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Chapter 5Chapter 5

GasesGases

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Matter exist in three physical states: gas, liquid, solid. Few substances exist in gaseous state under typical conditions. Gases are very important…we live at the bottom of an ocean of air

solution (atmosphere) whose composition by volume is roughly 78% N2, 21% O2, and 1% other gases including CO2.

The atmosphere both support life and act as a waste receptor and also shield us from harmful radiation from the sun.

In this chapter we will look carefully at the properties of gases which lead to various types of laws.

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The following table shows some substances that exist as gases

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Gases differ from solids and liquids in several aspects: fill any container uniformly and move freely within it,

so the volume of the gas = the volume of the container mixed completely with any other gases. when there

are several gases in a mixture, the volume of each

component is the same as the volume occupied by the entire mixturehighly compressible, when a gas is subjected to pressure, its volume decreases.The attraction forces between the molecules of gases are weak which allow rapid, independent movement of the molecules and cause the physical behavior of a gas to be nearly independent of its chemical composition.The behavior of a gas controlled by its volume, pressure, temperature, and the number of moles.

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Pressure of a Gas:Pressure of a Gas:Gases exert pressure on any surface with which they come in contact, because

gas molecules are constantly in motion.

Pressure is one of the most readily measurable properties of gas

Pressure define as force acting on an object per unit area.

P = F/A

F = mass x acceleration = kg. m/s = newton

P = (kg. m/s) / m2 = N/m2 = pascal

The SI unit of pressure is the pascal, which define as one newton per square meter.

The atmosphere surround the earth is a mixture of gases, it exert a pressure uniformly on everything on earth.

The pressure of the atmosphere is measured with a device called a barometer.

It was invented in 1643 by Torricilli.

The barometer is constructed by filling a glass tube (about 1m long) with liquid mercury (Hg) and inverting it in a dish of mercury as shown in the following figure.

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BarometerBarometer� The pressure of the atmosphere

at sea level will hold a column of mercury 760 mm Hg.

� 1 atm = 760 mm Hg =760torr� The SI unit of pressure is pascal

(pa): define as 1 Newton per square meter

1 pa = 1 N/m2

1 atm.= 101325pa=101.325kpa

1 atm Pressure

760 mm Hg

Vacuum

Pressure of Hg atm. Pressure = pressure of Hg

The Hg will stop get out from the column

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Measuring pressure of trapped gases

It is often desirable to measure the pressure of a gas present in closed system, the instrument used is called manometer.

Two types of manometers are used:

1) Open end manometer: it is simply a U tube containing Hg, one arm of the tube is connecting to a gas whose pressure is to be measured, while the other arm is remain open to the atmosphere.

2) Closed end manometer

it is used to measure low pressure gases, it consist of U-shaped tube with one arm sealed at the top.

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Gas

P gas = Patm. P gas < Patm. P gas > Patm.

Patm. Patm.

Patm.

Gas Gas

h

h

Open end manometer

P gas = Patm - h P gas = Patm + h

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gasgas

P gas = Patm.

P = 0

P gas << Patm.

PHg

Closed end manometervaccume

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Examples:Examples:a)a) Convert 0.527 atm to torr Convert 0.527 atm to torr P (torr) = 0.527 atm xP (torr) = 0.527 atm x

b)b) Convert 730 torr to kPa Convert 730 torr to kPaP (kPa) = 730 torr x P (kPa) = 730 torr x

c)c) Convert 147.2 kPa to (1) atm and (2) torr Convert 147.2 kPa to (1) atm and (2) torr1) P (atm.) = 147.2 Kpa x1) P (atm.) = 147.2 Kpa x

2) P (torr) = 147.2 Kpa x 2) P (torr) = 147.2 Kpa x

1 atm.760 torr = 400.52 torr

760 torr101.235 Kpa = 97.24 Kpa

101.235 Kpa1 atm. = 1.45 atm.

101.235 Kpa760 torr

= 1105.1 torr

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Using liquids other than Hg in barometer and manometer:Lab. Manometers contain Hg rather than other liquids because its high density allows the manometer to be convenient size.

If low pressures are to be measured, liquids other than Hg is used. This is because Hg with its large densities gives very small differences in the height of the columns in manometers.

If liquids of lower densities is used, the difference in the heights can be much larger.

1 torr Hg = 1 mmHg = 13.6 mmHg

There is a very simple relationship between the height of the columns of fluids in manometer (or barometer) and their densities

(height x density)liquid = (height x density)Hg

(h x d) liquid = (h x d)Hg

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Example:

A long glass tube, sealed at one end has an inner diameter of 10 mm. the tube is filled with water and inverted into a pail of water. If the atmospheric pressure is 760 mmHg. How high (in mmH2O) is the column of water in the tube. (density H2O = 1g/ml , density Hg = 13.5 g/ml)

(h x d) water = (h x d)Hg

h x 1 g/ml = 760 mmHg x 13.5 g/ml

h = 10260 mmH2O

h = 1026.0 cmH2O

h = 10.26 mH2O

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The Gas Laws:

Physical behavior (or state) of a gas can be describe by four variables will affect the state of a gas:

- Temperature, (T) - Volume, (V) -Pressure, (P)

- Quantity of gas present, n (moles)

These variables are interdependent : any one of them can be determine by measuring the other three.

We will consider several mathematical laws that relate these properties of gases which known as gas laws.

These laws derived from experiments involving careful measurements of the relevant gas properties.

From these experimental results, the mathematical relationship among the properties can be discovered.

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Boyle’s Law(1662): P-V relationship

He use J-shape tube closed at one end, he study the relationship between the pressure of the trapped gas and its volume. Representative values from Boyle’s law are given in the following table:

Volume (in3( P (inch Hg( P x V

.117 5 .12 0 .14 1 x 102

.87 2 .16 0 .14 0 x 102

.70 7 .20 0 .14 1 x 102

.58 8 .24 0 .14 1 x 102

.44 2 .32 0 .14 1 x 102

.35 3 .40 0 .14 1 x 102

.29 1 .48 0 .14 0 x 102

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These data show that the P x V of the trapped gas is constant:

PV= k , k: constant depend on (n, T)Boyle’s Law: the volume of a fixed quantity of gas is inversely proportional to its pressure.•Mathematically:

•A plot of V versus P is a hyperbola.•Similarly, a plot of V versus 1/P must be a straight line passing through the origin.•The Value of the constant depends on the temperature and quantity of gas in the sample.

V = constant x 1P

PV = constant

P1V1 = P2V2

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Slope = k

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PV

P (at constant T)

CO2

22.4

1 L

atm Ne

O2

CO2

Ideal

Boyle’s law holds precisely at very law pressures.Measurement at higher pressures reveals that PV is not constant, but varies as the pressure varies as shown in this figure. A gas that obey Boyle’s law is called ideal gas

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Example (1):

20.5 L of nitrogen at 25ºC and 742 torr are compressed to 9.8 atm at constant T. What is the new volume?

9.8 atm. = 9.8 atm. x

P1V1 = P2V2

742 torr x 20.5 L = 7448 torr x V2 V2 = 2.04 L

Example (2):

30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa?

P1V1 = P2V2 P =

P (kPa) = 30.19 torr x

1 atm.760 torr = 7448 torr

30.6 ml x 740 torr750 ml

= 30.19 torr

760 torr101.235 kPa = 4.02 kPa

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Charles’s law (1787): T-V relationshiphe found that the volume of a gas at constant pressure increase

linearly with the temperature of the gas.

A plot of the volume of a gas (at constant pressure) versus temp. (ºC) gives a straight line.

A very interesting feature of this plot is that the volume of all gases decrease until the gas occupy a theoritical zero volume at -273.15 ºC. This point represents the temp. at which all gases if they did not condense would have a volume zero

Kelvin used this linear relationship between gas volume and temp. to device the absolute temp. scale.

In this scale, the absolute zero (0 K or -273ºC) is the temp. at which an ideal gas would have zero volume.

K = ºC + 273

0 K is called absolute temp., it has never been reached

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V (

L)

T (ºC)

H2O

HeCH4

H2

-273.15ºC

A plot of volume versus temp.

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Charles’s law: the volume of a gas is directly proportional with the absolute temperature at constant pressure.

V = kT (T is in Kelvin)

V1

T1

=V2

T2

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Example (1):Example (1):

What would the final volume be if 247 mL of gas at 22ºC is What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant.heated to 98ºC , if the pressure is held constant.

Example(2):

=V1

T1

V2

T2

V2 =247 ml x 371 K

295 K= 310.6 ml

At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 L at constant pressure?

V1 V2=T1 T2 T2

=

T2 =80.0 L x 296.4 K

40.5 L296.4 K

80.0 L

40.5 L= 585.5 K

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Avogadro’s law:

Avogadro postulate that: equal volumes of gas at the same temperature and pressure will contain the same number of molecules.

V α n (n is the number of moles)

V = k n (k is constant)

Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.

V1

V2=

n1

n2

This relationship is obey closely by gases at low pressure

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The ideal gas lawWe have considered three laws that describe the behavior of gases.

Boyle’s law: V α 1/P (at constant T and n)

Charles’s law: V α T (at constant P and n)

Avogadro’s law: V α n (at constant T and P)

These relationships can be combine as follows:

V α V= R ( )

R is the combine proportionality constant called the universal gas constant. R = 0.0821 L. atm./K mol.

The preceding equation can be rearranged to the more familiar form of the ideal gas law:

PV = nRT

T nP

T nP

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The ideal gas law is an equation of state for a gas, where the state of the gas is its condition at a given time.

The gas that obeys this equation is said to behave ideally.

Ideal gas is a hypothetical gas that would obey this equation over all ranges of T and P, real gases deviate from ideal behavior but many of them come quite close to being ideal at low P(< 1atm.) and high T. Knowledge of any three of these properties is enough to define the state of a gas, since the fourth property can be determined from the equation. The ideal gas law is also used to calculate the changes that will occur when the conditions of the gas are changed

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Example (1):

A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature

PV = nRT T = PV/ nR

T =

Example (2):

Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr (molar mass = 83.8 g/mol).

PV = nRT n = PV / RT

n =

1.85 atm. x 47.3 L

1.62 mol x 0.0821 (L. atm./mol K)= 658 K

8.61 atm. x 18.5 L0.0821 (L. atm. /mol K) x 279.8 K

= 6.93 mol

Mass of K (g) =6.93 mol x 83.8 g/mol = 581 g

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Example (3):

Calculate the number of moles of a gas that occupy1.50 L at 37 oC and 725 torr;

P(atm) = 725 torr. x 760 torr1 atm. = 0.954 atm.

T (K) = 37 + 273 = 310 K

n = P VR T =

0.954 atm. x 1.5 L

0.0821 (L. atm./mol K) x 310 K

n = 0.056 mol

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The ideal gas law is also used to calculate the changes that will occur when the condition of the gas are changed.

In general, if we have a gas under two sets of conditions, In general, if we have a gas under two sets of conditions, then:then:

Example (1):

A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?

n: is constant (the same sample of a gas)

P1V1

P2V2

=n1T1

n2T2

P1V1

P2V2

T1

T2

= V2 =

P1V1T1

P2T2

V2 =732 torr. x 4.18 L x 303 K

756 torr. x 297.8 K= 4.12 L

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We define STP (standard temperature and pressure) = 0°C, 273.15 K, 1 atm.

Volume of 1 mol of gas at STP is:

This volume is called the molar volume of ideal gas because it is the volume of 1.0 mol under these conditions

V = n R TP

=1 mol x 0.0821 L. atm./mol K) x 273 K

1 atm.

V = 22.42 L

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The ideal gas equation can be used to determine: the density of a gas. the molar mass of the gas the volume of gases formed or consumed in chemical reaction.

Molar mass and the density of a gas:One very important use of ideal gas law is in the calculation of the molar mass of a gas from its density.

n (number of moles) = mass (g)Molar mass (g/mol)

=m

Molar mass (g/mol)

Substituting into the ideal gas equation

P = n RTV = (m/molar mass) RT

V=

m R TV x molar mass

Density (d) = mass (g)Volume (L)

= M (g)V(L)

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P = d =

molar mass = or molar mass =

Example (1):

What is the density of ammonia, NH3(molar mass = 17g/mol) at 23ºC and 735 torr.

P(atm.) = 735 torr. x

T (K) = 273 + 23ºC = 296 K

d =

d R Tmolar mass

P x molar massR T

d RTP

1 atm.760 torr.

= 0.967 atm.

P x molar mass

R T=

0.967 atm. x 17 g/mol0.0821 (L.atm./mol K) x 296 K

= 0.676 g/L

m RTP V

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Example (2):

A compound has the empirical formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains 0.80 g of the gaseous compound. What is the molecular formula?

P= 750 torr. x T(K) =273+100

Molar mass =

1 atm.760 torr.

= 0.987 atm.

m RTP V

0.8 g x 0.0821 L.atm/mol K x 373 K=

0.987 atm. x 0.256 L

Molar mass = 97 g/mol

X = molar mass compoundMolar mass E.F.(CHCl)

= 97 48.5

= 2

Molecular formula is 2(CHCl) , C2H2Cl2

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Gas stoichiometryMany chemical reactions involve gases, by assuming ideal behavior

for these gases we can carry out stoichiometric calculations if P, V, T of the gases are known.

� Reactions happen in moles� At Standard Temperature and Pressure (STP, 0ºC and 1 atm) 1

mole of gas occupies 22.42 L.� If not at STP, use the ideal gas law to calculate moles of reactant or

volume of product.

For the reaction:

2 H2(g) + O2(g) 2 H2O(g)

2 mol 1 mol 2 mol

2 x 22.4 L 1 x 22.42 L 2 x 22.42 L

2 x volume H2 1 x volume O2 2 x volume H2O

2 : 1 mol ratio of H2 : O2 mean 2 : 1 volume ratio of H2 : O2

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Example(1):

Consider the following reaction 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

A) What volume of H2O at 1.0 atm and 1000ºC can be produced from 10.0 L of NH3 and excess O2 at the same temperature and pressure?

B) What volume of O2 measured at STP will be consumed when 10.0 kg NH3 (molar mass = 17g/mol) is reacted?Volume of H2O = 10.0 L NH3 x

A)4 L NH3 6 L H2O = 15.0 L H2O

B) Mol of NH3 =10 x 103 g NH3

17 g/mol= 588.2 mol NH3

V(NH3) =588.2 mol x 0.0821 (L. atm/mol K) x 273 K1 atm.

= 1.32x104 L

Volume of O2= 1.32x104 L NH3 x5 L O2

4 L NH3

= 1.65 x 104 L

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Example(2):

Mercury can be achieved by the following reaction

2HgO 2Hg(ℓ) + O2(g)

A) What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide (molar mass = 216.5 g/mol) at STP.

B) At 400.ºC and 740 torr.

Heat

mol of HgO =4.1 g

216.5 g/mol= 0.019 mol HgO

V(O2) = 0.019 mol x 0.0821 (L. atm/mol K) x 273 K1 atm. = 0.426 L

A)

B) V(O2) = 0.0095 mol x 0.0821 (L. atm/mol K) x 673 K

(740/760) atm.= 0.539 L

mol O2 = 0.019 mol HgO x 1 mol O2

2 mol HgO= 0.0095 mol O2

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Dalton’s LawWhen two or more gases do not react chemically are placed in the same container, the pressure exert by each gas in the mixture is the same as it would be if it were the only gas in the container.

The pressure exerted by each gas in the mixture is called its partial pressure and observed by Dalton and called Dalton’s law of partial pressure

The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container.

The total pressure is the sum of the partial pressures.

PTotal = P1 + P2 + P3 + P4 + P5 ...

For each: P = nRT/V

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PT = n1 RT n2 RT n3 RT n4 RT+ + +V VV V

PT = (n1 + n2 + n3 + n4+….)

+ ………

RTV

PT = nT RTV

nT :sum of the number of moles of all gases

The pressure exert by an ideal gas is not affected by the identity (composition) of the gas particles.The partial pressure of a gas is related quantitatively to the total pressure by its mole fraction. Mol fraction (x): the ratio of the number of moles of a given compound in a mixture to the total number of mol of mixture

x = n1nT

=n1

n1 + n2 + n3 + n4+….)

x = P1V/RT

(P1V/RT) + (P2V/RT) + (P3V/RT) +….

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x =P1(V/RT)

(P1 + P2+ P3 + P4..)(V/RT)

x = P1

P1 + P2 + P3 + P4….= P1

PT

P1 = x PT

The partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure

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Example:

A) The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen?

B) What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

XN2 =PN2

PT

= 592 torr.752 torr.

= 0.787A)

B) PN2 = XN2 PT = 0.787 x 5.25 atm. = 4.133 atm.

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Example:The partial pressure of CH4 is 0.175 atm. and that of O2 is 0.25 atm. In a mixture of two gases, calculate:

1. The total pressure of the mixture.

2. The mol fraction of each gas in the mixture.

3. The mass of each gas in the mixture if it occupy 10.5 L at 65ºC.

1. PT = PCH4 + PO2 = 0.175 atm. + 0.25 atm. = 0.425 atm.

2. XCH4 = (0.175/0.425)= 0.412 XO2 = (0.25/0.425)= 0.588

nT =3. P VR T =

0.425 atm. x 10.5 L0.0821 (L. atm./mol K)x 338 K

= 0.161 mol

XCH4 =nCH4nT

nCH4 = 0.161 x 0.412 = 0.0663 mol

mass CH4 = 0.0663 mol x 16 g/mol = 1.061g

nO2 = 0.161 x 0.588 = 0.0947 mol

mass O2 = 0.0947 mol x 32 g/mol = 3.03 g

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3.50 L

O2

1.50 L

N2

2.70 atm

When these valves are opened, what is each partial pressure and the total pressure?

4.00 L

CH44.58 atm 0.752 atm

example

(Pi Vi)CH4 = (Pf Vf)CH4

2.7 atm. x 4 L = P x 9 LPfCH4 = 1.2 atm.

Pf N2 = 0.76 atm.

Pf O2 = 0.29 atm.

PT = 1.2 atm.+0.76 atm.+ 0.29 atm. = 2.25 atm.

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Collecting gas over water:

Water-insoluble gases prepared in lab. Are collected by displacement of water as shown

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A gas collected in this manner become “contaminated” with water molecules that evaporate into the gas, so the gas in the bottle is a mixture of water vapor and the gas evolved from the reaction.

Water vapor is present because water molecules escape from the surface of liquid, these water molecules also exert a pressure called the vapor pressure of water which depends on the temperature of liquid water.

So PT = P(gas) + P(H2O)

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Example:

Helium is collected over water at 25ºC and 1.0 atm. total pressure, what volume of gas must be collected to obtain 0.586 g He (molar mass= 4g/mol) . (PH2O = 23.8 torr.= 0.0313 atm. at 25ºC)

PT = P(He) + P(H2O)

P(He)= P(T) - P(H2O) P(He)= 1- 0.0313= 0.967 atm.

mol of He = (0.586g/4 g/mol) = 0.1465 mol

VHe = n RT/PVHe =

0.1465 mol x 0.0821 (L. atm./mol K) x 298 K= 3.69 L

0.967 atm

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Example:

N2O can be produced by the following reactionNH4NO3(s) N2O (g) + 2 H2O(ℓ)

what volume of N2O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH4NO3(molar mass = 80g/mol)? ( the vapor pressure of water

at 22ºC is 21 torr)

PT = 94 kPa = 0.93 atm. PH2O =21 torr. = 0.028 atm.

PN2O = 0.93 – 0.028 = 0.902 atm.

mol NH4NO3 = (2.6 g/80g/mol) = 0.0325 mol

mol N2O = 0.0325 mol

heat

V N2O = (n RT/P) = 0.0325 mol x 0.0821 (L. atm./mol K) x 295 K

0.902 atmV N2O = 0.873 L

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EffusionEffusion and Diffusion and Diffusion

If two gases are placed in the same container their molecules gradually mixed until the composition of the gas become uniform. This process of mixing is called Diffusion.

Diffusion: is the term used for mixing gases.

A process similar to diffusion called effusion

Effusion: Passage of gas under pressure through a small hole, into a vacuum.

The effusion rate measures how fast this happens.

Graham study the rate of effusion of various gases.

He found that the rate of effusion is inversely proportional to the square root densities of gases (d) at the same T & P .

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Rate of effusion α √ d

If you have two gases A and B, then

Rate of effusion of gas ARate of effusion of gas B

= √dB

dA

Molar mass = R TdP

Amolarmass

Bmolarmass

)(

)(

B gasfor effusion of Rate

A gasfor effusion of Rate =

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Example:

Which gas will effuse faster NH3 or CO2 , what are their relative rates of effusion.

molar mass NH3 = 17 g/mol, molar mass CO2 = 44 g/mol

NH3 effuse faster (smaller molar mass)

Rate of effusion of NH3 Rate of effusion of CO2

= √44/17 =1.6

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Example:

An unknown gas effuse at a rate that is 0.355 times that of O2 at the same T &P, what is the molar mass of this unknown gas

rate of effusion of gas A = 0.355 rate of effusion of O2

Rate of effusion of gas ARate of effusion of gas O2

= 0.355 = √32 g/mol

Molar mass (A)

Molar mass (A) = 254 g/mol