1. 10.3 The gas law The pressure-volume relationship: Boyle's law
Four variables are needed to define the physical condition or state of a gas, temperature, T, pressure, P, volume, V and amount of gas (in mole), n.
The equations that express the relationships among T, P, V and n are known as the gas laws.
The pressure-volume relationship: Boyle's law
Boyle’s law states that the volume (V) of a fixed quantity of gas maintained at constant temperature (T) is inversely proportional to the pressure (P).
V  1 P
V = 1 P
constant 
PV = constant (k)

2. Relationship between pressure and volume when mass and temperature are held constant.
Volume Pressure P  V
2 (V1) (P1)
3 (V2) (P2)
4 (V3) (P3)
P1V1 = k = P2V2
P1V1 = P2V2

3. The temperature-volume relationship: Charles's law
Charles’s law states that the volume (V) of a fixed amount of gas maintained at constant pressure (P) is directly proportional to its absolute temperature (T).
V  T
Relationship between volume and temperature when mass and pressure are held constant.
V = T
constant 
= constant (k) V T
= k V T
Volume Temperature V / T
2.5 (V1) (T1)
3.5 (V2) (T2)
4.5 (V3) (T3) T2 T1 V1 =
K = V2 T2 T1 V1 = V2

4. The quantity-volume-amount relationship: Avogadro’s law
Avogadro’s hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Avogadro’s law: The volume (V) of a gas maintained at constant temperature (T) and pressure (P) is directly proportional to the number of moles (n) of the gas.
V  n
V = constant  n
V = k n

5. 10.4 The ideal gas equation PV = nRT nT V  P nT V = R P V  1 P
Boyle’s law :
(constant n, T)
V  T
Chales’s law :
(constant n, P)
V  n
Avogadro’s law :
(constant P, T) P
Combine all the three to form a single
V  nT
If the proportionality constant is called "R", then it becomes:
V = R nT P
Rearranging to a more familiar form:
PV = nRT
ideal gas equation

6. An "ideal gas" is one whose physical behavior is accurately described by the ideal-gas equation
The constant R is called the gas constant
The value and units of R depend on the units used in determining P, V, n and T
Temperature, T, must always be expressed on an kelvin temperature scale (K)
The quantity of gas, n, is normally expressed in moles
The units chosen for pressure and volume are typically atmospheres (atm) and liters (l), however, other units may be chosen
273 K (0C) and 1 atm are called standard temperature and pressure (STP)

7. Example: If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2, what is the pressure inside the ball at 24C?
1000
144 cm3
=0.144 L V=
n = 0.33 g N2 
1 mol N2
28 g N2
= mol
T = =297 K
R = L-atom/mol-K
P = ? V
P =
nRT =
 (297 K)
( L )
( mol)
 ( L.atm / K. mol)
= 2 atm

8. = P1V1 T1 P2V2 T2 PV = nRT PV R = nT P1V1 P2V2 = R = n1T1 n2T2
Relating the ideal-gas equation and the gas laws
PV = nRT
R = PV nT
= R =
P1V1
n1T1
P2V2
n2T2
Usually amount of gas does not change, n1 = n2 =
P1V1 T1
P2V2 T2
Combined gas law

9. Example 1: A gas initially at 4. 0 L, 1
Example 1: A gas initially at 4.0 L, 1.2 atm, and 66C undergoes a change so that its final volume and temperature become 1.7 L and 42 C. What is its final pressure ? Assume the number of moles remains unchanged =
P1V1 T1
P2V2 T2
P1= 1.2 atm
V1= 4.0 L
T1= =339 K
V2= 1.7 L
T2= =315 K
P2= ? atm =
P1V1T2
T1V2 P2
1.2 atm  4.0 L  315 K
339 K  1.7 L =
= 2.62 atm

10. 10.5 Further applications of the ideal-gas equation
Gas densities and molar mass
Density is defined as mass per unit volume
d = m V
m = mass in gram
M = molar mass
number of mole, n = m M
Ideal gas equation, PV = nRT
n = PV RT m M = PV RT d = PM RT M =
dRT P V m = PM RT or

11. dRT M = P dRT M = P 1.17 g/L  0.08206 L.atm/K.mol  294 K = 0.974 atm
Example : Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21C and torr. M =
dRT P
d = 1.17 g/L
P = torr= 740.0/760 = atm
T = (21+273)= 294 K
R = L. atm / K. mol M =
dRT P
1.17 g/L  L.atm/K.mol  294 K
0.974 atm =
= g/mol

12. ( ( Volumes of gases in chemical reactions Mole NaN3 = Gram NaN3=
Example : Following the reaction if an air bag of has a volume of 36 L and is to be filled with nitrogen gas at a pressure of 1.15 atom at a temperature of 26.0 C, how many grams of NaN3 must be decomposed?
2NaN3(s) Na(s) + 3N2(g)
V = 36.0 L
P = 1.15 atm
T = (26+273)= 299 K
R = L. atm / K. mol RT
n = PV =
 36.0 L
(1.15 atm
( L.atm / K. mol)  299 K
= 1.69 mol N2 (
(1.69 mol N2)
2 mol NaN3
3 mol N2
Mole NaN3 =
= 1.12 mol NaN3 (
65 g NaN3
1 mol NaN3
(1.12 mol NaN3)
Gram NaN3=
= 72.8 g NaN3

13. Gas mixtures and partial pressure
Partial pressure of a gas in a mixture of gases is the pressure which the gas would exert if it were the only gas present in the container.
Total pressure in the container is proportional to the number of gas particles
If each dot represents 1 mole of gas particles, then there are 48 moles of gas particles
no particles of gas a = 12
no particles of gas b = 36
total pressure 4 atm.

14. 12 mol  4 atm 48 mol = 1 atm 36 mol  4 atm 48 mol = 3 atm
the container holds only 12 moles of gas a particles
12 mol
48 mol 
4 atm
pressure exerted by gas a =
= 1 atm
the container holds only 36 moles of gas a particles
36 mol
48 mol 
4 atm
pressure exerted by gas b =
= 3 atm

15. + Dalton’s Law of Partial Pressure Ptotal = Pgas a + Pgas b =
Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.
Ptotal = Pgas a Pgas b = +

16. ( ( ( ( ( ( P = RT V n RT V P = n RT V P = n RT V P = n RT V RT V n
If total pressure of a mixture of gases is Pt
and partial pressures of individual gases are P1, P2, P3…
Pt=P1+P2+P3….
Let n1, n2, n3…. be the number of moles of each gas
Total number of moles nt=n1+n2+n3….
P = RT V n ( ; RT V (
P = 1 n ; RT V (
P = 2 n RT V (
P = 3 n ;
Pt=(n1+n2+n3…. ) RT V ( = RT V ( n t

17. ( ( ( ( Mole H2 = Mole N2 = = 2.88 atm Pt= P + P = 2.24 + 0.64
Example : What is the total pressure exerted by a mixture of 2.0 g of H2 and 8.0 g of N2 at 273K in a 10.0 L velsel? (
(2.0 g H2)
1 mol H2
2 g H2
Mole H2 =
= 1 mol H2 (
(8.0 g N2)
1 mol N2
28 g N2
Mole N2 =
= mol N2 RT V (
P = H2 n =
(1 mol)
( L.atm / K. mol)  273 K
= 2.24 atm
10.0 L RT V (
P = N2 n =
(0.286 mol)
( L.atm / K. mol)  273 K
10.0 L
= 0.64 atm
Pt= P + P = N2 H2
= 2.88 atm

18. Collecting gases over water
2KClO3 (s) KCl (s)+ 3O2 (g)
PT = P P O2
H2O

19. Example 2: Oxygen gas generated by the decomposition of potassium chlorate is collected over water. The volume of oxygen collected at 24C and atmospheric pressure of 762 mmHg is 128 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapor at 24 C is 22.4 mmHg.
PT = P + P O2
H2O
P = PT - P
H2O O2
= 762 mmHg – 22.4 mmHg
= mmHg
= mmHg 
1 atm
760 mmHg
= atm
P = atm
T = = 297 K
V = 128 mL = L
R = L.atm /K.mol
number of mole, n = m M
M = molar mass = 32 g / mol
m =mass in gram ? m M = RT
PV = nRT
m =
PVM RT =
atm  L  32 g / mol
L.atm / K.mol  297 K
= g