Gas Laws

Gas Laws

1. Pressure and Volume (Boyle’s Law)

2.

3.

Pressure and VolumeExperiment Pressure Volume P x V

(atm) (L) (atm x L)

1 8.0 2.0 16

2 4.0 4.0 _____

3 2.0 8.0 _____

4 1.0 16 _____

Boyle's Law P x V = k (constant) when

T remains constant

P1V1= 8.0 atm x 2.0 L = 16 atm L

P2V2= 4.0 atm x 4.0 L = 16 atm L

P1V1 = P2V2 = k

Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes.

new vol. old vol. x Pfactor new P old P x Vfactor

V2 = V1 x P1 P2 = P1 x V1

P2 V2

P and V Changes

P1

P2

V1 V2

Boyle's Law

The pressure of a gas is inversely related to the volume when T does not change

Then the PV product remains constant

P1V1 = P2V2

P1V1= 8.0 atm x 2.0 L = 16 atm L

P2V2= 4.0 atm x 4.0 L = 16 atm L

Boyle’s Law Problem

Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

PV Calculation

Prepare a data table

DATA TABLE

Initial conditions Final conditions

P1 = 50 mm Hg P2 = 200 mm Hg

V1 = 1.6 L V2 = ?

Find New Volume (V2)

Solve for V2: P1V1 = P2V2

V2 = V1 x P1 /P2

V2 = 1.6 L x 50 mm Hg = 0.4 L

200 mm Hg

Check your understanding

A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain.

1) 3.2 L

2) 6.4 L

3) 12.8 L

Solution

A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant)

6.4 L x 0.70 atm = 3.2 L (1)

1.40 atm

Volume must decrease to cause an increase in the pressure

Check your understanding

A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.

1) 200. mmHg

2) 400. mmHg

3) 1200 mmHg

Solution

A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain.

600. mm Hg x 12.0 L = 200. mmHg (1)

36.0 L

Pressure decrease when volume increases.

Gas Laws

1. Pressure and Volume (Boyle’s Law)

2. Temperature and Volume (Charles’ Law)

3.

Charles’ Law

V = 125 mL V = 250 mL

T = 273 K T = 546 K

Observe the V and T of the balloons. How does volume change with temperature?

Charles’ Law: V and T

At constant pressure, the volume of a gas is

directly related to its absolute (K) temperature

V1 = V2

T1 T2

Understanding Charles’ Law

Use Charles’ Law to complete the statements

below:

1. If final T is higher than initial T, final V

is (greater, or less) than the initial V.

2. If final V is less than initial V, final T is

(higher, or lower) than the initial T.

Solution

V1 = V2

T1 T2

1. If final T is higher than initial T, final V

is (greater) than the initial V.

2. If final V is less than initial V, final T is (lower) than the initial T.

V and T Problem

A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

VT Calculation

Complete the following setup:

Initial conditions Final conditions

V1 = 785 mL V2 = ?

T1 = 21°C = 294 K T2 = 0°C = 273 K

V2 = _785__ mL x _273 K = _______ mL

V1 294 K

Check your answer: If temperature decreases,

V should decrease.

VT Calculation

Complete the following setup:

Initial conditions Final conditions

V1 = 785 mL V2 = ?

T1 = 21°C = 294 K T2 = 0°C = 273 K

V2 = _785__ mL x _273 K = _729___ mL

V1 294 K

Check your answer: If temperature decreases,

V should decrease.

Understanding Check

A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?

1) 443°C 2) 170°C 3) - 82°C

Solution

A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL?

2) 170°C

T2 = 291 K x 640 mL = 443 K

420 mL

= 443 K - 273 K = 170°C

Gas Laws

1. Pressure and Volume (Boyle’s Law)

2. Temperature and Volume (Charles’ Law)

3. Temperature and Pressure(Gay-Lussac’s Law)

Gay-Lussac’s Law: P and T

The pressure exerted by a confined gas

is directly related to the temperature

(Kelvin) at constant volume.

P (mm Hg) T (°C)

936 100

761 25

691 0

Understanding Gas Laws

Use Gay-Lussac’s law to complete the statements below:

1. When temperature decreases, the

pressure of a gas (decreases or increases).

2. When temperature increases, the pressure

of a gas (decreases or increases).

Solution

1. When temperature decreases, the

pressure of a gas (decreases).

2. When temperature increases, the

pressure of a gas (increases).

PT Problem

A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant)

T = 18°C T = 62°C

PT Calculation

P1 = 2.0 atm T1 = 18°C + 273 = 291 K

P2 = ? T2 = 62°C + 273 = 335 K

What happens to P when T increases?

P increases (directly related to T)

P2 = P1 x T2

T1

P2 = 2.0 atm x 335 K = 2.30 atm

291 K

?

Understanding Gas Laws

Complete with 1) Increases 2) Decreases

3) Does not change

A. Pressure _____, when V decreases

B. When T decreases, V _____.

C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T)

D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

Solution

A. Pressure 1) Increases, when V decreases

B. When T decreases, V 2) Decreases

C. Pressure 2) Decreases when V changes

from 12.0 L to 24.0 L (constant n and T)

D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)