## Combined Gas Law

equation

P V = n R T

P1 V1 =  P2 V2

T1       T2

explanation

one gas at one set of conditions

one gas that is changing conditions

when to use it

when the problem gives 3 of these:

P, V, n, T

more than one temperature, pressure, and/or volume in the problem

specific units req’d?

pressure = atm

volume = liters

quantity (n) = moles

temperature = Kelvins

temperature = Kelvins

pressure & volume can be any unit, but must be the same unit on both sides of the equation

## Gay-Lussac’s Law

equation

P1 V1 = P2 V2

V1 = V2

T1   T2

P1 = P2

T1   T2

explanation

pressure & volume are inversely proportional;

temperature is constant

volume & Kelvin temp of a gas are directly proportional; P is constant

pressure & Kelvin temp of a gas are directly proportional; V is constant

when to use it

given 2 difft pressures & 1 volume or given 2 difft volumes & 1 pressure

given 2 difft volumes & 1 temperature or given 2 difft temperatures & 1 volume

given 2 difft pressures & 1 temperature or given 2 difft temperatures & 1 pressure

specific units req’d?

any - but must be the same on both sides of equation

any unit for volume (same on both sides), Kelvin temperature

any unit for pressure (same on both sides), Kelvin temperature

## Graham’s Law

equation

Ptotal = Pgas1 + Pgas2 +...

Px = (moles gas X) . Ptotal

(total moles)

rate A = √MM B

rate B    √MM A

explanation

the sum of the pressures of the individual gases in a mixture equals the total pressure exerted by the mixture

amount of a gas in mixture is proportionate to the amount of its partial pressure

rate of gas A compared to the rate of gas B is equal to the square root of the inverse of their molar masses

when to use it

mixture of gases; only pressures given

mixture of gases; moles & total pressure given

when any form of the word “effusion” or “diffusion” is in the problem

specific units req’d?

any - but all values must have same unit of pressure

any - but all values must have same unit of pressure

no

1. Convert the following temperatures.

(A)  104 oC to K                  (B) -3 oC to K              (C) 67 K to oC               (D) 1671 K to oC
377 K                                    270 K                           -206 oC                         1398 oC

2. Convert the following pressures.

(A) 635 torr to atm            (B) 104.2 kPa to mm Hg              (C) 1.45 atm to Pa
0.836 atm                             781.8 mm Hg                                147,000 Pa (146885 before rounding)

3. A gas that effuses 1.19 times slower than nitrogen is added to light bulbs.  What is the

molecular mass of this unknown gas?

Rate N2        = √MM unknown              1.19 = √x            x = 39.7 g/mole
rate unknown     √MM N2                                 1     √28

4. (A) What is the molecular mass of a 0.2500 g sample of a gas at 99.8oC and 0.9131 atm

in a 100.0 cm3 container?  (B) What is the gas in the container?

MM = gRT          (0.2500 g) (0.0821 L.atm/mole.K) (372.8 K)       MM = 83.8 g/mole
PV                              (0.9131 atm) (0.1 L)                                      Krypton

5. A small 2.00 L fire extinguisher has an internal pressure of 506.6 kPa at 25oC.  What

volume of methyl bromide, the fire extinguisher’s main ingredient, is needed to fill an

empty fire extinguisher at standard pressure if the temperature remains constant?

P1V1 = P2V2                   (506.6 kPa) (2.00 L) = (101.3 kPa) V2

V2 = 10.0 L

6. If 45.0 g of propane gas burns completely in the following reaction:

C3H8(g)  +  5 O2(g)    3 CO2(g)  +  4 H2O(g)

then how many liters of carbon dioxide gas will be released if the system is at STP?

45.0 g C3H8 | 1 mole  = 1.023 moles C3H8                         1.023 moles C3H8 = x moles CO2
| 44.0 g                                                                        1                                  3

X = 3.069 moles CO2 | 22.4 L = 68.7 L
| 1 mole

7. Air in a closed cylinder is heated from 25°C to 36°C. If the initial pressure is 3.80 atm,

what is the final pressure?

P1 = P2             3.80 atm =   P2         P2 = 3.94 atm
T1   T2             298 K      309 K

8. At what temperature Celsius will 19.4 g of molecular oxygen, O2, exert a pressure of

1820 mm Hg in a 5.12 L cylinder?

PV = nRT                    19.4 g O2 | 1 mole  = 0.60625 moles                     1820 mm Hg | 1 atm         = 2.395 atm
| 32 g                                                                             | 760 mm Hg

(2.395 atm) (5.12 L) = (0.60625 moles) (0.0821 L.atm/mole.K) T
T = 246K        -->
-27 oC

9. To what temperature must 32.0 ft3 of a gas at 2.0 °C be heated for it to occupy

1.00 x 102 ft3 at the same pressure? (ft3 is a unit of volume)

V1 = V2                32 ft3 = 100 ft3                   T2 = 859 K or 586 oC
T1   T2                      275 K         T2

10. Determine the molar mass of a gas that has a density of 2.18 g/L at 66°C and 720

mm Hg.

D = MM P                     2.18 g/L =           MM (0.947 atm)                       MM = 64 g/mole
RT                                         (0.0821 L.atm/mole.K) (339 K)

11. A 3.10 mL bubble of methane gas forms at the bottom of a bog where the temp-

erature is 12oC and the pressure is 8.5 atm. The bubble rises to the surface where the

temperature is 35oC and the pressure is 1.18 atm.  What is the new volume of the

methane bubble?

P1V1 = P2V2                   (8.5 atm) (3.10 mL) = (1.18 atm) V2             V2 = 24 mL
T1       T2                              285 K                           295 K

12. A mixture of 2.00 moles of H2, 2.00 moles of NH3, 4.00 moles of CO2 and 5.00 moles

of N2 exerts a total pressure of 800. torr.  What is the partial pressure of each gas?

Px =   moles x    . Ptotal        H2 and NH3 = 2.00 moles . 800 torr = 123 torr
total moles                                       13.00 moles

CO2 = 4.00 moles . 800 torr = 246 torr                N2 =  5.00 moles . 800 torr = 308 torr
13.00 moles                                                              13.00 moles

13. For the reaction 2 H2(g) + O2(g) 2 H2O(g), how many liters of water can be made from

5.0 L of oxygen gas and an excess of hydrogen?

5.0 L O2  =  x L H2O          x = 10. L
1                   2