Pg 66-74
Also advanced material not found in text

 At a constant temperature and pressure, a given volume of gas always has the same number of particles.
 The coefficients of a balanced reaction is the same ratio as the volumes of reactants and products

2

2
For the above example, it is understood that half the volume of oxygen is needed to react with a given volume of carbon monoxide.
This can be used to carry out calculations about volume of gaseous product and the volume of any excess reagents.

 10 mL of ethyne (C
2
H
2
) is reacted with 50 mL of hydrogen to produce ethane (C
2
H
6
), calculate the total volume and composition of the remaining gas mixture, assuming constant T and P.
1 st get balanced equation: C
2
H
2
(g) + 2H
2
(g)  C
2
H
6
(g)
2 nd look at the volume ratios: 1 mol ethyne to 2 mol of hydrogen, therefore 1 vol to 2 vol
3 rd analyse: If all 10 mL of ethyne is used, it needs only 20 mL of hydrogen, therefore hydrogen is in excess by 50 mL20 mL = 30 mL. In the end you’ll have 10 mL
Ethane and the leftover 30 mL hydrogen

 The temperature and pressure are specified and used to calculate the volume of one mole of gas.
 Standard temperature and pressure (STP) is at sea level 1 atm = 101.3 kPa and 0 o C = 273 K this volume is 22.4 L
Molar gas volume, V molecules of gas m
.
It contains 6.02 x 10 23

(a) Calculate how many moles of oxygen molecules are there in 5.00 L at STP n= V
STP
= 5.00 L = 0.223 mol
22.4 L/mol 22.4 L/mol
(b) How many oxygen molecules are there in 5.00L at
STP?
0.223 mol x 6.02 x 10 23 molecules = 1.34 x 10 23 molecules mol

 Mathematical relationship between the volume of a gas (V) and the number of moles of gas present (n).
n1 = n2
V1 V2
One mole of gas occupies the same volume as one mole of another gas at the same temp and pressure
Molar volume’s unit is L/mol

 (a) How many moles are present in 44.8 L of methane gas (CH
4
) at STP? (b) What is the mass of the gas? (c) How many molecules are present?
22.4 L = 1 mol of any gas at STP
So 44.8L x 1mol = 2.00 mol
22.4L

 2.00 mol CH
4 x 16.05 g = 32.10 g CH
4 mol CH
4
(c) How many molecules are present?
 2.00 mol x 6.02 x10 23 molecules
= 1.20 x10 24 molecules mol

 Boyle noticed that the product of the volume of air times the pressure exerted on it was very nearly a constant, or PV=constant.
 If V increases, P decreases proportionately and vice versa. (Inverse proportions)


Temperature must be constant.
Example : A balloon under normal pressure is blown up (1 atm), if we put it under water and exert more pressure on it (2 atm), the volume of the balloon will be smaller (1/2 its original size)
 P
1
V
1
=P
2
V
2

 Gas expands (volume increases) when heated and contracts (volume decreases) when cooled.
 The volume of a fixed mass of gas varies directly with the Kelvin temperature provided the pressure is constant. V= constant x T
 V
1
= V
2
T
1
T
2

 The pressure of a gas increases as its temperature increases.
 As a gas is heated, its molecules move more quickly, hitting up against the walls of the container more often, causing increased pressure.
 P
1
= P
2
T
1
T
2

 P
1
V
1
= P
2
V
2
T
1
T
2
T must be in Kelvins, but P and V can be any proper unit provided they are consistently used throughout the calculation

 If a given mass of gas occupies a volume of
8.50 L at a pressure of 95.0 kPa and 35 o C, what volume will it occupy at a pressure of
75.0 kPa and a temperature of 150 o C?
1 st convert o C to K:
35 + 273 = 308 K 150 + 273 = 423 K
2 nd rearrange equation and solve problem:
V
2
= V
1 x P
1 x T
2
= 8.50 x 95.0 x 423 = 14.8 L
P
2 x T
1
75.0 x 308

 Kelvin temperature is proportional to the average kinetic energy of the gas molecules.
 It is a measure of random motion of the gas molecules
 More motion = higher temperature

 Ideal behaviour is when a gas obeys Boyle’s,
Charles’ and Gay-Lussac’s laws well
 At ordinary temperature and pressures, but there is deviation at low temperature and high pressures

 where all collisions between molecules are perfectly elastic and in which there are no intermolecular attractive forces.
 Its like hard spheres bouncing around, but
NO interaction.

 PV = nRT
 P= pressure (kPa)
 Volume = (L)
 n= number of moles
 R=universal gas constant =8.3145 J mol -1 K -1
 T= temperature (K)

3.376 g of a gas occupies 2.368 L at 17.6 o C and a pressure of 96.73 kPa, determine its molar mass.
PV= nRT rearrange equation for n
Temp = 17.6 o C + 273 = 290.6 K n= PV/RT
= (96.73 x 2.368) / (8.314 x 290.6)
= 0.09481 mol
Molar mass = mass/ mole
= 3.376 g / 0.09481 mol
= 35.61 g/mol

Visit this flash to see how temp, pressure and volume are related

 Pg 73 # 38 – 43