Key
HOMEWORK for Chapter 11: Gases
11.8 The Ideal Gas Law (Read pgs. 348-351 in the chemistry textbook)
11.10 Partial Pressures (Dalton’s Law) (Read pgs. 352-355 in the chemistry textbook)
11.8 Questions about the Ideal Gas Law
1. (a) What is the Ideal Gas Law?
it combines all 4 gas properties and the 4 gas laws that cover
those properties, into one law and accounts for changes from STP
values for all 4 properties
I.
Boyle’s
=
II.
Charles’s
=
III.
Gay-Lussac’s
=
IV.
Gay-Lussac’s
=
P×V
𝐕
𝐓
𝐏
𝐓
𝐧
𝐕
(b) What are the STP values of all 4 properties?
PROPERTY
Pressure
STP VALUE
1 atm or 760 mmHg or 101.325 kPa, etc.
Volume
22.4 L
Temperatue
273 Kelvin
Amount
1 mole
(c)What is the mathematical equation for the Ideal Gas Law?
PV = nRT
1
2.
(a) What is “R” in the Ideal Gas Law?
R = the gas law constant
it is a number which takes into account all 4 gas properties
T, P, V, n and their relationship to one another (direct or
inverse)
R has a constant value for any given pressure measurement.
(b) How is “R” determined?
Rearrange the Ideal Gas Law equation to solve for R
𝐏𝐕
𝒏𝐓
=R
Then insert the STP values for the 4 gas properties
Example: R value if pressure is given in atm:
R=
𝑷𝑽
𝒏𝑻
(𝟏.𝟎𝟎 𝒂𝒕𝒎)(𝟐𝟐.𝟒 𝑳)
→ R = (𝟏.𝟎𝟎
𝒎𝒐𝒍)(𝟐𝟕𝟑 𝑲𝒆𝒍𝒗𝒊𝒏)
( 𝒂𝒕𝒎)(𝑳)
→ R = 0.0821 (
𝒎𝒐𝒍)( 𝑲)
Example: R value if pressure is given in mmHg or torr:
R=
𝑷𝑽
𝒏𝑻
(𝟕𝟔𝟎 𝒎𝒎𝑯𝒈)(𝟐𝟐.𝟒 𝑳)
→ R = (𝟏.𝟎𝟎
𝒎𝒐𝒍)(𝟐𝟕𝟑 𝑲𝒆𝒍𝒗𝒊𝒏)
→ R = 62.4
( 𝒎𝒎𝑯𝒈)( 𝑳)
( 𝒎𝒐𝒍)( 𝑲)
Example: R value if pressure is given in kPa:
R=
𝑷𝑽
𝒏𝑻
(𝟏𝟎𝟏.𝟑𝟐𝟓 𝒌𝑷𝒂)(𝟐𝟐.𝟒 𝑳)
→ R = (𝟏.𝟎𝟎
𝒎𝒐𝒍)(𝟐𝟕𝟑 𝑲𝒆𝒍𝒗𝒊𝒏)
( 𝒌𝑷𝒂)(𝑳)
→ R = 8.31 (
𝒎𝒐𝒍)( 𝑲)
NOTE that each R constant has 4 units!!!!
2
Using the Ideal Gas Law
3. Calculate the pressure, in atmospheres, of 2.00 mole of helium gas in a 1.00 x 104 mL container at 27oC.
STEP 1: Identify the “Unknown”
P=?
STEP 2: Identify all the variables which have been given and convert
them to the proper units (if needed)
V = 1.0 x 104 mL = 10.0-L
n = 2.00 mol
T = 270C = 300. K
STEP 3: Choose the correct R-Value for the problem
( 𝒂𝒕𝒎)(𝑳)
R = 0.0821 ( 𝒎𝒐𝒍)( 𝑲)
STEP 4: Plug all the values from Steps 2 & 3 into the Ideal Gas Law
Equation and solve
( 𝒂𝒕𝒎)(𝑳)
PV = nRT → x(10.0-L) = (2.00 mol)( 0.0821 (
→P=
𝒎𝒐𝒍)( 𝑲)
)(300.K)
4.93 atm
3
4. What is the volume, in liters, of 4.00 mol of methane, CH4, at 18oC and 1.40 atm?
( 𝒂𝒕𝒎)(𝑳)
PV = nRT → 1.40 atm(x) = (4.00 mol)( 0.0821 ( 𝒎𝒐𝒍)( 𝑲) )(291K)
→V=
68.3 L
5. A tank of oxygen gas has a volume of 20.0 L. How many moles of oxygen are in the container if the gas
has a pressure of 845 mmHg at 22oC?
PV = nRT → (845 mmHg)(20.0 L) = x( 62.4
→n=
( 𝒎𝒎𝑯𝒈)(𝑳)
( 𝒎𝒐𝒍)( 𝑲)
)(295K)
0.918 mol
6. A 0.892 mol sample of nitrogen gas has a volume of 50.0 L and a pressure of 630. mmHg. What is the
temperature, in degrees Celsius, of the gas?
PV = nRT → (630. mmHg)(50.0 L) = (0.892 mol)( 62.4
→ T = 566 K =
( 𝒎𝒎𝑯𝒈)(𝑳)
( 𝒎𝒐𝒍)( 𝑲)
)x
293 oC
4
Dalton’s Law
7. State Dalton’s Law
The total pressure of a gas mixture is the sum of the partial
pressure of the gases in the mixture.
8. What is the partial pressure of a gas?
the pressure a gas would exert if it were the ONLY gas in the
container
9. What is the equation for Dalton’s Law?
PT = P1 + P2 + P3 + ……
10. What is the typical composition of air?
GAS
PARTIAL
PRESSURE
(mmHg)
PERCENTAGE
(%)
Nitrogen, N2
594
78.2
Oxygen, O2
160.
21.0
Carbon Dioxide, CO2
Argon, Ar
Water, H2O
6
0.8
Total Air
760.
100
.
5
11. An anesthetic consists of a mixture of cyclopropane gas, C3H6, and oxygen gas, O2. If the mixture has a
total pressure of 1.09 atm, and the partial pressure of the cyclopropane is 73 mmHg, what is the partial
pressure, in atm, of the oxygen in the anesthetic?
PT = Pcyclo + Poxygen
→ 1.09 atm = 0.0961 atm + x → 0.994 atm
12. In a gas mixture, the partial pressures are nitrogen 425 torr, oxygen 115 torr, and helium 225 torr. What is
the total pressure (torr) exerted by the gas mixture?
PT = Pnitrogen + Poxygen + Phelium
→ 425 torr + 115 torr + 225 torr → 765 torr
13.
In a gas mixture, the partial pressures are argon 415 mmHg, neon 75 mmHg, and nitrogen 125 mmHg.
What is the total pressure (atm) exerted by the gas mixture?
PT = Pargon + Pneon + Pnitrogen
→ 415 mmHg + 75 mmHg + 125 mmHg → 615 mmHg
→
𝟔𝟏𝟓 𝒎𝒎𝑯𝒈
𝟕𝟔𝟎 𝒎𝒎𝑯𝒈
=
0.809 atm
6
14. A gas mixture containing oxygen, nitrogen, and helium exerts a total pressure of 925 torr. If the partial
pressures are oxygen 425 torr and helium 75 torr, what is the partial pressure (torr) of the nitrogen in the
mixture?
PT = Poxygen + Pnitrogen + Phelium
→ 925 torr = 425 torr + x
+ 75 torr →
425 torr
Gases Collected Over Water
This reaction, an active metal with a strong acid, creates hydrogen
gas.
Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq)
The collected H2 gas is mixed with water vapor, so…..
PT = Phydrogen + Pwater vapor
7
Use the table of Vapor Pressure of Water above to help answer the following questions.
15.
(a) When magnesium reacts with HCl, a volume of 355 mL of hydrogen gas is collected over water at
26oC. If the total pressure is 96.7 kPa, what was the partial pressure, in kPa, of the hydrogen?
PT = Phydrogen + Pwater vapor
→ 96.7 kPa = x + 3.4 kPa → 93.3 kPa
(b) How many moles of hydrogen gas were collected?
PV = nRT
=
( 𝒌𝑷𝒂)(𝑳)
→ (93.3 kPa) (0.355 L) = x(8.31 ( 𝒎𝒐𝒍)( 𝑲))(299 K)
0.0133 moles H2
8
16. (a) A 456-mL sample of oxygen gas was collected over water at a pressure of 744 mmHg and a
temperature of 20.oC. What was the partial pressure (mmHg) of the oxygen gas?
PT = Poxygen+ Pwater vapor
→ 744 mmHg = x + 17 mmHg →
727 mmHg
(b) How many moles of oxygen gas were collected?
PV = nRT
=
→ (727 mmHg) (0.456 L) = x(62.4
( 𝒎𝒎𝑯𝒈)(𝑳)
( 𝒎𝒐𝒍)( 𝑲)
)(293 K)
0.0181 moles O2
17. When solid KClO3 is heated, it decomposes to give solid KCl and O2 gas. A volume of 256 mL of gas is
collected over water at a total pressure of 765 mmHg and 24oC.
Δ
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
(a) What was the partial pressure of the O2 gas?
PT = Poxygen+ Pwater vapor
→ 765 mmHg = x + 23 mmHg
→
742 mmHg
(b) How many moles of O2 gas were in the gas sample?
PV = nRT
=
→ (742 mmHg) (0.256 L) = x(62.4
( 𝒎𝒎𝑯𝒈)(𝑳)
( 𝒎𝒐𝒍)( 𝑲)
)(297 K)
0.0102 moles O2
9
18. When solid CaCO3 is heated, it decomposed to give solid CaO and CO2 gas. A volume of 425 mL of gas is
collected over water at a total pressure of 758 mmHg and 16oC.
Δ
CaCO3(s) → CaO(s) + CO2(g)
(a) What was the partial pressure of the CO2 gas?
PT = Pcarbon dioxide + Pwater vapor
→ 758 mmHg = x + 14 mmHg →
744 mmHg
(b) How many moles of CO2 gas were in the CO2 sample?
PV = nRT
=
→ (744 mmHg) (0.425 L) = x(62.4
( 𝒎𝒎𝑯𝒈)(𝑳)
( 𝒎𝒐𝒍)( 𝑲)
)(289 K)
0.0175moles CO2
10