1 Gases solid liquidgas Condensed phases (s, l) High density. Particles close to each other. Strong attraction forces between particles. Motion.

1

Gases

solid liquid gas

Condensed phases (s, l)High density.Particles close to each other.Strong attraction forces between particles.Motion of particles limited.

Gas phase (g)Low density.Particles far from each other.Weak attraction forces between particles.Particles move very fast.Volume of gas phase is volume of container.

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2

a. Particles move fast in all directions (randomly).b. Particles collide with each other.c. Particles collide with the walls. V/

3

Ideal gas

Hypothetical model1) Gas particles have zero volume.

Just a point (radius =0).

2) No attraction/repulsion between gas particles.Potential energy is zero.

Space between particles

V

Particle volume

b

Volume of gas= Space between particles + volume of particles

Vg = V + bVg ≈ V V/

4

Ideal Gas Law

TRnVp

Can be derived theoretically under above assumptions. Can be arrived to experimentally at relatively low pressures and high temperatures (Boyle, Charles, Guy Lussac, Avogadro, Amonton).

Gas volumeGas pressureAmount of gas

in molesGeneral gas

constant

temperature

Temperature (Kelvin, K): A measure of the kinetic energy of the gas particle

T α Ekinetic

T α ½ m v2 T α v2

v α √T

R8.314 J mol-1K-1

p in Pa, V in m3

0.0821 atm. L. mol-1 K-1

p in atm., V in LV/

5

Gas pressure Particles collide with the wall. Force exerted on wall. Pressure = Force / A Pa = N / m2

1 bar = 100 000 Pa 1 atm. = 101 325 Pa = 760 mmHg 1 mmHg = 1 torr

p = ρ g h

height

density

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6

Boyle’s Law

constant temperature

constant temperature, constant amount of gasconstant T and n

Vp

V

constp

VpVpconstVp

constTnTRnVp

1.

.

.,

2211

T3

T1

p

V

T3 > T

2 > T

1

T2

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7

Charles’ Law

TVTconstV

T

V

T

Vconst

p

Rn

T

V

constpnTRnVp

.

.

.,

2

2

1

1

constant pressure, constant amount of gasconstant p and n

A

gm

A

Fpp opposite

oppositegas

m

m

V

T

p1p2

p3

p1 < p2 < p3

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8

Amonton’s Law

TPTconstP

T

P

T

Pconst

V

Rn

T

P

constVnTRnVp

.

.

.,

2

2

1

1

constant volume, constant amount of gasconstant V and n

P

T

V1

V2

V3

V1 < V2 < V3

V/

9

Combined Gas Law

2

22

1

11.

.

T

Vp

T

Vpn

T

VpconstRn

T

Vp

constnTRnVp

constant amount of gasconstant n

Problem: What is the volume of a gas at STP if its volume at room temperature and at 300 mmHg was 250 cm3?

STP: Standard Temperature and Pressure T=0ºC=273 K p=1 atm.=760 torr

Room temperature25ºC=298 K

T / K = t /ºC + 273

298 K300 mmHg

250 cm3

T1 , p1 , V1

273 K760 mmHg

? cm3

T2 , p2 , V2

32

3

2

22

2

1

11

4.90

760

273

298

250300

cmV

mmHg

K

K

cmmmHgV

Vp

T

T

Vp

V/

10

Avogadro’s Law

nVnconstV

n

V

n

Vconst

p

TR

n

V

constpTTRnVp

.

.

.,

2

2

1

1

constant temperature and pressureconstant T and p

At constant temperature and pressure, the gas volume is proportional to its amount (number of moles).

Molar volume: volume of 1 mole of gas

Latm

KKmolLatmmolV

nKTatmpSTPAt

p

TRnV

4.22.1

273..0821.01

1273.1:11

At 1 atm. and 25ºC: molar volume is 24 L V/

11

Mwt

TRdp

V

mdTR

MwtV

mp

TRMwt

mVp

Mwt

mnTRnVp

Latm

KKmolLatm

molg

gV

p

TR

Mwt

mV

3.21.88.0

293..0821.0

32

25 11

1

What volume will 25 g of O2 occupy at 20ºC and 0.88 atm.?

Calculate the density of oxygen at STP?

LgKKmolLatm

molgatmd

TR

Mwtpd

Mwt

TRdp

/308.1298..0821.0

32.111

1

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12

Dalton’s Law of Partial PressureThe pressure of a gas mixture is the sum of the partial pressures of all gases in the mixture.

...321 pppppi

itotal

The partial pressure of a gas in a mixture is the pressure of that gas if it were alone.

V

TRnp AA

V

TRnp CC

V

TRnp BB

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13

V

TRnp

V

TRnnnp

V

TRn

V

TRn

V

TRnp

pppp

totaltotal

CBAtotal

CBAtotal

CBAtotal

Example: 200 mL of N2 at 25ºC and a pressure of 250 torr are mixed with 350 mL of O2 at 25ºC and a pressure of 300 torr so that the resulting volume is 350 mL. What would be the final pressure of the mixture?

mol

KKmolLatm

Latm

KKmolLatm

mLtorr

TR

Vpn

molKKmolLatm

Latm

KKmolLatm

mLtorr

TR

Vpn

O

OOO

N

NNN

00565.0298..0821.0

1000/350.760/300

298..0821.0

350300

00269.0298..0821.0

1000/200.760/250

298..0821.0

200250

1111

1111

2

22

2

2

22

2

torrtorratmatmLV

TRnp totaltotal 517760.680.0.680.0

300.0

2980821.000565.000269.0

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N2 O2

5 bar1 bar

Calculate the partial pressures of oxygen and nitrogen after opening the stopcock. Calculate the total pressure!

const. temperature!!!

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2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Pressure of water vapor

A student collects 245 mL of O2 at 25ºC and 758 mmHg. If the vapor pressure of water at 25ºC equals 23.76 mmHg, calculate the partial pressure of oxygen and the volume of dry oxygen at STP!

mmHgp

p

ppp

ppp

O

O

OHtotalO

OHOtotal

734

76.23758

2

2

22

22

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A student collects 245 mL of O2 at 25ºC and 758 mmHg. If the vapor pressure of water at 25ºC equals 23.76 mmHg, calculate the partial pressure of oxygen and the volume of dry oxygen at STP!

298 K734 mmHg

245 cm3

T1 , p1 , V1

273 K760 mmHg

? cm3

T2 , p2 , V2

STP

2

2

1

112

2

22

1

11

p

T

T

VpV

T

Vp

T

Vp

32

3

2

217

760

273

298

245734

cmV

mmHg

K

K

cmmmHgV

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Graham’s Law of Effusion

• Escape of gas molecules through a tiny hole into an evacuated space.

• Diffusion:spread of 1 substance throughout a space of another substance.

• Effusion rate of a gas is inversely proportional to the square root of its density (molar mass).

1

2

1

2

1

2

2

1

t

t

d

d

M

M

r

r

dr

Mr

1

1

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• Compare the effusion rates of helium and molecular oxygen at the same temperature and pressure.

827.2/0003.4

/00.32

2

molg

molg

r

r

O

He

He effuses 2.827 times faster than oxygen.

• A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 sec. It will take __??__ seconds for the same amount of H2 gas to effuse under the same conditions.

• A sample of HI gas (MW = 128) effuses at 0.0962 cm/sec. A sample of butylamine gas effuses at 0.126 cm/sec. What is the molecular weight of butylamine?

Laws of effusion apply also for Diffusion.

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19

The glass tube shown above has cotton plugs inserted at either end. The plug on the left is moistened with a few drops of aqueous ammonia, from which NH3 gas slowly escapes. The plug on the right is similarly moistened with a strong solution of hydrochloric acid, from which gaseous HCl escapes. The gases diffuse in opposite directions within the tube; at the point where they meet, they combine to form solid ammonium chloride, which appears first as a white fog and then begins to coat the inside of the tube.

NH3(g) + HCl(g) → NH4Cl(s)

a) In what part of the tube (left, right, center) will the NH4Cl first be observed?b) If the distance between the two ends of the tube is 100 cm, how many cm from the left end of the tube will the NH4Cl first form?

.tanconstistdr

t

x

time

cedisr

1

2

2

1

M

M

x

x

x1 x2

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