1. Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

2. Volume and Moles
How does adding more molecules of a gas change the volume of the air in your glove? If the glove has a leak, how does the loss of air molecules change the volume?

3. Avogadro’s Law Remember you must convert units to moles and liters.
We can calculate volume or number of moles using this equation.
Volume of gas is Directly proportional to Number of moles (n) of gas
V = an (at constant T and P) V = a n
(V) is volume of gas in Liters
(a) is a constant V = V2
(n) is the number of moles n n2
Initial Final
Remember you must convert units to moles and liters.

4. Learning Check #1
True (1) or False(2) 1.___The Pressure exerted by a gas at constant Volume is not affected by the Temp. of the gas. 2.___ At constant Pressure, the Volume of a gas is directly proportional to the absolute Temp. 3.___ At constant Temperature, doubling the Pressure will cause the Volume of a gas sample to decrease to one-half its original V.

5. Solution #1
True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of a gas sample to decrease to one-half its original V.

6. Learning Check #2
If 3.20 g of O2 gas occupies a volume of 2.24 L at 0°C and 1.0 atm,
what volume would be occupied by 32.0 g of O2 gas under same conditions?
Remember you must convert units to moles and liters.
Prepare a data table and convert units:
DATA TABLE
Initial conditions Final conditions
V1 = L V2 = ? L
n1 = g O2 (1 mol/ 32g) = .1 mol O n2 = g (1 mol/ 32g) = 1 mol O2
V = V2 V2 = V1 n2
n n n1
Initial Final

7. Solution #2 V2 = 2.24 L 1.0 mol O2 22.4 L O2 or 2.24 x 101 L O2
DATA TABLE
Initial conditions Final conditions
V1 = L V2 = ? L
n1 = g O2 (1 mol/ 32g) = .1 mol O n2 = g (1 mol/ 32g) = 1 mol O2
V = V2 V2 = V1 n2
n n n1
Initial Final
V2 = L 1.0 mol O L O2 or 2.24 x 101 L O2
.1 mol O2

8. Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

9. Ideal Gas Law Can be combined to form PV = nRT
The equality for the 3 variables involved in
Boyle’s, Charles’, and Avogadro’s law.
Can be combined to form PV = nRT
You must convert all units to moles, liters, Kelvin and atm.
(P) is the Pressure in atm
(V) is volume in Liters
(n) is the number of particles in moles
(T) is Temperature in Kelvin
(R) is the universal gas constant;
R = L atm / mol K

10. Ideal Gases PV = nRT R is known as the Universal gas constant STP
Describes how most real gases behave at
STP
273 K (0°C) and pressures of 1 atm
1mol of any gas at STP takes up 22.4 L of volume
Using STP conditions:
(P) (V)
R = PV = (1.00 atm) (22.4 L) = L- atm
nT (1mol) (273K) mol - K
(n) (T)

11. Learning Check #1
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as
an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23.0°C,
what is the pressure (atm) in the tank in the dentist office?
Rearrange Ideal Gas Law Equation to solve for P:
PV = nRT P = nRT V
Remember to convert all units, Kelvin, liters, mol, and atm.
Prepare a data table:
DATA TABLE
P = ? atm V = L n = mol
R = L atm T = 23 °C + 273= 296 K
mol K

12. Solution #1 V Substitute values of n, R, T and V and solve for P n R T
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic.
If 2.86 mol of gas occupies a 20.0 L tank at 23.0°C,
what is the pressure (atm) in the tank in the dentist office?
DATA TABLE
P = ? atm V = L n = mol R = L atm T = 23 °C + 273= 296 K
K mol
Rearrange ideal gas law for unknown P: P = nRT V
Substitute values of n, R, T and V and solve for P
n R T
P = mol L atm K
20.0 L K-mol
= atm

13. Learning Check #2
A 5.00 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg.
How many grams of oxygen are in the cylinder?
Rearrange Ideal Gas Law Equation to solve for n:
PV = nRT n = PV RT
Remember to convert all units, Kelvin, liters, mol, and atm.
Prepare a data table:
DATA TABLE
P = mm Hg (1 atm/760 mm Hg) = atm V = L
R = L atm T = 20°C = 293 K n = ? Mol O2
mol K

14. Solution #2
DATA TABLE
P = mm Hg (1 atm/760 mm Hg) = atm V = L
R = L atm T = 20°C + 273= 293 K n = ? mol O2
mol K
Solve ideal gas equation for n (moles) n = PV RT
= (.967 atm) (5.00 L) (mol K) = mol
(293 K) ( L atm)
= mol O2 x g O2 = g O2
1 mol O2

15. Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

16. Combined Gas Laws
Boyle’s - P1V1 = P2V2 Charles’ - V1 = V2 T1 T2 Combined Gas Law P1V1 = P2V2 T1 T2 You can rearrange the equation to solve for pressure, volume or temperature. Example: Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

17. Combined Gas Laws Problem #1
A sample of He2 has a volume of L, pressure of atm
and a temperature of 29.0°C. What is the new temperature (K) of
the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
Rearrange Combined Gas Law Equation to solve for T2:
P1V1T2 = P2V2T1 T2 = P2V2T1
P1V1
Remember to convert all units, Kelvin, liters, mol, and atm:
DATA TABLE
P1 = atm P2 = atm
V1 = L V2 = mL = L
T2 = ? K T = °C = K

18. Solution #1
DATA TABLE P1 = atm P2 = 3.20 atm V1 = L V2 = 90.0 mL = .090 L T2 = ? K T1 = 29°C = 302 K Solve for T2 Enter data T2 = P2V2T1 T2 = 3.20 atm x .090 L x 302 K = 604 K P1V atm .180 L T2 = 604 K or 6.04 x 102 K

19. Learning Check #2
A gas has a volume of 675 mL at 35°C and atm pressure.
What is the temp. in K when the gas has a volume of L
and a pressure of 802 mm Hg?
Rearrange Combined Gas Law Equation to solve for T2:
P1V1T2 = P2V2T1 T2 = P2V2T1
P1V1
Remember to convert all units, Kelvin, liters, mol, and atm:
DATA TABLE
P1 = ____atm P2 = ____atm
V1 = ____L V2 = ______L
T2 = ? K T = _____K

20. Solution #2 T2 = 180 K DATA TABLE
P1 = atm P2 = mm Hg (1atm/760 mm Hg) = atm
V1 = L V2 = L
T2 = ? K T = °C = K
Solve for T2 Enter data
T2 = P2V2T T2 = atm x .315 L x 308 K = K
P1V atm L
T2 = K

21. Gas Laws
Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP and Molar Volume

22. STP and Molar volume
The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

23. P1 = _____ V1 = _____ T1 = _____ K
Learning Check #1
A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
Data Table:
P1 = _____ V1 = _____ T1 = _____ K
P2 = _____ V2 = ?? T2 = _____K
P1V1T2 = P2V2T1 V = P1V1T2
P2T1
V2 = 15 L x atm x K = _____L
atm K

24. Solution #1 Data Table: P1 = 1.0 atm V1 = 15 L T1 = 273 K
P2 = 2.0 atm V2 = ?? T2 = 248 K
V2 = 15 L x atm x K = 6.8 L
2.0 atm K

25. Molar Volume At STP 2.0 g H2 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole
(STP) (STP) (STP)
V = 22.4 L V = 22.4 L V = 22.4 L
It does not matter what gas you are dealing with, if you have exactly
one mol of that gas you will have L of that gas.
1 mole of a gas at STP = L
22.4 L and mole
1 mole L

26. Learning Check #1 A. What is the volume at STP of 4.00 g of CH4?
1) L 2) L 3) 44.8 L
4.00 g CH L CH4
B. How many grams of He2 are present in 8.00 L of gas at STP?
1) g 2) g 3) 1.43 g
8.00 L He g He2

27. Solution #1 A. What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L CH4(STP) = L CH4
16.0 g CH mole CH4
B. How many grams of He2 are present in 8.00 L of gas at STP?
8.00 L He2 x 1 mole He2 x g He2 = g He L He mole He2