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Combined and ideal gas laws

Combined gas law

- If we combine all of the relationships from the 3

laws covered thus far (Boyles, Charless, and

Gay-Lussacs) we can develop a mathematical

equation that can solve for a situation where 3

variables change

PVk1

V/Tk2

P/Tk3

Combined gas law

- Amount is held constant
- Is used when you have a change in volume,

pressure, or temperature

P1V1T2 P2V2T1

Example problem

A gas with a volume of 4.0L at STP. What is its

volume at 2.0atm and at 30C?

1atm

2.0 atm

?

4.0 L

273K

30C 273

303K

Example problem

(1 atm)

(4.0L)

(2 atm)

( V )

2

(273K)

(303K)

2.22L V2

Classroom Practice

P1 V1 T1 P2 V2 T2

1.5 atm 3.0 L 20?C 2.5 atm 30?C

720mmHg 256 ml 240 ml

2.5 L 22?C 1.8 L

95 kPa 4.0L 101 kPa 6.0 L 198?C

850mmHg 15?C 30?C

1.00 atm 150 ml 0.762 atm

125 kPa 100?C 100kPa 100 ml 75?C

Avogadros Law

- So far weve compared all the variables except

the amount of a gas (n). - There is a lesser known law called Avogadros Law

which relates V n. - It turns out that they are directly related to

each other. - As of moles increases then V increases.

V/n k

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Ideal Gas Law

- Which leads us to the ideal gas law
- The fourth and final variable is amount
- We have been holding it constant.
- We can set up a much more powerful eqn, which can

be derived by combining the proportions expressed

by the previous laws.

Ideal Gas Law

- If we combine all of the laws together including

Avogadros Law mentioned earlier we get

Where R is the universal gas constant

Normally written as

PV nRT

Ideal Gas Constant (R)

- R is a constant that connects the 4 variables
- R is dependent on the units of the variables for

P, V, T - Temp is always in Kelvin
- Volume is in liters
- Pressure is in either atm or mmHg or kPa

Ideal Gas Constant

- Because of the different pressure units there are

3 possibilities for our ideal gas constant

- If pressure is given in atm

- If pressure is given in mmHg

- If pressure is given in kPa

Using the Ideal Gas Law

What volume does 9.45g of C2H2 occupy at STP?

1atm

P ?

R ?

?

V ?

T ?

273K

n ?

.3635 mol

(1.0atm)

(V)

(.3635mol)

(273K)

V 8.15L

A camping stove propane tank holds 3000g of C3H8.

How large a container would be needed to hold

the same amount of propane as a gas at 25C and a

pressure of 303 kpa?

303kPa

P ?

R ?

?

V ?

T ?

298K

n ?

68.2 mol

(303kPa)

(V)

(298K)

(68.2 mol)

V 557.7L

Classroom Practice

- Use the Ideal Gas Law to complete the following

table for ammonia gas (NH3).

Pressure Volume Temp Moles Grams

2.50 atm 0?C 32.0

75.0 ml 30?C 0.385

768 mmHg 6.0 L 100?C

195 kPa 58.7 L 19.8

Ideal Gas Law Stoichiometry

What volume of hydrogen gas must be burned to

form 1.00 L of water vapor at 1.00 atm

pressure and 300C?

(1.00 atm)

(1.00 L)

nH2O

(573K)

(.0821L atm/mol K)

nH2O .021257 mols

Ideal Gas Law Stoichiometry

2H2 O2 ? 2H2O

.021257 mol

.476 L H2

Classroom Practice

To find the formula of a transition metal

carbonyl, one of a family of compounds having the

general formula Mx(CO)y, you can heat the solid

compound in a vacuum to produce solid metal and

CO gas. You heat 0.112 g of Crx(CO)y Crx(CO)y(s)

? x Cr(s) y CO(g) and find that the CO

evolved has a pressure of 369 mmHg in a 155 ml

flask at 27?C. What is the empirical formula of

Crx(CO)y?

Variations of the Ideal Gas Law

- We can use the ideal gas law to derive a version

to solve for MM. - We need to know that the unit mole is equal to m

MM, where m is the mass of the gas sample

PV nRT

n m/MM

Variations of the Ideal Gas Law

- We can then use the MM equation to derive a

version that solves for the density of a gas. - Remember that D m/V

Classroom Practice 1

- A gas consisting of only carbon and hydrogen has

an empirical formula of CH2. The gas has a

density of 1.65 g/L at 27?C and 734 mmHg.

Determine the molar mass and the molecular

formula of the gas. - Silicon tetrachloride (SiCl4) and trichlorosilane

(SiHCl3) are both starting materials for the

production of electro-nics-grade silicon.

Calculate the densities of pure SiCl4 and pure

SiHCl3 vapor at 85?C and 758 mmHg.

Real Vs. Ideal

- All of our calculations with gases have been

assuming ideal conditions and behaviors. - We assumed that there was no attraction

established between particles. - We assumed that each particle has no volume of

its own - Under normal atmospheric conditions gases tend to

behave as we expect and as predicted by the KMT.

Real Vs. Ideal

- However, under high pressures and low

temperatures, gases tend to deviate from ideal

behaviors. - Under extreme conditions we tend to see a

tendency of gases to not behave as independently

as the ideal gas law predicts. - Attractive forces between gas particles under

high pressures or low temperature cause the gas

not to behave predictably.

Loose Ends of Gases

- There are a couple more laws that we need to

address dealing with gases. - Daltons Law of Partial Pressures
- Grahams Law of Diffusion and Effusion.

Daltons Law of Partial Pressure

- States that the total pressure of a mixture of

gases is equal to the sum of the partial

pressures of the component gases.

PTP1P2P3

- What that means is that each gas involved in a

mixture exerts an independent pressure on its

containers walls

Simple Daltons Law Calculation

- Three of the primary components of air are CO2,

N2, and O2. In a sample containing a mixture of

these gases at exactly 760 mmHg, the partial

pressures of CO2 and N2 are given as PCO2 0.285

mmHg and PN2 593.525 mmHg. What is the partial

pressure of O2?

Simple Daltons Law Calculation

PT PCO2 PN2 PO2

760mmHg .285mmHg 593.525mmHg PO2

PO2 167mmHg

Daltons Law of Partial Pressure

- Partial pressures are also important when a gas

is collected through water. - Any time a gas is collected through water the gas

is contaminated with water vapor. - You can determine the pressure of the dry gas by

subtracting out the water vapor

Atmospheric Pressure

Ptot Patmospheric pressure Pgas PH2O

- The waters vapor pressure can be determined from

a list and subtract-ed from the atmospheric

pressure

WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES

Temp (C) (mmHg) (kPa)

0.0 4.6 .61

5.0 6.5 .87

10.0 9.2 1.23

15.0 12.8 1.71

15.5 13.2 1.76

16.0 13.6 1.82

16.5 14.1 1.88

17.0 14.5 1.94

17.5 15.0 2.00

18.0 15.5 2.06

18.5 16.0 2.13

WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES

Temp (C) (mmHg) (kPa)

19.0 16.5 2.19

19.5 17.0 2.27

20.0 17.5 2.34

20.5 18.1 2.41

21.0 18.6 2.49

21.5 19.2 2.57

22.0 19.8 2.64

22.5 20.4 2.72

23.0 21.1 2.81

23.5 21.7 2.90

24.0 22.4 2.98

WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES

Temp (C) (mmHg) (kPa)

24.5 23.1 3.10

25.0 23.8 3.17

26.0 25.2 3.36

27.0 26.7 3.57

28.0 28.3 3.78

29.0 30.0 4.01

30.0 31.8 4.25

35.0 42.2 5.63

40.0 55.3 7.38

50.0 92.5 12.34

60.0 149.4 19.93

WATER VAPOR PRESSURES WATER VAPOR PRESSURES WATER VAPOR PRESSURES

Temp (C) (mmHg) (kPa)

70.0 233.7 31.18

80.0 355.1 47.37

90.0 525.8 70.12

95.0 633.9 84.53

100.0 760.0 101.32

Simple Daltons Law Calculation

- Determine the partial pressure of oxygen

collected by water displace-ment if the water

temperature is 20.0C and the total pressure of

the gases in the collection bottle is 730 mmHg.

PH2O at 20.0C 17.5 mmHg

Simple Daltons Law Calculation

PT PH2O PO2

PH2O 17.5 mmHg

PT 730 mmHg

730mmHg 17.5468 PO2

PO2 712.5 mmHg

Your Turn

- A mixture of 1.00 g H2 and 1.00 g He is placed in

a 1.00 L container at 27?C. Calculate the

partial pressure of each gas and the total

pressure in atm. - Helium is collected over water _at_ 25?C and 1.00

atm total pressure. What total volume of He must

be collected to obtain 0.586 g of He?

1) PH2 12.3 atm PHe 6.16 atm 2) 3.7 L

Grahams Law

- Thomas Graham studied the effusion and diffusion

of gases. - Diffusion is the mixing of gases through each

other. - Effusion is the process whereby the molecules of

a gas escape from its container through a tiny

hole

Diffusion

Effusion

Grahams Law

- Grahams Law states that the rates of effusion

and diffusion of gases at the same temperature

and pressure is dependent on the size of the

molecule. - The bigger the molecule the slower it moves the

slower it mixes and escapes.

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Grahams Law

- Kinetic energy can be calculated with the

equation ½ mv2 - m is the mass of the object
- v is the velocity.
- If we work with two different at the same

temperature their energies would be equal and the

equation can be rewritten as

½ MAvA2 ½ MBvB2

- M represents molar mass
- v represents molecular velocity
- A is one gas
- B is another gas

- If we want to compare both gases velocities, to

determine which gas moves faster, we could write

a ratio of their velocities. - Rearranging things and taking the square root

would give the eqn

vA

MB

vB

MA

- This shows that the velocities of two different

gases are inversely propor-tional to the square

roots of their molar masses. - This can be expanded to deal with rates of

diffusion or effusion

Grahams Law

- The way you can interpret the equation is that

the number of times faster A moves than B, is the

square root of the ratio of the molar mass of B

divided by the Molar mass of A - So if A is half the size of B than it effuses or

diffuses 1.4 times faster.

Grahams Law Example Calc.

If equal amounts of helium and argon are placed

in a porous container and allowed to escape,

which gas will escape faster and how much faster?

Grahams Law Example Calc.

Rate of effusion of He

Rate of effusion of Ar

Helium is 3.16 times faster than Argon.

Your Turn

- Calculate the average rate of effusion of a H2

molecule at 0 C if the average rate of effusion

of an O2 molecule at this temperature is 500 m/s. - The rate of effusion of a gas was meas-ured to be

24.0 ml/min. Under the same conditions, the rate

of effusion of pure CH4 gas is 47.8 ml/min. What

is the molar mass of the unknown gas?

1) 2000 m/s 2) 63.5 g/mol