Gas Laws

Elements that exist as gases at 250C and 1 atmosphere

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.

Force
Pressure = Area

Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Barometer

10 miles

4 miles
Sea level

0.2 atm

0.5 atm
1 atm

As P (h) increases

V decreases

Boyle’s Law

P  1/V
P x V = constant
P1 x V1 = P2 x V2

Constant temperature
Constant amount of gas

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?

P1 x V1 = P2 x V2
P1 = 726 mmHg

P2 = ?

V1 = 946 mL

V2 = 154 mL

P1 x V1
726 mmHg x 946 mL
P2 =
=
= 4460 mmHg
154 mL
V2

As T increases

V increases

Variation of gas volume with temperature at constant pressure.

Charles’ &
Gay-Lussac’s
Law

VT
V = constant x T
V1/T1 = V2/T2

Temperature must be in Kelvin
T (K) = t (0C) + 273.15

A sample of carbon monoxide gas occupies 3.20 L at
125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1/T1 = V2/T2
V1 = 3.20 L

V2 = 1.54 L

T1 = 398.15 K

T2 = ?

V2 x T1
T2 =
=
V1

1.54 L x 398.15 K
3.20 L

= 192 K

Avogadro’s Law
V  number of moles (n)
V = constant x n
V1/n1 = V2/n2

Constant temperature
Constant pressure

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2
1 mole NH3

4NO + 6H2O
1 mole NO

At constant T and P
1 volume NH3

1 volume NO

Ideal Gas Equation
1
Boyle’s law: V (at constant n and T)
P
Charles’ law: V  T(at constant n and P)
Avogadro’s law: V n(at constant P and T) nT V
P

nT nT V = constant x
=R
P
P

R is the gas constant

PV = nRT

The conditions 0 0C 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
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)

What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = 273.15 K

PV = nRT nRT V=
P

P = 1 atm
1 mol HCl n = 49.8 g x
= 1.37 mol
36.45 g HCl

1.37 mol x 0.0821
V=
V = 30.6 L

L•atm mol•K 1 atm

x 273.15 K

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to
85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT

n, V and R are constant

nR
= P = constant
T
V
P1
P2
=
T1
T2

P1 = 1.20 atm
T1 = 291 K

P2 = ?
T2 = 358 K

T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1

Density (d) Calculations
PM
m d= =
V
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 dRT M=
P

d is the density of the gas in g/L

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g)

6CO2 (g) + 6H2O (l)

g C6H12O6

mol C6H12O6

5.60 g C6H12O6 x

6 mol CO2
1 mol C6H12O6 x = 0.187 mol CO2
180 g C6H12O6
1 mol C6H12O6

V=

nRT
=
P

mol CO2

V CO2

L•atm x 310.15 K mol•K 1.00 atm

0.187 mol x 0.0821

= 4.76 L

Dalton’s Law of Partial Pressures

V and T are constant

P1

P2

Ptotal = P1 + P2

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

nART
PA =
V

nA is the number of moles of A

nBRT
PB =
V

nB is the number of moles of B

PT = PA + PB
PA = XA PT

nA
XA = nA + nB
PB = XB PT
Pi = Xi PT

nB
XB = nA + nB

A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37