1. IDEAL GAS LAW

2. Ideal Gas Law Derivation
Recall that
P1V1 = P2V2
n1T1 n2T2
Also we learned that
At OoC, and 1 atm (101.3 kPa) that 1 mole of gas occupies 22.4 L (molar volume)

3. Ideal Gas Law Derivation
We can set one side of the equation under standard conditions ie.
n = 1 mole
V = 22.4 L
P = kPa
T = 0oC

4. Ideal Gas Law Derivation
P1V1 = P2V2
n1T1 n2T2
So….
(101.3kPa) (22.4L) = P2V2
(1mole) (273K) n2T2

5. Ideal Gas Law Derivation
8.314 kPa x L = P V
mol x K n T
Overall….
PV = nRT
8.314 kPaL/mol K = R
= Ideal Gas Law Constant

6. Ideal Gases
Behave as described by the ideal gas equation; no real gas is actually ideal
Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less
In real gases, particles attract each other reducing the pressure
Real gases behave more like ideal gases as pressure approaches zero.

7. PV = nRT R is known as the universal gas constant
Using other STP conditions
P V
R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K)
n T
= L-atm
mol-K

8. Additional R Values
What is the value of R when the STP value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L)
nT (1mol) (273K)
= L-mm Hg mol-K

9. Learning Check G16
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°C, what is the pressure (kPa) in the tank in the dentist office?

10. Solution G16 Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = L L
T = 23°C K
n = mol 2.86 mol
P = ? ?

11. Rearrange ideal gas law for unknown P
P = nRT V
Substitute values of n, R, T and V and solve for P
P = (2.86 mol)(8.314)(296 K)
(20.0 L)
= kPa

12. What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = kPa
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
V = 30.6 L

13. Learning Check G17
A 5.0 L cylinder contains oxygen gas at 20.0°C and 98 kPa. How many grams of oxygen are in the cylinder?

14. Solution G17 Solve ideal gas equation for n (moles) n = PV RT
= (98 kPa)(5.0 L)(mol K)
(62.4 mmHg L)(293 K)
= mol O2 x g O2 = 6.4 g O2
1 mol O2

15. Molar Mass of a gas n = PV = (0.813 atm) (0.215 L) = 0.00703 mol
What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C?
n = PV = (0.813 atm) (0.215 L) = mol
RT ( L-atm/molK) (303K)
Molar mass = g = g = g/mol
mol mol

16. Density of a Gas
Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.
P = atm T = 273 K
Rearrange the ideal gas equation for moles/L
PV = nRT PV = nRT P = n
RTV RTV RT V

17. Substitute
(1.00 atm ) mol-K = mol O2/L
( L-atm) (273 K)
Change moles/L to g/L
mol O2 x g O2 = g/L
1 L mol O2
Therefore the density of O2 gas at STP is
1.43 grams per liter

18. Formulas of Gases
A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas?

19. Formulas of Gases Calculate Empirical formula
85.7 g C x 1 mol C = mol C/7.14 = 1 C
12.0 g C
14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H
1.0 g H
Empirical formula = CH2
EF mass = (1.0) = 14.0 g/EF

20. Using STP and density ( 1 L = 2.50 g)
2.50 g x L = g/mol
1 L mol
n = EF/ mol = g/mol = 4
14.0 g/EF
molecular formula
CH2 x = C4H8