1. 11. What are the basic assumptions of KMT?
1. Gases consist of tiny (submicroscopic) particles.
The distance between particles is large compared with the size
of the particles themselves. The volume occupied by a gas
consists mostly of empty space.
3. Gas particles have no attraction for one another.
Gas particles move in a straight line in all directions, colliding
frequently with one another and with the walls of the container.
No energy is lost by the collision of a gas particle with another
gas particle or with the walls of the container. All collisions are
perfectly elastic.
6. The average kinetic energy for particles is the same for all gases
at the same temperature, and its value is directly proportional to
the Kelvin temperature.

2. 15. What are the characteristics of an ideal gas?
An “ideal gas” behaves according to KMT.
No ideal gases exist, but under certain
conditions of temperature and pressure,
real gases approach ideal behavior, or
at least show only small deviation from it.
At very high pressure, low temperature
real gases deviate greatly from ideal behavior.

3. The barometer reads 715 mm Hg.
Calculate the corresponding pressure:
a) 715 mm Hg x 1 atm = atm
760 mm Hg
b) 715 mm Hg x 29.9 in Hg = 28.1 in Hg
760 mm Hg
c) 715 mm Hg x 14.7 psi = 13.8 psi
760 mm Hg

4. 27. Express the following pressures in atm:
a) 28 mm Hg x 1 atm = atm
760 mm Hg
b) cm Hg x 1 atm = atm
76 cm Hg
c) 795 torr x 1 atm = 1.05 atm
760 torr
d) 5.00 kPa x 1 atm = atm
101.3 kPa

5. P1V1 = P2V2 or V2 = P1V1 P2 (500. mm Hg)(400.mL) 760 mm Hg
29. A gas occupies a volume of 400. mL at 500. mm Hg pressure. What will be its volume, at constant temperature, if pressure is changed to (a) 760 mm Hg? (b) 250 torr?
P1V1 = P2V2 or V2 = P1V1 P2
(500. mm Hg)(400.mL)
760 mm Hg
250 torr
V2 = 2.6 x 102 mL
V2 = 8.0 x 102 mL

6. 31. P1 V1 = P2 V2 P2 = P1 V1 = (640. mm Hg)(500. mL) V2 855 mL
31. A 5oo. –mL sample of a gas is at a pressure of 640. mm Hg. What must be the pressure, at constant temperature, if volume is changed to 855 mL?
P2 = P1 V1 = (640. mm Hg)(500. mL)
V mL
P2 = mm Hg
31. P1 V1 = P2 V2

7. 33. Given 6.00 L of N2 gas at -25C, what volume will the nitrogen occupy at (a) 0.0C (b) 100. K? (Assume constant pressure)
V1 = V2 or V2 = V1T2
T1 T T1
(a) (6.00 L)(273 K) = 6.60 L
248 K
(b) (6.00 L)( 100. K) = 2.42 L

8. Use the combined gas law P1V1 = P2V2 V2 = P1V1 T2 T1 T2 P2T1
35. A gas occupies a volume of 410 mL at 17C and 740 mm Hg pressure. Calculate the volume the gas would occupy at STP.
Use the combined gas law
P1V1 = P2V2 V2 = P1V1 T2
T T P2T1
V2 = (740 mm Hg)(410 ml)(273 K)
(760 mm Hg) (300.K)
V2 = 360 mL

9. A sample of propane gas, C3H8, was
collected over water at 22.5 oC and 745 torr.
The volume of the wet gas is 1.25 L. What
will be the volume of the dry propane at
standard pressure?
P Total = PC3H8 + PH2O
@ 22.5 oC PH2O = 20.5 torr
PC3H8 = 745 – 20.5 = 725 torr

10. To calculate the volume of the dry propane,
note that the temperature is constant, so …
P1V1 = P2V2
P1V1 = V2 P2
(725 torr)(1.25 L) = L C3H8
760 torr

11. Given the equation:
4 NH3 + 5 O2 → 4 NO + 6 H2O
How many moles of NH3 are required to
produce 5.5 mol of NO?
4 NH3 = X mol NH3
4 NO mol NO
X = 5.5 mol NH3

12. Given the equation:
4 NH3 + 5 O2 → 4 NO + 6 H2O
b) How many liters of NO can be made from
12 L of O2 and 10 L of NH3 at STP?
4 NH3 = 10 L NH3
4 NO X NO
X = 10 L NO
5 O2 = 12 L O2
4 NO X NO
X = 9.6 L NO

13. Given the equation:
4 NH3 + 5 O2 → 4 NO + 6 H2O
c) At constant temperature and pressure, what is
the maximum volume, in Liters, of NO can be
made from 3.0 L of NH3 and 3.0 L of O2?
4 NH3 = 3.0 L NH3
4 NO X NO
X = 3.0 L NO
5 O2 = 3.0 L O2
4 NO X NO
X = 2.4 L NO

14. 63) Given the equation:
4 FeS (s) + 7 O2 (g) → 2 Fe2O3 (s) + 4 SO2 (g)
How many liters of O2, measured at STP, will
react with kg of FeS?
600. g FeS 1 mol = mol FeS
87.9 g
6.83 mol FeS 7 O2 = mol O2
4 FeS
12.0 mol O L = 269 L O2
1 mol

15. Sketch a graph to show each of the following
relationships: V P V T P V T n

16. At 27 0C and 750 torr pressure, what will
be the volume of 2.3 mol Ne?
V = nRT P
PV = nRT
27 0C = 300. K
750 torr 1atm = atm
760 torr
V = (2.3)(0.0821)(300.) = 57 L Ne
0.99

17. 91a) Using the ideal gas equation, calculate the
volume of mol of H2 at 47 0C and 1.6 atm
pressure
PV = nRT
V = nRT P
47 0C = 320. K
V = (0.510)(0.0821)(320.) = 8.4 L H2
1.6

18. What is the Kelvin temperature of a system
in which 4.50 mol of a gas occupy L at
4.15 atm?
PV = nRT
T = PV nR
T = (4.15)(0.250) = K
(4.50)(0.0821)

19. How many moles of N2 gas occupy 5.20 L
at 250 K and atm?
PV = nRT
n = PV RT
n = (0.500 atm)(5.20 L) = mol N2
(0.0821)(250 K)