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 = P × V

II. Charles’s = VT

III. Gay-Lussac’s = PT

IV. Gay-Lussac’s = nV

(b) What are the STP values of all 4 properties?

PROPERTY STP VALUEPressure 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 = nRT2. (a) What is “R” in the Ideal Gas Law?

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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 RPVnT = R

Then insert the STP values for the 4 gas properties

Example: R value if pressure is given in atm:

R = PVnT → R = (1.00atm )(22.4 L)(1.00mol )(273Kelvin) → R = 0.0821 (atm )(L)

(mol )(K )

Example: R value if pressure is given in mmHg or torr:

R = PVnT → R = (760mmHg )(22.4 L)(1.00mol )(273Kelvin) → R = 62.4 (mmHg )(L)

(mol )(K )

Example: R value if pressure is given in kPa:

R = PVnT → R = (101.325 kPa )(22.4 L)(1.00mol )(273Kelvin) → R = 8.31 (kPa )(L)

(mol )(K )

NOTE that each R constant has 4 units!!!!Using the Ideal Gas Law

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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

R = 0.0821 (atm )(L)(mol )(K )

STEP 4: Plug all the values from Steps 2 & 3 into the Ideal Gas Law Equation and solve

PV = nRT → x(10.0-L) = (2.00 mol)( 0.0821 (atm )(L)(mol )(K ) )(300.K)

→ P = 4.93 atm

4. What is the volume, in liters, of 4.00 mol of methane, CH4, at 18oC and 1.40 atm?

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PV = nRT → 1.40 atm(x) = (4.00 mol)( 0.0821 (atm )(L)(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 (mmHg )(L)

(mol )(K ) )(295K)

→ n = 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 (mmHg )(L)

(mol )(K ) )x

→ T = 566 K = 293 oC

Dalton’s Law

7. State Dalton’s Law

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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?

PARTIALPRESSURE PERCENTAGE

GAS (mmHg) (%) .

Nitrogen, N2 594 78.2

Oxygen, O2 160. 21.0

Carbon Dioxide, CO2

Argon, Ar 6 0.8Water, H2O

Total Air 760. 100

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?

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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

→ 615mmHg760mmHg = 0.809 atm

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

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→ 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

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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 → (93.3 kPa) (0.355 L) = x(8.31 (kPa )(L)(mol )(K ))(299 K)

= 0.0133 moles H2

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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.

Δ2 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

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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

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