10.A – Gases: Partial Pressure
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
115 kPa 1 atm
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
93,500 Pa 1 kPa
760 mm Hg
1 atm
=
760 mm Hg
12. 0.490 atm to kPa
= 0.923 atm
1 atm
950. torr 1 atm
760 torr
13. 120 ˚C to Kelvin
=
1.25 atm
TK = TC + 273.15 = 120 ˚C + 273.15 = 390 K
9. 298.98 K to ˚C
TK = TC + 273.15 = 35.82 ˚C + 273.15 = 308.97 K
5. 100. K to ˚C
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
1 atm
8. 950. torr to atm
0.658 atm
4. 35.82˚C to Kelvin
TC = TK - 273.15 = 100. K - 273.15
0.490 atm 101.325 kPa =
1000 Pa 101.325 kPa
3. 500. mm Hg to atm
500. mm Hg
=
101.325 kPa
2. 1800. mm Hg to kPa
1800. mm Hg
11. 3.5 x 104 torr to mm Hg
15. 5 Kelvin to ˚C
TK = TC + 273.15 = -227.1 ˚C + 273.15 =
46.05 K TC = TK - 273.15 = 5 K - 273.15 = ―268 ˚C
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!!!)
16. A mixture of 2.00 moles of H2, 3.00 moles of NH3, 4.00 moles of CO2 and 5.00
18. A tank containing ammonia and argon has a total pressure equal to 1.8 atm. The
moles of N2 exerts a total pressure of 800. torr. What is the partial pressure of
pressure of the ammonia is 1.2 atm. What is the pressure of the argon gas?
each gas?
Base Equation:
Base Equation:
Pmol = PT ÷ (total # of mol)
Req` Work:
Pmol = 800. torr ÷ (2.00 mol + 3.00 mol + 4.00 mol + 5.00 mol)
Pmol = 57.1 torr/mol
PH2 = 57.1 torr/mol • 2.00 mol =
114 torr
PNH3 = 57.1 torr/mol • 3.00 mol =
171 torr
PCO2 = 57.1 torr/mol • 4.00 mol =
228 torr
PO2 = 57.1 torr/mol • 5/00 mol =
286 torr
17. The partial pressure of F2 is 300. torr in a mixture of gases where the total
pressure is 1.00 atm. What is the mole fraction of F2?
PF2 =
300. torr 1 atm _
760 torr
Molfract = PF2 ÷ PT
= 0.395 atm ÷ 1.00 atm
PT = P1 + P2 + P3
Required Work to Show:
Answer: 0.6 atm Ar
Algebraic Rearrangement: PAr = PT ― PNH3
PAr
= 1.8 atm ― 1.2 atm
19. 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
Algebraic Rearrangement: PT = PCO2 + PN2 + PO2
PT =
425 kPa + 750 kPa + 525 kPa
= 0.395 atm
= 0.395 F2
10.B – Gases: KMT
1. 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 _
2. “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.
10.C – Gases: Variable ID
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.
n1 = constant
V1 = constant
P1 = 786 mm Hg
T1 = 360. K
1. If 2.00 mol of gas occupies 4.50L at STP. How much of the same gas will occupy
3.00L at STP?
n2 = constant
V2 = constant
P2 = 1811 mm Hg
T2 = X K
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
2. 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
8. 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
3. 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
P1 = constant
T1 = constant
n2 = 1.80 mol
V2 = X L
P2 = constant
T2 = constant
4. 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.
n1 = constant
n2 = constant
V1 = 34
V2 = X
dm3
P1 = 500 kPa
T1 = constant
dm3
P2 = 340 kPa
T2 = constant
5. 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
7. 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
9. 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
10. 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
11. 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
12. 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
6. 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)?
10.D – 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 =
n2 = _ n1 V2 _
V1
2.00 mol 3.00 L
4.50 L
Answer: 1.33 mol gas
Base Equation:
V1 =
n1
Required Work to Show:
V2
n2
Algebraic Rearrangement:
V2 =
V2 = _ V1 n2 _
n1
5.00 L 1.80 mol
0.965 mol
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
Answer: 20. L gas
5. 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:
P1V2
T2
Algebraic Rearrangement:
V2 =
V2 = _ P1 V1 _
P2
34 dm3 500 kPa
340 kPa
Answer: 50 dm3 O2
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
P2 = _ P1 T2 _
T1
6. 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:
Answer: 2.0 L N2
P1V2
T2
Algebraic Rearrangement:
V2 =
1.4 L 315 kPa
220 kPa
V2 = _ P1 V1 _
P2
7. 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 =
Answer: 829 K
360. K 1811 mm Hg
786 mm Hg
T2 = _ T1 P2 _
P1
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T2 =
T2 = _ T1 V2 _
V1
312.24 K 7.92 L
12.23 L
9. 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
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
V2 = _ P1 V1 _
P2
22.5 L 748 mm Hg
725 mm Hg
10. 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
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
P2 =
P2 = _ P1 V1 _
V2
205 kPa 4.0 L
12.0 L
Answer: 68.3 L N2
11. 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:
Answer: 7.54 atm air
P1V2
T2
Algebraic Rearrangement:
P2 =
1.00 atm 196.0 L
226.0 L
Algebraic Rearrangement:
V2 =
100.0 K 5.00 L
20.0 L
T2 = _ T1 V2 _
V1
13. 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
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
V2 = _ V1 T2 _
T1
900.0 mL 132.0 K
300.2 K
Answer: 295.7 mL gas
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
T2 = _ T1 V2 _
V1
298.2 K 45.0 L
15.0 L
Answer: 895 K
Base Equation:
P1V1 =
T1
Required Work to Show:
P2 = _ P1 V1 _
V2
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
16. 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?
Base Equation:
Base Equation:
P1V2
T2
15. 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?
Answer: 23.2 L N2
Base Equation:
P1V1 =
T1
Required Work to Show:
14. 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:
Base Equation:
Answer: 25.0 K
8. 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:
12. 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
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
T1 =
Answer: 300. K
300. K 150. kPa 14 L
100. kPa 21 L
T1 = _ T2 P1 V1 _
P2 V2
17. 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?
Base Equation:
P1V1 =
T1
Required Work to Show:
P1V2
T2
Algebraic Rearrangement:
V2 =
V2 = _V1 P1 T2 _
T1 P2
18 dm3 280 kPa 350 K
500 K 380 kPa
Answer: 9 dm3 Ar
18. 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
10.E – 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 =
L∙kPa
mol∙K
3.7 mol 8.314
154.2 kPa
V = _ n RkPa T _
P
301 K
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