1As P (h) increases V decreases

Apparatus for Studying the Relationship BetweenPressure and Volume of a Gas

2

P 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

3

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

P x V = constant

4As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

5

Variation of Gas Volume with Temperatureat Constant Pressure

V T

V = constant x T

V1/T1 = V2 /T2T (K) = t (0C) + 273.15

Charles’ & Gay-Lussac’s Law

Temperature must bein Kelvin

6

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1 /T1 = V2 /T2

T1 = 125 (0C) + 273.15 (K) = 398.15 K

7

Avogadro’s Law

V number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

Constant temperatureConstant pressure

8

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

9

Summary of Gas Laws

Boyle’s Law

10

Charles Law

11

Avogadro’s Law

12

Ideal Gas Equation

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: P (at constant n and T)1V

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

13

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

14

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl

36.45 g HCl= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.7 L

15

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

Sample Problem 5.2 Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?

PLAN: SOLUTION:

V1 in cm3

V1 in mL

V1 in L

V2 in L

unit conversion

gas law calculation

P1 = 1.12 atm P2 = 2.64 atm

V1 = 24.8 cm3 V2 = unknown

P and T are constant

24.8 cm3 1 mL

1 cm3

L

103 mL= 0.0248 L

P1V1

n1T1

P2V2

n2T2

=P1V1 = P2V2

P1V1

P2

V2 = = 0.0248 L1.12 atm

2.46 atm= 0.0105 L

1cm3=1mL

103 mL=1L

xP1/P2

Sample Problem 5.3 Applying the Pressure-Temperature Relationship

PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open?

PLAN: SOLUTION:

P1(atm) T1 and T2(0C)

P1(torr) T1 and T2(K)

P1 = 0.991atm P2 = unknown

T1 = 230C T2 = 1000C

P2(torr)

1atm=760torr

x T2/T1

K=0C+273.15

P1V1

n1T1

P2V2

n2T2

=P1

T1

P2

T2

=

0.991 atm1 atm

760 torr = 753 torr

P2 = P1 T2

T1 = 753 torr

373K

296K= 949 torr

Sample Problem 5.4 Applying the Volume-Amount Relationship

PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.

PLAN:

SOLUTION:

We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.

n1(mol) of He

n2(mol) of He

mol to be added

g to be added

x V2/V1

x M

subtract n1

n1 = 1.10 mol n2 = unknown

V1 = 26.2 dm3 V2 = 55.0 dm3

P and T are constant

P1V1

n1T1

P2V2

n2T2

=

V1

n1

V2

n2

= n2 = n1 V2

V1

n2 = 1.10 mol55.0 dm3

26.2 dm3= 2.31 mol

4.003 g He

mol He= 9.24 g He

Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions

PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C.

PLAN:

SOLUTION:

V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.

V = 438 L T = 210C (convert to K)

n = 0.885 kg (convert to mol) P = unknown

210C + 273.15 = 294.15K0.885kg103 g

kg

mol O2

32.00 g O2

= 27.7 mol O2

P = nRT

V=

24.7 mol 294.15Katm*L

mol*K0.0821x x

438 L= 1.53 atm

Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation

PROBLEM: The piston-cylinders below depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K.

PLAN:

SOLUTION:

We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume.

Which of the following balanced equations describes the reaction?

(1) A2 + B2 2AB (2) 2AB + B2 2AB2

(4) 2AB2 A2 + 2B2(3) A + B2 AB2

Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).

New figures go here.