1. The Gas Laws

2. Apparatus for Studying the Relationship Between Pressure and Volume of a Gas
As P (h) increases
V decreases

3. Boyle’s Law P ∝ 1/V Constant temperature P x V = constant
Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2

4. Example 1
A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kPA, what will the pressure be at 2.5L?
Given:
V1 = 4.0L V2 = 2.5L
P1= 210 kPa P2 = ?
P1V1 = P2V2
(210 kPa) × (4.0L) = P2 × (2.5L)
P2 = 336 kPa ≈ 340 kPa

5. Example 2
A sample of neon gas occupies 0.200L at atm. What will be its volume at 29.2 kPa pressure?
P1V1 = P2V2
Given:
V1 = 0.200L V2 = ?
P1= atm P2 = 29.2 kPa
(87.1 kPa) × (0.200L) = (29.2 kPa) × V2
V2 = 0.597L
**Units must match for each variable (doesn’t matter which one is converted)
0.860 atm kPa
1 atm
= 87.1 kPa

6. Day 2
Comparing volume and temperature
Keep pressure constant

7. Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases

8. Charles’s Law 𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 **Temperature must be
in Kelvin
K = 0C + 273
𝑉 1 𝑇 1 = 𝑉 2 𝑇 2
𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Constant pressure
Constant amount of gas
𝑉∝𝑇

9. Example 1
A gas sample at 40.0°C occupies a volume of 2.32L. If the temperature is raised to 75.0°C, what will the volume be, assuming the pressure remains constant?
Given:
T1 = 40.0°C = 313K T2 = 75.0°C = 348K
V1= 2.32L V2 = ?
V V2
T T2 =
2.32L V2 =
V2 = 2.58L

10. Example 2
A gas sample at 55.0°C occupies a volume of 3.50L. At what new temperature in Celcius will the volume increase to 8.00L?
Given:
T1 = 55.0°C = 328.0K T2 = ?
V1= 3.50L V2 = 8.00L
V V2
T T2 =
3.50L T2 =
T2 = 750K-273=477°C

11. Day 3

12. Gay-Lussac’s Law 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃∝𝑇 Constant volume
𝑃 1 𝑇 1 = 𝑃 2 𝑇 2
𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Constant volume
Constant amount of gas
**Temperature must be
in Kelvin
K = 0C + 273

13. Example 1
The pressure of a gas in a tank is 3.20 atm at 22.0°C. If the temperature rises to 60.0°C, what will be the gas pressure in the tank?
Given:
P1 = 3.20 atm P2 = ?
T1= 22.0°C = 295.0K T2 = 60.0°C =333.0K
P P2
T T2 =
3.20 atm P2
295.0K K =
P2 = 3.61 atm

14. Example 2
A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C?
Given:
P1 = 665 torr P2 = ?
T1= 22.0°C = 295K T2 = 44.6°C =317.6K
P P2
T T2 =
665 torr P2
295K K =
P2 = 716 torr

15. STOP

16. Avogadro’s Law and Combine Gas Law

17. What is the relationship between number of moles and volume?

18. What is the same about these gases? What is different about these gases?

19. Avogadro’s Law 𝑉 𝑛 = constant 𝑉 1 𝑛 1 = 𝑉 2 𝑛 2
V ∝ n (number of moles)
Constant temperature
Constant pressure
𝑉 𝑛 = constant
𝑉 1 𝑛 1 = 𝑉 2 𝑛 2
1 mol of gas at STP = 22.4L

20. Gas Stoichiometry
When gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts (mole ratios) but also relative volumes (volume ratios).

21. Example 1
5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and pressure.
Given:
V1 = 5.00 L V2 = ?
n1= mol n2 = 1.80 mol
V V2
n n2 =
5.00 L V2
0.965 mol mol =
V2 = 9.33 L

22. Calculate the volume that 0.881 mol of gas at STP will occupy.
Example 2
Calculate the volume that mol of gas at STP will occupy.
0.881 mol 22.4 L
1 mol
= 19.7 L

23. How many grams of carbon dioxide gas are in a 0.75 L balloon at STP?
Example 3
How many grams of carbon dioxide gas are in a 0.75 L balloon at STP?
0.75 L CO mol CO g CO L CO2 1 mol CO2
= 1.5 g CO2

24. Example 4
What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C3H8)? Assume constant temperature and pressure.
𝐶 3 𝐻 𝑂 2 →3 𝐶 𝑂 𝐻 2 𝑂
4.00 L C3H8 5 L O2
1 L C3H8
= 20.0 L

25. Gas Law Summary

26. Charles’s Law
Gay-Lussac’s Law

28. Combined Gas Law Combine all 4 to make one master equation:
𝐵𝑜𝑦𝑙 𝑒 ′ 𝑠 𝑙𝑎𝑤:𝑃×𝑉=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑇
𝐶ℎ𝑎𝑟𝑙 𝑒 ′ 𝑠 𝑙𝑎𝑤: 𝑉 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑃
𝐺𝑎𝑦−𝐿𝑢𝑠𝑠𝑎 𝑐 ′ 𝑠 𝑙𝑎𝑤: 𝑃 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛 𝑎𝑛𝑑 𝑉
𝐴𝑣𝑜𝑔𝑎𝑑𝑟 𝑜 ′ 𝑠 𝑙𝑎𝑤: 𝑉 𝑛 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃 𝑎𝑛𝑑 𝑇)
Combine all 4 to make one master equation:
Putting all of our gas laws together we get…Because we will typically not change the number of moles in our reactions, we can say that n1=n2 so then we can now take out the number of moles and get a combined gas law
𝑃 1 𝑉 1 𝑛 1 𝑇 1 = 𝑃 2 𝑉 2 𝑛 2 𝑇 2
𝑃𝑉 𝑛𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2

29. Example 1
A gas at 110 kPa and 30.0˚C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0˚C and the pressure increased to 440 kPa, what is the new volume?
Given:
P1 = 110 kPa P2 = 440 kPa
T1 = 30.0˚C =303K T2 = 80.0˚C = 353K
V1 = 2.00L V2 = ?
110 (2.00) 440 (V2) =
V2 = 0.583L ≈ 0.58L
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2

30. Example 2
An unopened bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm and temperature of 5.00˚C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and temperature changes to 1.52 atm and 2.90˚C, what will be the volume of gas in the bottle?
1.30 (46.0) (V2)
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 =
Given:
V1 = 46.0 mL V2 = ?
P1 = 1.30 atm P2 = 1.52 atm
T1 = 5.00˚C = 278K T2 = 2.90˚C = 275.9K
V2 = 39.0mL

31. STOP

32. Warm Up
A sample of gas of unknown pressure occupies 3.2 L at a temperature of 302 K. The sample of gas is then tested under known condition and has a pressure of 22.6 kPa and occupies 1.2 L at 310 K. What was the original pressure of the gas?

34. The Ideal Gas Law Ideal Gas Real Gas
Follows some gas laws under some conditions of temperature and pressure.
Does not conform to the Kinetic Molecular Theory. Real gases have a volume and attractive and repulsive forces.
A real gas differs from an ideal gas the most at low temperature, high pressure, and high concentration.
Low T gas is moving less
High P gas is more attractive
High Conc has more collisions
Ideal Gas
Follows all gas laws under all conditions of temperature and pressure.
Follows all conditions of the Kinetic Molecular Theory (KMT)
An ideal gas does not exist in real life

35. (R is the universal gas constant)
Ideal Gas law
𝑃𝑉 𝑛𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(R is the universal gas constant)
PV = nRT
R = PV nT =
(1 atm)(22.4L)
(1 mol)(273K)
𝑅 = 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝐾 = 8.314𝐿 𝑘𝑃𝑎 𝑚𝑜𝑙 𝐾 = 62.4𝐿 𝑚𝑚𝐻𝑔 𝑚𝑜𝑙 𝐾 = 62.4𝐿 𝑡𝑜𝑟𝑟 𝑚𝑜𝑙 𝐾

38. Example 1
Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm.
PV=nRT
V = 3.00L
T = 298K
P = 1.50atm
n = ?
R = L atm
mol K
(1.50)(3.00) = n (0.0821) (298)
n= mol

39. Example 2
What will the pressure (in kPa) be when there are mol of gas in a 5.00L container at 17.0˚C?
V = 5.00L
T = 17.0˚C = 290K
P = ? kPa
n = mol
R = L kPa
mol K
PV=nRT
P(5.00) = (0.400)(8.314)(290)
P= 193 kPa

41. STOP

42. Stoichiometry

43. (3.00 atm)(5.00 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K)
Example 1
If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced.
𝑁 2 + 3𝐻 2 →2𝑁 𝐻 3
PV=nRT
(3.00 atm)(5.00 L) = (n)( 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K)
n= mol N2
0.613 mol N mol NH g NH mol N mol NH3
= 20.9 g NH3

44. (2.00 atm)(6.75 L) = (n)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K)
Example 2
How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a pressure of 2.00 atm and 298K? CaCO3(s)  CO2(g) + CaO(s)
PV=nRT
(2.00 atm)(6.75 L) = (n)( 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(298 K)
n= mol CO2
0.552 mol CO2 1 mol CaCO g CaCO mol CO mol CaCO3
= 55.2 g CaCO3

45. (3.50 atm)(V) = (1.13 mol)(0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(225 K)
Example 3
How many liters of chlorine will be needed to make 95.0 grams of C2H2Cl4 at 3.50 atm and 225K? 2Cl2(g) + C2H2(g)  C2H2Cl4(l)
95.0 g C2H2Cl4 1 mol C2H2Cl mol Cl g C2H2Cl4 1 mol C2H2Cl4
= 1.13mol Cl2
PV=nRT
(3.50 atm)(V) = (1.13 mol)( 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 𝑘 )(225 K)
V= 5.97 L Cl2

46. Graphic Organizer for PV=nRT and Stoichiometry

47. d is the density of the gas in g/L
Density (d) Calculations
d = PM RT n V = P RT m
M V = P RT so
n = m M
m is the mass of the gas in g
d = m V
M is the molar mass of the gas
d is the density of the gas in g/L
Molar Mass (M ) of a Gaseous Substance
dRT P
M =

48. Example 1
What is the molar mass of a pure gas that has a density of 1.40gL at STP?
d = PM RT

49. Example 2
Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.

50. Test Review
Using the principles behind Graham’s Law, knowing that M H2 = 2.02 g/mol and M N2 = g/mol which gas would diffuse faster in a closed container?
Hydrogen
2. Why do real gases deviate from ideal gas behavior at high pressure, low temperature and high concentration ?
High P-increased collisions, increase temp
High C- particles collisions increase and so doespressure
Low T-slow particles down, less collisions and pressure
3. When can you use 22.4 L = 1 mole and when can you use volume to volume ratios?
At STP = 1 atm and 273 K

51. Test Review
4. Given the two problems below, which problem starts with stoichiometry and then uses PV=nRt, and which problem starts with PV=nRt and then uses stoich?
N2(g) + 3H2 (g) 2NH3(g)
A. What is the mass of NH3 gas is produced at 55.5oC and 670 torr, when 1L of H2 reacts with an excess of N2?
PV=nRT, then Stoich
B. What volume of NH3 gas is produced at 55.5oC and 670 torr when g of H2 reacts with N2?
Stoich, then PV=nRT