2. 1. BOYLE’S LAW
In 1662 Robert Boyle studied the relationship
between volume and pressure of a gas of
fixed amount at constant temperature.
Boyle's law was published in the same
year.
3. 1. BOYLE’S LAW
o The inverse relationship between pressure and volume.
o It states that the volume of a sample of gas changes
inversely with the pressure of the gas as long as the
temperature and the amount of the gas remain
constant.
o When the pressure of the gas decreases, the volume of
the gas increases.
o When the pressure of a gas increases, the volume of
the gas decreases.
5. BOYLE’S LAW- SAMPLE PROBLEMS
1. A tank of nitrogen has a volume of 14.0 L and a
pressure of 760.0 mmHg. Find the volume of the
nitrogen when its pressure is changed to 400.0 mmHg
while the temperature is held constant.
V1= 14.0 L P1= 760 mmHg
V2= ? P2= 400 mmHg
P1V1=P2V2----- V2= P1V1--- (760mmHg)(14.0L)
P2 400mmHg
Answer: 26.6 L
6. BOYLE’S LAW- SAMPLE PROBLEMS
2. A gas occupies 11.2 liters at 0.860 atm. What is the
pressure if the volume becomes 15.0 L?
V1= 11.2 L P1= 0.860 atm
V2= 15.0 L P2= ?
P1V1=P2V2----- P2= P1V1--- (0.860atm)(11.2L)
V2 15.0L
Answer: 0.642 atm
8. 2. CHARLE’S LAW
o Charles's law, or the relationship between volume
and temperature, was found in 1787 by Jacques
Charles, a French physicist and balloonist .
o The statement of Charles's law is : the volume (V) of
a given mass of a gas, at constant pressure (P), is
directly proportional to its temperature (T).
10. CHARLE’S LAW- SAMPLE PROBLEMS
1. A container holds 50.0 mL of nitrogen at 25° C and a
pressure of 736 mm Hg. What will be its volume if the
temperature increases to 35° C?
V1 = 50.0 mL T1 = 25° C + 273 = 298 K
V2 = ? T2 = 35° C + 273= 308 K
V1= V2 ----- V2 = V1 x T2/T1
T1 T2 V2 = 50.0 mL x 308 K/298 K
Answer: 55.9 mL
11. CHARLE’S LAW- SAMPLE PROBLEMS
2. Find the final temperature of a 2.00 L gas sample at 20.0 ° C
cooled until it occupies a volume of 5.00 L.
V1 = 2.00 L T1 = 20° C + 273 = 293 K
V2 = 5.00 L T2 = ?
V1= V2 ----- T2 = T1 x V2/V1
T1 T2 V2 = 293 K x 5.00 L /2.00 L
Answer: 732.5 K
13. 3. GAY- LUSSAC’S LAW
o Gay-Lussac's law, Amontons' law or the pressure law was
found by Joseph Louis Gay-Lussac in 1808.
o Temperature and pressure relationship.
o It states that the pressure of a gas is directly related to its
Kelvin temperature.
15. 3. GAY- LUSSAC’S LAW- SAMPLE PROBLEMS
1. Determine the pressure change when a constant volume
of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.
Solution: P1= 1 atm T1=20 °C +273= 293K
P2= ? T2=30 °C +273=303 K
---- P2= P1T2/T1 ----1.00 atm x 303/293
Answer= 1.03 atm
16. 3. GAY- LUSSAC’S LAW- SAMPLE PROBLEMS
2. A gas has a pressure of 699.0 mmHg at 40.0 °C. What is
the temperature at standard pressure?
Solution: P1= 699.0 mmHg T1=40 °C +273= 313K
P2= 760mmHg T2= ?
---- T2= P2T1/P2----760mmHg x 313/699mmHg
Answer= 340.3 K
18. 4. AVOGADRO’S LAW
o Avogadro's law (hypothesized in 1811) states that
the volume occupied by an ideal gas is directly
proportional to the number of molecules of the gas
present in the container.
o This gives rise to the molar volume of a gas, which
at STP (273.15 K, 1 atm) is about 22.4 L.
20. 4. AVOGADRO’S LAW-SAMPLE PROBLEMS
1. 5.00 L of a gas is known to contain 0.965 mol. If the
amount of gas is increased to 1.80 mol, what new volume
will result (at an unchanged temperature and pressure)?
Solution: V1= 5.00 L n1= o.965 mol
V2= ? n2= 1.80 mol
--V2=V1n2/n1-- V2=5Lx1.80 mol
0.965 mol
Answer: 9.33L
21. 4. AVOGADRO’S LAW-SAMPLE PROBLEMS
2. A 6.0 L sample at 25°C and 2.00 atm of pressure contains 0.5
mole of a gas. If an additional 0.25 mole of gas at the same
pressure and temperature are added, what is the final total
volume of the gas?
Solution: V1= 6.0 L n1= 0.5 mol
V2= ? n2= 0.5+0.25 mol=0.75 mol
--V2=V1n2/n1-- V2=6.0Lx0.75 mol
0. 5 mol
Answer: 9L
23. 5. COMBINED GAS LAW
o The Combined gas law or General Gas Equation is
obtained by combining Boyle's Law, Charles's law,
and Gay-Lussac's Law. It shows the relationship
between the pressure, volume, and temperature for
a fixed mass (quantity) of gas.
25. 5. COMBINED GAS LAW
1. A gas has a volume of 800.0 mL at −23.0 °C and 300.0 torr.
What would the volume of the gas be at 227.0 °C and 600.0 torr
of pressure?
Solution:
V2= P1V1T2 --- V2= (300 torr)(800mL)(500K)
T1P2 (250 K)(600 torr)
Answer: = 800 mL
P1 = 300.0 torr P2 = 600.0 torr
V1 = 800.0 mL V2 = x
T1 = 250. K T2 = 500. K
26. 5. COMBINED GAS LAW
1. At conditions of 785.0 torr of pressure and 15.0 °C temperature, a
gas occupies a volume of 45.5 mL. What will be the volume of the
same gas at 745.0 torr and 30.0 °C?
Solution:
V2= P1V1T2 --- V2= (785.5 torr)(45.5mL)(303K)
T1P2 (288 K)(745.0 torr)
Answer: = 50.38 mL
P1 = 785.0 torr P2 = 745.0 torr
V1 = 45.5 mL V2 = ?
T1 = 288K T2 = 303K
27. 6. IDEAL GAS LAW
o The law that describes the pressure, volume,
temperature, and number of moles of a gas.
o Avogadro's law and the combined gas law
develops into the ideal gas law
28. 6. IDEAL GAS LAW
PV= nRT
Where:
P= pressure of the gas
V= volume of the gas
n= amount of gas
R= ideal gas constant
T= temperature of the gas
29. 6. IDEAL GAS LAW
1. Calculate the pressure exerted by a 0.25 mole sulfur
hexafluoride in a steel vessel having a capacity of
1,250 at 70.0°C.
Solution: n=0.25 mole T= 70.0+273=343K
R=0.0821L.atm/mole.K V=1.25L
PV= nRT---(0.25mol)(0.0821 L.atm/mol.K)(343K)
1.25 L
Answer= 5.63 atm
30. 6. IDEAL GAS LAW
2. Fermentation of glucose produce gas in the form of
carbon dioxide. How many moles of carbon dioxide is
produced if 0.78L of carbon dioxide at 20.1 °C and 1.00
atm was collected during the process?
Solution: P= 1.00atm T= 20.1+273=293.1K
R=0.0821L.atm/mole.K V=0.78L
PV= nRT---n=PV -- (1.00atm)(0.78K)
RT (0.0821L.atm/mol.K)(293.1K)
Answer= 0.032 mol
