Gases

solid liquid gas

Condensed phases (s, l) Gas phase (g)

➢High density. ➢Low density.
➢Particles close to each other. ➢Particles far from each other.
➢Strong attraction forces ➢Weak attraction forces between
between particles. particles.
➢Motion of particles limited. ➢Particles move very fast.
➢Volume of gas phase is volume of
container. V/ 1
a. Particles move fast in all directions (randomly).
b. Particles collide with each other.
c. Particles collide with the walls. V/ 2
 Ideal gas

Hypothetical model
1) Gas particles have zero volume.
Just a point (radius =0).

2) No attraction/repulsion between gas

particles.
Potential energy is zero.

Particle volume Space between

particles
b
V
Volume of gas= Space between particles + volume of particles
Vg = V + b
Vg ≈ V
V/ 3
 Ideal Gas Law
❖ 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).

pV = n R T
Gas pressure temperature
Gas volume Amount of gas General gas
in moles constant

Temperature (Kelvin, K):

A measure of the kinetic energy of the
R
gas particle 8.314 J mol-1K-1
p in Pa, V in m3
T α Ekinetic
T α ½ m v2 T α v2 0.0821 atm. L. mol-1 K-1
p in atm., V in L V/ 4
v α √T
Gas pressure
❑ Particles collide with the wall. height
❑ Force exerted on wall.
❑ Pressure = Force / A p=ρgh
❑ Pa = N / m2
❑ 1 bar = 100 000 Pa density
❑ 1 atm. = 101 325 Pa = 760 mmHg
❑ 1 mmHg = 1 torr

V/ 5
 Boyle’s Law
constant temperature, constant amount of gas
constant T and n

pV = n R T n, T = const.
p V = const. p1V1 = p2V2
const. 1
p= p
V V

p T3 > T 2 > T 1

T3
T2
T1

constant temperature
V/ 6
V
 Charles’ Law
constant pressure, constant amount of gas
constant p and n

Fopposite
mg
p gas = popposite = =
A A
pV = n R T n, p = const.
m V nR V1 V2
= = const. =
T p T1 T2
V = const.  T V T
m
V
p1 < p2 < p3
p1


p2
p3

V/ 7
T
 Amonton’s Law
constant volume, constant amount of gas
constant V and n

pV = n R T n,V = const.
P nR P1 P2
= = const. =
T V T1 T2
P = const.  T P T

V1 < V2 < V3
P V1
V2

 V3

V/ 8
T
 Combined Gas Law
constant amount of gas
constant n

pV = n R T n = const.
pV pV p1 V1 p2 V2
= n R = const. n =
T T T1 T2
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 Room temperature

T=0ºC=273 K p=1 atm.=760 torr 25ºC=298 K

T / K = t /ºC + 273
298 K 273 K p1V1 T2
 = V2
300 mmHg 760 mmHg T1 p2
250 cm3 ? cm3
300 mmHg  250 cm3 273 K
T 1 , p 1 , V1 T 2 , p 2 , V2 V2 = 
298 K 760 mmHg
V2 = 90.4 cm3 V/ 9
 Avogadro’s Law
constant temperature and pressure
constant T and p
pV = n R T T , p = const.
V RT V1 V2
= = const. =
n p n1 n2
V = const.  n V n
✓ At constant temperature and pressure, the gas volume is
proportional to its amount (number of moles).

Molar volume: volume of 1 mole of gas

n RT
V=
p
At STP : p = 1 atm. T = 273 K n =1
1 mol  0.0821atm.L.mol −1 K −1  273 K
V= = 22.4 L
1 atm.
V/10
At 1 atm. and 25ºC: molar volume is 24 L
 m m
pV = n R T n= pV =  RT
Mwt Mwt
m m RT
p=  RT d= p= d 
V Mwt V Mwt
What volume will 25 g of O2 occupy at 20ºC and 0.88 atm.?
m RT
V= 
Mwt p
25 g 0.0821atm.L.mol −1 K −1  293 K
V= −1
 = 21.3 L
32 g  mol 0.88 atm.

Calculate the density of oxygen at STP?

RT p  Mwt
p= d  d=
Mwt RT

1atm.  32 g  mol −1
d= = 1.428 g / L
0.0821atm.L.mol K  273 K
−1 −1

V/11
 Dalton’s Law of Partial Pressure
The pressure of a gas mixture is the sum of the partial pressures of all
gases in the mixture.

ptotal =  pi = p1 + p2 + p3 + ...
i
The partial pressure of a gas in a mixture is the pressure of that gas if
it were alone.

RT RT RT
p A = nA  p B = nB  pC = nC 
V V V V/12
 ptotal = p A + pB + pC
RT RT RT
ptotal = n A  + nB  + nC 
V V V
= (n A + nB + nC ) 
RT
ptotal
V
RT
ptotal = ntotal 
V
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
300 mL. What would be the final pressure of the mixture?

nN 2 =
p N 2  VN 2
=
250 torr  200 mL
=
(250 / 760) atm.  (200 / 1000) L = 0.00269 mol
R TN 2 0.0821atm.L.mol −1 K −1  298 K 0.0821atm.L.mol −1 K −1  298 K

nO2 =
pO2  VO2
=
300 torr  350 mL
=
(300 / 760) atm.  (350 / 1000) L = 0.00565 mol
R TO2 0.0821atm.L.mol −1 K −1  298 K 0.0821atm.L.mol −1 K −1  298 K

0.0821 298
= (0.00269 + 0.00565) 
RT
ptotal = ntotal  = 0.680 atm. = 0.680 atm.  760 torr = 517 torr
V 0.300 L
V/13
 N2
O2
5 bar
1 bar

Calculate the partial pressures of oxygen and nitrogen

after opening the stopcock. Calculate the total pressure!

const. temperature!!!

V/14
 Pressure
of water
vapor
2 KClO3(s) → 2 KCl(s) + 3 O2(g)

A student collects 245 mL of O2 at

ptotal = pO2 + pH 2O
25ºC and 758 mmHg. If the vapor pO2 = ptotal − pH 2O
pressure of water at 25ºC equals
23.76 mmHg, calculate the partial pO2 = 758 − 23.76
pressure of oxygen and the volume
of dry oxygen at STP! pO2 = 734 mmHg V/15
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 K 273 K
734 mmHg 760 mmHg
245 cm3 ? cm3
T 1 , p1 , V 1 T 2 , p 2 , V2
STP

p1 V1 p2 V2 p1 V1 T2
= V2 = 
T1 T2 T1 p2

734 mmHg  245 cm3 273 K

V2 = 
298 K 760 mmHg
V2 = 217 cm3
V/16
 Effusion
• Escape of gas molecules through a tiny
hole into an evacuated space.
• Diffusion:
spread of 1 substance throughout a
space of another substance.

Graham’s Law of Effusion

• Effusion rate of a gas is inversely proportional to the
square root of its density (molar mass).
1
r
M r1 M2 d2 t2
= = =
1 r2 M1 d1 t1
r
d
V/17
 • Compare the effusion rates of helium and molecular oxygen at
the same temperature and pressure.

rHe 32.00 g / mol

= = 2.827
rO 2 4.0003g / mol
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? V/18

❖ Laws of effusion apply

also for Diffusion. 18
 x1 x2

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?

r=
dis tan ce x
= rd (t is const.)
time t
x1 M2
=
x2 M1 V/19