1. 1.4: Gas Properties, Basic Gas Laws
1/17/17

2. can expand and contract
Part 1: Gas Properties
gases have many properties that are very different from solids and liquids:
The first three rows in the table above sum up what we covered in our notes on Phases of Matter (1.3). The last two rows are explained next:
unlike liquids and solids, gases have what is known as an “indefinite” volume.
this means that the volume of a gas is dependent upon more than just how many molecules are present in the sample of gas.
Property
Solids & Liquids
Gases
particle spacing
particle movement
kinetic energy
volume
exhibit pressure?
close together
vibrational or fluid
low to medium
definite amount no
very spread out
very fast, random
high to very high
can expand and contract
yes

3. this means that the volume of a gas is dependent upon more than just how many molecules are present in the sample of gas.
since gas particles move very quickly and spread out in all directions (a process called diffusion), gases can occupy many different volumes for the same amount of gas particles. They will take the shape and volume of any container they are placed in.
the two other variables that determine the volume of a gas are temperature (T) and pressure (P).
pressure = the amount of force exerted per unit of area.
according to the Kinetic Molecular Theory (KMT), gas particles are continuously moving and colliding with any object in their path, including the sides of their containers.
these collisions result in pressure inside containers, such as a balloon. The arrows in the balloon show these forces.

4. pressure = the amount of force exerted per unit of area.
according to the Kinetic Molecular Theory (KMT), gas particles are continuously moving and colliding with any object in their path, including the sides of their containers.
these collisions result in pressure inside containers, such as a balloon. The arrows in the balloon show these forces.
expandable containers, like balloons, relieve pressure by increasing in size. Fixed-volume containers, such as spray cans, decrease internal pressure as gas molecules are released.
pressure is measure in Pascals (Pa), or kilopascals (kPa).
gas laws help show the effect of pressure on the volume and temperature of a gas.

5. millimeters of mercury
Part 1I: What Is a Gas Law?
the gas laws are simple, mathematical relationships between the pressure (P), volume (V), and temperature (T) of a gas.
the 3 basic gas laws:
Boyle’s law and Gay-Lussac’s law involve pressure. Most people are not familiar with the many units pressure can be measured in (except maybe psi). So here they are:
remember: units of volume = milliliters (mL), liters (L), and cubic centimeters (cm3).
Charles’s Law
V1 = V2
T1 T2
Gay-Lussac Law
P1 = P2
T1 T2
Boyle’s Law
P1V1 = P2V2
Unit
Abbr.
atmospheres
atm
millimeters of mercury
mmHg
pounds per square inch
psi
kilopascals
kPa

6. Variables involved Which gas law? Prediction of answer
Part III: Boyle’s Law (1662)
Boyle’s Law states that the pressure of a fixed mass of gas varies inversely with the volume at a constant temperature.
this means if you compare the initial volume and pressure of a gas with the new conditions of the gas, you will get an inverse relationship every time
P1 and V1 indicate initial (or starting) conditions
P2 and V2 indicate new (or final) conditions
for these problems, this is what’s important:
Boyle’s Law
P1V1 = P2V2 P V
Variables involved
Which gas law?
Prediction of answer

7. P1V1 = P2V2 V2 = (150 kPa)(14.3 L) = ____ ____ 250 kPa P2 P2
Ex1: Using 14.3 L of N2 as the initial volume, calculate the volume that would result if the pressure was raised from 150 kPa to 250 kPa.
P1 = __________
V1 = __________
P2 = __________
V2 = __________
Ex2: A 5.6 L weather balloon rises into the atmosphere where it volume increases to 7.9 L. If the initial pressure in the balloon was 13.5 psi, what is the new pressure inside the balloon?
150 kPa
14.3 L
250 kPa ?
P1V1 = P2V2
____ ____
P P2
V2 = (150 kPa)(14.3 L) =
250 kPa
V2 = 150 14.3 ÷ 250 =
V2 = P1V1 P2
V2 = L
P2 = (13.5 psi)(5.6 L) =
7.9 L
13.5 psi
5.6 L ?
7.9 L
____ ____
V V2
P1V1 = P2V2
P2 = 13.5 5.6 ÷ 7.9 =
P2 = P1V1 V2
P2 = psi

8. V1T2 = V2T1 ____ ____ V2 = (5.0 L)(316 K) = T1 T1 310 K
Part IV: Charles’s Law (1787)
Charles’s Law states that the volume of a fixed mass of gas varies directly with the temperature at a constant pressure.
this means that as the volume of gas increases, so does the temperature
Ex3: A sample of gas occupied a volume of 5.0L at a temp of 37.0C. If the temp were to increase by 6C, what would be the volume of the gas under this new condition?
V1 = __________
T1 = ____C____K
V2 = __________
T2 = ____C____K
Charles’s Law
V1 = V2
T1 T2 V T
5.0 L 37 ? 43
____ ____
T T1
V1T2 = V2T1
Cross-multiplied version of Charles’s Law (easier to rearrange than original)
V2 = (5.0 L)(316 K) =
310 K
310
316
V2 = 5.0 316 ÷ 310 =
V2 = V1T2 T1
V2 = L

9. V2 = V1T2 V2 = (4000 L)(379 K) = T1 312 K V2 = 4000 379 ÷ 312 =
Ex4: The gas in a 4000 L hot-air balloon is heated from C to 106C. What new volume does this hot-air balloon have after heating?
V1 = __________
T1 = ____C____K
V2 = __________
T2 = ____C____K
Part V: Gay-Lussac’s Law (1802)
Gay-Lussac’s Law states that the pressure of a fixed mass of gas varies directly with the temperature at a constant volume.
this means that as the pressure of gas increases, so does the temperature
Ex5: The gas left in a used aerosol can is at a pressure of kPa at 17C. If the can is thrown into a fire, what will the pressure be inside the can at 1045C?
Charles’s Law
V1T2 = V2T1
4000 L 39 ?
106
V2 = V1T2 T1
V2 = (4000 L)(379 K) =
312 K
312
379
V2 = 4000 379 ÷ 312 =
V2 = L
Gay-Lussac Law
P1 = P2
T1 T2 P T

10. P1T2 = P2T1 ____ ____ T1 T1 P2 = (125.3 kPa)(1318 K) = 290 K P2 = P1T2
Ex5: The gas left in a used aerosol can is at a pressure of kPa at 17C. If the can is thrown into a fire, what will the pressure be inside the can at 1045C?
P1 = __________
T1 = ____C____K
P2 = __________
T2 = ____C____K
Ex6: What temperature is required to raise the pressure in a helium gas tank from 3200 mmHg to 9800 mmHg if the original temp. of the gas was 28C?
P2 = _________
Gay-Lussac Law
P1 T2= P2T1
P1T2 = P2T1
125.3 kpa 17 ?
1045
____ ____
T T1
P2 = (125.3 kPa)(1318 K) =
290 K
290
1318
P2 = P1T2 T1
P2 = 125.3 1318 ÷ 290 =
P2 = kPa
3200 mmHg 28
9800 mmHg ?
P1T2 = P2T1
T2 = (9800 mmHg)(301 K) =
3200 mmHg
____ ____
P P1
301 ?
T2 = 9800 301 ÷ 3200 =
T2 = P2T1 P1
T2 = K = C