressure? Kinetic Molecular Theory (KMT) _
2. Complete the followign statements in relation to the KMT.
a. Gases consist of large numbers of tiny particles that are far apart relative to their size. This means that indiviual particle volume of the gas I negliabe_ (i.e. a gas is
considered as a whole group rather than as individual interacting particles)
b.Collisions between gas particles and between particles and container walls are elastic collisions. This means that the will leave with the same amount of energy
they collided with (No loss of KE/conversion to PE due to collisions)
c. Gas particles are in constant, rapid, random motion. This can be inferred because that molecules travel in straight l ine pather between the collision at the wall of
their container and collisions with each other
d. There are no forces of attraction or repulsion between gas particles. This means that that no energy is required to separate the particles of a gas as there is no
mutual attraction (No IMF’s)
e.The average kinetic energy of gas particles depends on the temperature of the gas. This means that temperature is proportional to the KE of the particles of a gas
(But only when temperature is expressed in Kelvins)
3. Real _ gasses conforms to all postulates of the KMT.
4. In order to fully describe a gas, 4 _ measurable quantities must be stated.
a. Define PRESSURE: amount of force exerted per unit of area _
units: 1 atm (“atmosphere”) = 760 _ mm Hg (“millimeters mercury”) = 760 _ torr = 101.325 _ kPa (“kilopascals”)
measured with a barometer _
b. Define TEMPERATURE: measure of the average KE of the particles of a substance _
units: degrees Celsius ( °C _) or Kelvin ( K _)
how to convert from ˚C to K? K = °C + 273.15 K _
c. Define VOLUME: measure of 3-D space occupied by a sample of matter_
units: 1 Liter (L) = 1000 _ mL = 1000 _ cm3 = 100 _ dm3
d. Define QUANTITY: amount of substance_
units: mole_
Convert from grams to moles using molar mass _ abbreviated MM _
5. “STP” stands for “ Standard Temperature and Pressure _”. The conditions at STP are exactly 1_ atm of pressure and a temperature of exactly 273.15 K _ and any
gas at STP will occupy a volume of 22.4 _ L.
9.H – Gases: Variables of Gas Laws
Instructions: ON A SEPARATE SHEET OF PAPER, preform the following temperature and pressure conversions. Remember, Sig Figs, NW = NC, Boxed Answers and N3
(this includes labeling and in your set-up!!!)
1. 2.00 atm to mm Hg
2.00 atm
760 mm Hg =
6. 115 kPa to atm
1520 mmHg
1 atm
101.325 kPa =
=
1.13 atm
3.50x104 torr
760 mm Hg = 3.50 x 104 mmHg
760 torr
7. 93,500 Pa to atm
240.0 mmHg
760 mm Hg
1 atm
=
760 mm Hg
1 atm
= 0.923 atm
950. torr 1 atm
760 torr
5. 100. K to ˚C
1.25 atm
TK = TC + 273.15 = 120 ˚C + 273.15 = 390 K
14. -25.2 ˚C to Kelvin
TC = TK - 273.15 = 298.98 K - 273.15= 25.83 ˚C TK = TC + 273.15 = -25.2 ˚C + 273.15 = 248.0 K
10. -227.1 ˚C to Kelvin
= ―173 ˚C
49.6 kPa
13. 120 ˚C to Kelvin
=
9. 298.98 K to ˚C
TK = TC + 273.15 = 35.82 ˚C + 273.15 = 308.97 K
0.490 atm 101.325 kPa =
1 atm
8. 950. torr to atm
0.658 atm
4. 35.82˚C to Kelvin
TC = TK - 273.15 = 100. K - 273.15
93,500 Pa 1 kPa
12. 0.490 atm to kPa
1000 Pa 101.325 kPa
3. 500. mm Hg to atm
500. mm Hg
115 kPa 1 atm
101.325 kPa
2. 1800. mm Hg to kPa
1800. mm Hg
11. 3.5 x 104 torr to mm Hg
TK = TC + 273.15 = -227.1 ˚C + 273.15 =
15. 5 Kelvin to ˚C
46.05 K TC = TK - 273.15 = 5 K - 273.15 = ―268 ˚C
Instructions: ON A SEPARATE SHEET OF PAPER, identify the values and symbols of all variables present within the following data. Be sure to denote the unknown
variable, what you need to solve for, as = x. Variables to denote in include but are not limited to include n1, V1, P1, T1, n2, V2, P2, and T2. Remember, temperature for Gas
Laws must be reported in K. No Work must be shown.
16. If 2.00 mol of gas occupies 4.50L at STP. How much of the same gas will
occupy 3.00L at STP?
n1 = 2.00 mol
V1 = 4.50 L
P1 = 1 atm
T1 = 273.15 K
n2 = X mol
V2 = 3.00 L
P2 = 1 atm
T2 = 273.15 K
17. A gas has an initial volume of 15 L. If the temperature increases from 330 K to
450 K, find the new volume.
n1 = constant
V1 = 15 L
P1 = constant
T1 = 330 K
n2 = constant
V2 = X L
P2 = constant
T2 = 450 K
18. A gas exerts 1.2 atm of pressure. If the temperature is raised from 225 K to
325 K, find the new pressure.
n1 = constant
V1 = constant
P1 = 1.2 atm
T1 = 225 K
n2 = constant
V2 = constant
P2 = X atm
T2 = 325 K
19. Suppose 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is
increased to 1.80 mol, what new volume will result?
n1 = 0.965 mol
V1 = 5.00 L
n2 = 1.80 mol
V2 = X L
P1 = constant
T1 = constant
P2 = constant
T2 = constant
dm3
20. A sample of oxygen takes up 34
of space when it is under 500 kPa of
pressure. When the pressure is changed to 340 kPa, find the new volume.
n1 = constant
V1 = 34 dm3
P1 = 500 kPa
T1 = constant
n2 = constant
V2 = X dm3
P2 = 340 kPa
T2 = constant
21. The pressure of some N2 drops from 315 kPa to 220 kPa. If the initial volume is
1.4 L, find the new volume.
n1 = constant
V1 = 1.4 L
P1 = 315 kPa
T1 = constant
n2 = constant
V2 = X L
P2 = 220 kPa
T2 = constant
n1 = constant
V1 = constant
P1 = 786 mm Hg
T1 = 360. K
n2 = constant
V2 = constant
P2 = 1811 mm Hg
T2 = X K
23. When the temperature of a gas changes, its volume decreases from 12.23 L to
7.92 L. If the final temperature is measured to be 312.24 K, what was the initial
temperature (in K)?
n1 = constant
V1 = 12.23 L
P1 = constant
T1 = X K
n2 = constant
V2 = 7.92 L
P2 = constant
T2 = 312.24 K
24. If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant
temperature. What is the new volume?
n1 = constant
V1 = 22.5 L
P1 = 748 mm Hg
T1 = constant
n2 = constant
V2 = X L
P2 = 725 mm Hg
T2 = constant
25. A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a
volume of 12.0L. What is the pressure in the container if the temperature
remains constant?
n1 = constant
V1 = 4.0 L
P1 = 205 kPa
T1 = constant
n2 = constant
V2 = 12.0 L
P2 = X kPa
T2 = constant
26. What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into
a cylinder whose volume is 26.0 liters?
n1 = constant
V1 = 196.0 L
P1 = 1.00 atm
T1 = constant
n2 = constant
V2 = 26.0 L
P2 = X atm
T2 = constant
27. A 40.0 L tank of ammonia has a pressure of 12.7 kPa. Calculate the volume of
the ammonia if its pressure is changed to 8.4 kPa while its temperature remains
constant.
n1 = constant
V1 = 40.0 L
P1 = 12.7 kPa
T1 = constant
n2 = constant
V2 = X L
P2 = 8.4 kPa
T2 = constant
22. The pressure of neon changes from 786 mm Hg to 1811 mm Hg. If the initial
temperature 87oC, what is the new temperature (in K)?
9.I – Gases: A,B,C and D of Gas Laws
Instructions: ON A SEPARATE SHEET OF PAPER, preform the following basic gas law calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N3 (this includes
labeling and in your set-up!!!)
1. If 2.00 mol of gas occupies 4.50L at STP. How much of the same gas will occupy
4. Suppose 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is
3.00L at STP?
increased to 1.80 mol, what new volume will result (at an unchanged
temperature and pressure)?
Base Equation:
V1 =
n1
Required Work to Show:
V2
n2
Algebraic Rearrangement:
n2 =
Answer: 1.33 mol gas
2.00 mol 3.00 L
4.50 L
n2 = _ n1 V2 _
V1
Base Equation:
V1 =
n1
Required Work to Show:
V2
n2
Algebraic Rearrangement:
V2 =
Answer: 9.33 L gas
2. A gas has an initial volume of 15 L. If the temperature increases from 330 K to
450 K, find the new volume. Pressure is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
V2 = _ V1 T2 _
T1
15 L 330 K
450 K
3. A gas exerts 1.2 atm of pressure. If the temperature is raised from 225 K to 325
K, find the new pressure. Volume is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
Answer: 1.7 atm gas
P1V2
T2
Algebraic Rearrangement:
P2 =
1.2 atm 325 K
225 K
5. A canister contains 425 kPa of carbon dioxide, 750 kPa of nitrogen, and 525 kPa
of oxygen. What is the total pressure of the container?
Base Equation:
PT = P1 + P2 + P3
Required Work to Show:
Answer: 1.70 x 103 kPa
Answer: 20. L gas
P2 = _ P1 T2 _
T1
5.00 L 1.80 mol
0.965 mol
V2 = _ V1 n2 _
n1
Algebraic Rearrangement: PT = PCO2 + PN2 + PO2
PT =
425 kPa + 750 kPa + 525 kPa
6. A sample of oxygen takes up 34 dm3 of space when it is under 500 kPa of
pressure. When the pressure is changed to 340 kPa, find the new volume.
Temperature is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
Answer: 50 dm3 O2
P1V2
T2
Algebraic Rearrangement:
V2 =
34 dm3 500 kPa
340 kPa
V2 = _ P1 V1 _
P2
7. The pressure of some N2 drops from 315 kPa to 220 kPa. If the initial volume is
1.4 L, find the new volume. Temperature is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
Answer: 2.0 L N2
1.4 L 315 kPa
220 kPa
V2 = _ P1 V1 _
P2
14. A 40.0 L tank of ammonia has a pressure of 12.7 kPa. Calculate the volume of
the ammonia if its pressure is changed to 8.4 kPa while its temperature remains
constant. Temperature is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
Answer: 60. L NH3
8. The pressure of neon changes from 786 mm Hg to 1811 mm Hg. If the initial
temperature 87oC, what is the new temperature (in K)? Volume is Constant, so
cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T2 =
T2 = _ T1 P2 _
P1
360. K 1811 mm Hg
786 mm Hg
Answer: 829 K
15. A container containing 5.00 L of a gas is collected at 100.0 K and then allowed to
expand to 20.0 L. What must the new temperature be in order to maintain the
same pressure? Pressure is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
T2 = _ T1 V2 _
V1
100.0 K 5.00 L
20.0 L
Answer: 25.0 K
9. When the temperature of a gas changes, its volume decreases from 12.23 L to
7.92 L. If the final temperature is measured to be 312.24 K, what was the initial
temperature (in K)? Pressure is Constant, so cancel it out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T2 =
T2 = _ T1 V2 _
V1
312.24 K 7.92 L
12.23 L
10. If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant
temperature. What is the new volume? Temperature is Constant, so cancel it
out
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
Answer: 23.2 L N2
22.5 L 748 mm Hg
725 mm Hg
V2 = _ P1 V1 _
P2
Required Work to Show:
Answer: 0.6 atm Ar
Algebraic Rearrangement: PAr = PT ― PNH3
PAr
= 1.8 atm ― 1.2 atm
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
P2 =
Answer: 68.3 L N2
205 kPa 4.0 L
12.0 L
P1V1 =
T1
Required Work to Show:
P2 = _ P1 V1 _
V2
Answer: 7.54 atm air
P1V2
T2
Algebraic Rearrangement:
P2 =
1.00 atm 196.0 L
226.0 L
V2 =
V2 = _ V1 T2 _
T1
900.0 mL 132.0 K
300.2 K
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
Answer: 895 K
298.2 K 45.0 L
15.0 L
T2 = _ T1 V2 _
V1
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T2 =
Answer: 100 K
353.4 K 50.0 kPa 40 L
120.0 kPa 45 L
T2 = _ T1 P2 V2 _
P1 V1
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T1 =
T1 = _ T2 P1 V1 _
P2 V2
300. K 150. kPa 14 L
100. kPa 21 L
Answer: 300. K
20. A sample of argon goes from 500 K to 350 K and its pressure changes from 280
kPa to 380 kPa. If the initial volume is 18 dm3, what is the final volume?
13. What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into
a cylinder whose volume is 26.0 liters? Temperature is Constant, so cancel it
out
P1V1 =
T1
Required Work to Show:
Algebraic Rearrangement:
Answer: 295.7 mL gas
Base Equation:
Base Equation:
P1V2
T2
19. A sample of nitrogen goes from 21 L to 14 L and its pressure increases from
100. kPa to 150. kPa. The final temperature is 300. K. What was the initial
temperature in Kelvins?
12. A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a
volume of 12.0L. What is the pressure in the container if the temperature
remains constant? Temperature is Constant, so cancel it out
Base Equation:
Base Equation:
18. The pressure of a gas changes from 120.0 kPa to 50.0 kPa. The volume
changes from 45 L to 40 L. If the initial temperature is 353.4 K, what is the final
temperature in K?
11. A tank containing ammonia and argon has a total pressure equal to 1.8 atm. The
pressure of the ammonia is 1.2 atm. What is the pressure of the argon gas?
PT = P1 + P2 + P3
16. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at
132.0 K? Pressure is Constant, so cancel it out
17. If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the
new temperature be to maintain constant pressure? Pressure is Constant, so
cancel it out
Answer: 202 K
Base Equation:
12.7 kPa 40.0 L
8.4 kPa
V2 = _ P1 V1 _
P2
P2 = _ P1 V1 _
V2
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
Answer: 9 dm3 Ar
18 dm3 280 kPa 350 K
500 K 380 kPa
V2 = _V1 P1 T2 _
T1 P2
21. A sample of neon experiences a pressure drop from 75 kPa to 53 kPa. The
temperature increases from 327.4 K to 521.5 K. If the initial volume is 12 L,
what is the final volume?
Base Equation:
P1V1 =
T1
Required Work to Show:
Answer: 27 L Ne
P1V2
T2
Algebraic Rearrangement:
V2 =
12 L 75 kPa 521.5 K
327.4 K 53 kPa
V2 = _V1 P1 T2 _
T1 P2
9.J – Gases: Ideal Gas Law
1.What are the differnces between an ideal gas and a real gas?
Ideal = no individual particle volume or mass = elastic collisions = no IMF’s = ALL moleules KE proprtional to temperature_
Real = individual particles with inelastic collision influnced by IMF’s possessing nonuniform KE _
2. REAL GASES BEHAVE NEARLY IDEALLY UNDER CONDITIONS of high _ temperature, low_ pressure, & low_ molar mass.
Instructions: Complete the following statements to remind yourself of the requirments of the Ideal Gas law before completing the calualtions that follow.
In P V = n R T:
"P" stands for pressure_ , must be in units of mmHg / torr / atm / kPa_
"V" stands for volume_ , must be in units of liters_
"n" stands for mole_ , must be in units of moles_
"T" stands for temperature_ , must be in units of Kelvins_
"R" stands for the Ideal Gas Constant , has a value that varies_ dependent on unit of pressure used_
Instructions: ON A SEPARATE SHEET OF PAPER, preform the following basic gas law calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N3 (this includes
labeling and in your set-up!!!)
3. If 3.7 moles of propane are at a temperature of 28 ˚C and are under 154.2 kPa of
8. Given 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is
pressure, what volume does the sample occupy?
the temperature?
Base Equation:
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
V =
V = _ n RkPa T _
P
301 K
L∙kPa
mol∙K
3.7 mol 8.314
154.2 kPa
Base Equation:
PV = nRT
Required Work to Show:
Answer: 60. L Propane
4. A sample of carbon monoxide at 57 ˚C and under 0.67 atm of pressure takes up
85.3 L of space. What mass of carbon monoxide is present in the sample?
Base Equation:
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
n= _P V _
Ratm T
n = 0.67 atm
85.3 L
L∙atm 330. K
0.0821 mol∙K
2.1 mol CO
= 2.1 mol CO
Answer: 59 g CO
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
71 g F2
P =
P = _ n RkPa T _
V
1 mol F2 _ = 1.9 mol F2
38.00 g F2
1.9 mol 8.314
6.843 L
L∙kPa
mol∙K
228 K
Answer: 530 kPa F2
6. At 971 mm Hg, 145 g of carbon dioxide have a volume of 34.13 L. What is the
temperature of the sample, in ˚C?
Base Equation:
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
145 g CO
T =
T= _P V_
n RmmHg
1 mol CO _ = 5.18 mol CO
28.01 g CO
971 mm Hg
5.18 mol
62.4
L∙mmHg
mol∙K
0.0821
_
L∙atm
mol∙K
9. An unknown quantity of gas at a pressure of 1.2 atm, a volume of 31 liters, and a
temperature of 87 0C, how many moles of gas are present?
Base Equation:
PV = nRT
Algebraic Rearrangement:
n = 0.0821
1.2 atm
L∙atm
mol∙K
n = _ Ratm T _
P V
360. K
31 L
10. A vessel contains 3.21 moles of gas with a volume of 60.9 liters and at a
temperature of 400.1 K, what is the pressure inside the container?
Base Equation:
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
L∙atm
n = 0.0821 mol∙K
3.11 atm
410. K
13.46 L
67.3 g X _ = 83.7 g/mol X
0.804 mol X
PV = nRT
Required Work to Show:
Algebraic Rearrangement:
P = 3.21 mol 8.314
60.9 L
L∙kPa
mol∙K
P = _ n RkPa T _
V
400.1 K
Answer: 175 kPa or 1310 torr/mmHg or 1.73 atm
11. A vessel contains 7.7 moles of gas at a pressure of 0.09 atm and at a
temperature of 56 0C, what is the volume of the container that the gas is in mL?
Base Equation:
PV = nRT
Required Work to Show:
Answer: 2310000 mL
Base Equation:
7. At 137oC and under a pressure of 3.11 atm, a 67.3 g sample of an unknown noble
gas occupies 13.46 L of space. What is the gas?
Answer: Kr
12 L
4 mol
Algebraic Rearrangement:
V =
7.7 mol 0.0821
0.09 atm
L∙atm
mol∙K
V = _ n RkPa T _
P
329 K
= 2310 L
or 2.31 x 106 mL
12. A vessel contains 1.37 moles of gas at a temperature of 67.2 0C, and a volume
of 88.89 liters, what is the pressure of the gas in atm?
34.13 L
Answer: 103 K
Base Equation:
5.6 atm
Answer: 0.79 mol Gas
5. At – 45 ˚C, 71 g of fluorine gas take up 6843 mL of space. What is the pressure
of the gas, in kPa?
Base Equation:
T=
T= _P V_
n RmmHg
Answer: 200 K
Required Work to Show:
28.01 g CO _
1 mol CO
Algebraic Rearrangement:
n = _ Ratm T _
P V
= 0.804 mol X
PV = nRT
Required Work to Show:
Answer: 0.430 atm
Algebraic Rearrangement:
P = 1.37 mol 0.0821
88.9 L
L∙atm
mol∙K
P = _ n RkPa T _
V
340. K