Gases

Chapter 5 Problems: 14, 19, 20, 21, 26, 32, 34, 36, 40, 42, 43, 44, 47, 48, 50, 52, 54, 59, 60, 63, 66, 67, 76, 84, 88, 102, Graham’s law problems

Elements that exist as gases at 25 0 C and 1 atmosphere 5.1

5.1

• • • • Physical Characteristics of Gases Gases assume the volume and shape of their containers.

Gases are the most compressible state of matter.

Gases will mix evenly and completely when confined to the same container.

Gases have much lower densities than liquids and solids.

5.1

Pressure = Force Area ( force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mm Hg = 760 torr = 101,325 Pa = 14.7 psi = 29.92 in. Hg Barometer 5.2

10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

5.2

Boyle’s Law

P α 1/V

This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.

P

1

V

1

= P

2

V

2 Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

Charles’s Law

If n and P are constant, then V α T V and T are directly proportional.

V

1

V

2 =

T

1

T

2 • If one temperature goes up, the volume goes up!

Jacques Charles (1746 1823). Isolated boron and studied gases. Balloonist.

Gay-Lussac’s Law

If n and V are constant, then P α T P and T are directly proportional.

P

1

P

2 =

T

1

T

2

Joseph Louis Gay Lussac (1778-1850)

If one temperature goes up, the pressure goes up!

Combined Gas Law

• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!

P 1 V 1 P 2 V 2 = T 1 T 2

No, it’s not related to R2D2

And now, we pause for this commercial message from STP

OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

Avogadro’s Law

V

a number of moles (

n

)

V

= constant x

n V

1 /

n

1 =

V

2 /

n

2 Constant temperature Constant pressure 5.3

Ideal Gas Equation

Boyle’s law: V a (at constant

n P

Charles’ law:

V

a

T

(at constant

n

and

T

and

P

) ) Avogadro’s law: V a

n

(at constant

P

and

T

)

V V

a

nT P nT

= constant x =

R P nT P R

is the

gas constant PV = nRT

5.4

The conditions 0 0 C and 1 atm are called

standard temperature and pressure (STP).

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

PV = nRT R = PV nT

(1 atm)(22.414L) = (1 mol)(273.15 K)

R

= 0.082057 L • atm / (mol • K) 5.4

Density (

d

) Calculations

d = m V

=

P M RT m

is the mass of the gas in g

M

is the molar mass of the gas Molar Mass (

M

) of a Gaseous Substance

Solve for n, which = mass/molar mass Or,

M

=

dRT P d

is the density of the gas in g/L 5.4

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0

0 C. What is the molar mass of the gas?

PV = nRT

n

= 2.10 L X 1 atm L •atm 0.0821 x 300.15 K n = 0.0852 mol g n =

M

M

= 54.6 g/mol 5.3

Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (

s

) + 6O 2 (

g

) 6CO 2 (

g

) + 6H 2 O (

l

) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2

V

CO 2 5.60 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 = 0.187 mol CO 2

V

=

nRT P

= L •atm 0.187 mol x 0.0821 x 310.15 K mol •K 1.00 atm = 4.76 L 5.5

P

1 Dalton’s Law of Partial Pressures

V

and

T

are

constant

P

2

P

total

= P

1 +

P

2 5.6

Consider a case in which two gases, A and B , are in a container of volume V.

P

A =

n A

RT

V P

B =

n B

RT

V P

T =

P

A +

P

B

P

A =

X

A

P

T

n

A is the number of moles of A

n

B is the number of moles of B

X

A =

n

A

n

A +

n

B

X

B =

n

A

n

B +

n

B

P

B =

X

B

P

T

P i

= X

i P

T

mole fraction (X i ) =

n i n T

5.6

A sample of natural gas contains 8.24 moles of CH 4 , 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8 . If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )?

P i

=

X i P

T

P

T = 1.37 atm

X

propane = 0.116

8.24 + 0.421 + 0.116

= 0.0132

P

propane = 0.0132 x 1.37 atm = 0.0181 atm 5.6

Bottle full of oxygen gas and water vapor

2KClO 3 (

s

) 2KCl (

s

) + 3O 2 (

g

)

P

T =

P

O +

P

H O 2 2 5.6

5.6

P Depth (ft) 0

Chemistry in Action:

Scuba Diving and the Gas Laws Pressure (atm) 1 33 66 2 3 V 5.6

Kinetic Molecular Theory of Gases 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be

points

; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy 5.7

Kinetic theory of gases and … • Compressibility of Gases • Boyle’s Law

P

a collision rate with wall Collision rate a number density Number density a

P

a 1/

V

1/

V

• Charles’ Law

P

a collision rate with wall Collision rate a average kinetic energy of gas molecules Average kinetic energy a

P

a

T T

5.7

Kinetic theory of gases and … • Avogadro’s Law

P

a collision rate with wall Collision rate a number density Number density a

n P

a

n

• Dalton’s Law of Partial Pressures Molecules do not attract or repel one another

P P

exerted by one type of molecule is unaffected by the presence of another gas total = S

P

i 5.7

Deviations from Ideal Behavior 1 mole of ideal gas

PV

=

nRT n = PV RT

= 1.0

Repulsive Forces Attractive Forces 5.8

Effect of intermolecular forces on the pressure exerted by a gas.

5.8

Van der Waals equation

nonideal gas

(

P V

2 2

)

V

–

nb

) =

nRT

corrected pressure corrected volume 5.8

The distribution of speeds for nitrogen gas molecules at three different temperatures

u

rms =  3

RT M

Velocity of a Gas

The distribution of speeds of three different gases at the same temperature 5.7

Gas diffusion

is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH 4 Cl NH 3 17 g/mol HCl 36 g/mol 5.7

Gas Diffusion

relation of mass to rate of diffusion • HCl and NH 3 diffuse from opposite ends of tube. • Gases meet to form NH 4 Cl • HCl heavier than NH 3 • Therefore, NH 4 Cl forms closer to HCl end of tube.

GAS DIFFUSION AND EFFUSION

• diffusion is the gradual mixing of molecules of different gases.

• effusion is the movement of molecules through a small hole into an empty container.

GAS DIFFUSION AND EFFUSION

Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv 2

Rate for A Rate for B M of B M of A Rate of effusion is inversely proportional to its molar mass.

Thomas Graham, 1805-1869. Professor in Glasgow and London.

GAS DIFFUSION AND EFFUSION

Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is • proportional to T • inversely proportional to M.

He

Therefore, He effuses more rapidly than O 2 at same T.

Graham’s Law Problem 1

1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850 degrees celsius according to the equation: 4 NH 3 (g) + 5 O 2 (g) --> 4 NO(g) + 6 H 2 O(g) Using Graham's Law, what is the ratio of the effusion rates of NH 3 (g) to O 2 (g)?

Graham’s Law Problem 2

What is the rate of effusion for H 2 if 15.00 ml of CO 2 takes 4.55 sec to effuse out of a container?

Graham’s Law Problem 3

What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2 (g)?