Boyle’s Law P1V1 2 - · PDF fileBoyle’s Law At constant temperature (in Kelvin)...

Boyle’s Law

At constant temperature (in Kelvin) and constant

moles of gas, the pressure of a gas is inversely

proportional to the gas volume.

1

Final condition:

P2 at V2

P1V1 = P2V2 = k

Initial condition:

P1 at V1

Boyle’s Law

At constant temperature (in Kelvin) and constant

moles of gas, the pressure of a gas is inversely

proportional to the gas volume.

2

P1V1 = P2V2 = k

Practice Problem

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?

3

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 mL

154 mL = = 4460 mmHg

Charles’s Law

At constant pressure and moles of gas, the volume

of a gas is directly proportional to the absolute

temperature in kelvin.

4

Hg

Hg

gas

gas

𝑽𝟏

𝑻𝟏=

𝑽𝟐

𝑻𝟐= 𝒌

Practice Problem

A sample of carbon monoxide gas occupies

3.20 L at 125 0C. At what temperature (in

Kelvin) will the gas occupy a volume of

1.54 L if the pressure remains constant?

5

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K

3.20 L = = 192 K

V1/T1 = V2/T2

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

Avogadro’s Law

At a specified pressure

and temperature, the

volume of a gas is

proportional to its

number of moles.

6

𝑽𝟏

𝒏𝟏=

𝑽𝟐

𝒏𝟐= 𝒌

Breathe in, breathe out

7

Ears Popping

8

http://abetterchemtext.com/gases/ear_pop.htm

Under normal conditions:

pressure inside the Eustachian

tube is equal to the pressure

on the ear canal.

At higher altitudes: pressure

inside the Eustachian tube is

greater than the pressure on

the ear canal.

Up, up, and away

9

What is STP?

Scientists defined a standard laboratory

temperature and pressure called STP.

What is the volume occupied by 1 mole of

any gas at STP?

10

Temperature = 273.15 K (0 °C)

Pressure = 1 atm = 760 torr

𝑉 =𝑛𝑅𝑇

𝑃=

1 𝑚𝑜𝑙 0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 273.15 𝐾

1 𝑎𝑡𝑚

𝑉 = 22.4 𝐿 standard molar volume

Standard Molar Volume

11

12

Ideal Gas Law

13

PV = nRT

R = PV

nT =

1 atm x 22.414 L

1 mol x 273.15 K =

0.0821 atm·L

mol·K

R is the universal gas constant;

the numerical value of R depends on the units used.

Ideal Gas Law and Gas Density

14

PV = nRT and d = m

V

PV = m

MW

RT

MW x P

RT V

m = d =

, n = m

MW

Learning Check

Rate the gases according to density from

lowest to highest assuming all of the gases

are maintained at the same pressure and

temperature.

15

MW x P

RT d =

16

Use gas laws to determine a balanced

equation

The piston-cylinder is depicted before and after

a gaseous reaction that is carried out at

constant pressure. The temperature is 150 K

before the reaction and 300 K after the

reaction. (Assume the cylinder is insulated.)

17

Which of the following balanced equations describes the reaction?

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

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

Dalton’s Law of Partial Pressure

The total pressure of a mixture of gases is the

sum of the pressure of the individual gases.

V and T

are

constant

P1 P2 Ptotal = P1 + P2 18

Type equation here.

20

We can express the partial pressure Pi in a mixture

of gases in terms of the total gas pressure, PT, and

the mole fraction of each gas in the mixture.

𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵 =𝑛𝐴𝑅𝑇

𝑉𝑇+

𝑛𝐵𝑅𝑇

𝑉𝑇

𝑃𝑇 = 𝑛𝐴 + 𝑛𝐵

𝑅𝑇

𝑉𝑇

𝑃𝐴

𝑃𝑇=

𝑛𝐴𝑅𝑇𝑉𝑇

𝑛𝑇𝑅𝑇𝑉𝑇

=𝑛𝐴

𝑛𝑇=

𝑛𝐴

𝑛𝐴 + 𝑛𝐵

𝑃𝐴 =𝑛𝐴

𝑛𝑇𝑃𝑇

𝑃𝐵 =𝑛𝐵

𝑛𝑇𝑃𝑇

𝑷𝒊 = 𝒙𝒊𝑷𝑻

Dalton’s Law

Factor out RT/V

Get ratio of PA/PT

Ptotal = P1 + P2