1. Ideal Gas Law (Equation):
P = Pressure
PV = nRT
V = Volume
n = number of moles
R = ideal gas constant
T = Temperature (K)

2. R = PV = (1 atm )(22.4 L) R = 0.0821 L * atm
We can calculate R by looking at 1 mol of a gas
at STP:
STP = 273 K and 1 atm
@ STP, 1 mol of a gas occupies 22.4 L
R = PV = (1 atm )(22.4 L)
nT (1 mol)(273 K)
R = L * atm
mol * K

3. Use ideal gas law, PV=nRT.
Practice Problem
A 23.8-L cylinder contains oxygen gas at C
and 732 torr. How many moles of oxygen are in the
cylinder?
PV = n RT
Use ideal gas law, PV=nRT.
Pressure must be in atm, Temperature in K:
732 torr x 1 atm = atm
760 torr
(0.963 atm)(23.8 L) = n
( L atm/mole K)(293 K )
= mol

4. Gas Quiz - Rd 1
PV = nRT
R = L atm
mol K
P1V1 = P2V2
T T2
1 atm = 760 mm Hg
A 5.8 L sample of gas at 5.0oC and 750 mm Hg
is cooled so that the volume becomes 9.6 L and
the pressure 380 mmHg. What is the temperature
in oC?
What is the volume of mole of oxygen gas
at 25.0 oC and 725 torr?

5. P1V1 = P2V2 T1 T2 (750 mm Hg)(5.8 L) = (9.6 L)(380 mm Hg) 278 K T2
A 5.8 L sample of gas at 5.0 oC and 750 mm Hg is
cooled so that the volume becomes 9.6 L and the
pressure 380 mm Hg. What is the temp in oC?
P1V1 = P2V2
T T2
(750 mm Hg)(5.8 L) = (9.6 L)(380 mm Hg)
278 K T2
T2 = 233 (230) K
= -40oC (-43oC)

6. PV = nRT V = nRT P P = 725 torr x 1 atm = 0.954 atm 760 torr
What is the volume of mole of oxygen gas at
25.0 oC and 725 torr?
PV = nRT
V = nRT P
P = 725 torr x 1 atm = atm
760 torr
V = (0.750 mole)( L* atm/mol*K)(298 K)
0.954 atm
V = 19.2 L