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Chemistry. States of matter – Session 1. Session Opener. Session Objectives. Session Objectives. 1. Definition and differences between solids, liquids and gases. 2. Measurable properties of gas: Mass, volume and temperature 3. Boyle’s law 4. Charles’ law 5. Avogadro’s hypothesis

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Session objectives
Session Objectives

  • 1. Definition and differences between solids, liquids and gases.

  • 2. Measurable properties of gas: Mass, volume and temperature

  • 3. Boyle’s law

  • 4. Charles’ law

  • 5. Avogadro’s hypothesis

  • Ideal gas equation

  • Dalton’s law

  • 8. Amagat’s law of partial volume

  • 9. Molecular mass of mixture of gases



Gaseous state assumptions

(2) Intermolecular forces negligible.

Gaseous state-assumptions

(1)The molecules are very loosely packed

voids

(3) Molecules move very rapidly in all directions in a random manner.

(4) The molecules collide with one another and also with walls of the container perfectly elastic collisions.

(5) The mass, volume and pressure of gases can be calculated.


Measurable properties of gases
Measurable Properties of Gases

Mass

Volume

Volume of a gas is the space occupied by itsmolecules and is equal to the volume of the container. 1m = 10 dm = 100 cm

1m3 = 103 dm3 = 106cm3

1L = 1 dm3 = 103cm3


Measurable properties of gases1
Measurable Properties of Gases

Pressure

Pressure is measured in terms ofatmospheric pressure, which is the pressure exerted by the atmospheric gases on the surface of the earth.

1 atm = 76 cm of mercury = 760 mm of mercury

1 atm = 101.325 kPa


Measurable properties of gases2
Measurable Properties of Gases

Centigrade or Celsius scale (°C)

Temperature

Fahrenheit Scale (°F)

Kelvin scale (K)

100° C = 180° F

–273.15° C = 0K or 0° C = 273.15K and t° C = (t + 273.15) K

STP or NTP

Temperature = 0° C = 273.15 K = 273 K

Pressure = 1 atm = 76 cm = 760 mm = 101.325 kPa



Illustrative example 1
Illustrative example 1

The temperature which is same on Fahrenheit and Celsiusscale is

(a) 40 (b) 90(c) 30 (d) none

Solution:

Let that temperature be x


Boyle s law
Boyle’s Law

Boyle’s law :

(at constant temperature and fixed mass)

P1V1=P2V2=constant [at a constant temperature]



Illustrative example 2

So,

Illustrative example 2

A sample of gas occupies 2 L under apressure of 800 atmosphere. What will be its volume if the pressure is decreased to 500 atmosphere. Assume that the temperature of the gas sample does not change?

Solution:

V2=?

V1=2 L

P2=500 atm

P1=800 atm

P1V1=P2V2

According to Boyle’s law


Illustrative example 3
Illustrative example 3

A gas–filled freely collapsible balloon is pushed from the surface level of a laketo a depth of 100m. Approximately what percent of its original volume will balloonfinally have? Assuming that the gas behaves ideally.

Solution:



Charle s law

Charle’s law:

(at constant pressure and fixed mass)

Charle’s law


Avogadro s law
Avogadro’s law

(at constant pressure and temperature)

“equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure”.


Do you know
Do you know?

Loschmidt number.

It is the number of molecules present in 1cc of a gas or vapour at STP. Its value is2.617 x 1019 per cc


Ideal gas equation
Ideal gas equation

Boyle’s law,

Charles’ law,

Avogadro’s hypothesis,

Combining these, we get


Ideal gas equation1
Ideal gas equation

The gas constant, RUnits of gas constant (R)

R = 0.0821 L atm K–1 mole–1

R = 8.314 107 ergs K–1 mole–1

= 8.314 JK–1 mole–1

R = 1.987 cal K–1 mole–1

= 5.1891019 eVK–1mol–1


Do you know1
Do you know?

Gas constant per molecule is known as Boltzmann constant (K)


Density and molar mass relation

If Wg of a gas of molecular mass M, occupies a volume

V, under pressure P at temperature T, then

Density

Density and molar mass relation

Number of moles of gas :


Vapour density
Vapour density

[Since molecular weight of H2 is 2]

= vapour density of a gas


Dalton s law
Dalton’s Law

a moles of He


Dalton s law1
Dalton’s Law

b moles of O2


Dalton s law2
Dalton’s Law

c moles of CO2


Dalton s law3
Dalton’s Law

Total gas pressure,

Dalton’s law of partial pressure



Illustrative example 4
Illustrative example 4

At 27° C a cylinder of 20 litres capacity contains three gases He, O2 and N2 0.502 g, 0.250 g and 1.00 g respectively. If all these gases behave ideally, calculate partial pressure of each gas as well as total pressure.

Solution:

Let number of moles of He, O2 and N2 be n1, n2, n3respectively.


Solution1
Solution

n2 = 0.25 g/32.0 g mole–1 = 0.0078 moles of O2

n3 = 1.00 g/28.0 mole–1 = 0.0357 moles of N2

Total number of moles in the gaseous mixture

n = 0.1255 + 0.0078 + 0.0357 = 0.169 moles

From gas equation,

= 0.208 atm


Solution2
Solution

Partial pressure of helium

= 0.1545 atm

= 0.0096atm

= 0.044 atm


Applications of dalton s law of partial pressure
Applications of Dalton’s law of partial pressure

a. To determine the pressure of a dry gas

Pdry gas = pmoist gas – Aqueous Tension (at t° C)

b. To calculate partial pressure

In a mixture of non-reacting gases,

Partial pressure = mole fraction x total pressure



Illustrative example 5
Illustrative example 5

A certain quantity of a gas occupies100 mL when collected over waterat 150C and 750mm pressure. It occupies 91.9 mL in dry state at NTP. Find the aqueous vapour pressure at 150C.

Solution:

P= aqueous vapour pressure P1(dry gas)=750-p

P2=760 mm V1=100 ml

V2=91.9 ml T1=15+273=288 K



Amagat s law of partial volume

= mole fraction

Amagat’s Law of Partial Volume

V = VA + VB + VC + . . . . . . (in mixture of gases)

According to Amagat’s law,


Molecular mass of the mixture of gases
Molecular mass of the mixture of gases

For example, air contains 79% nitrogen and 21% oxygen approximately.Now mass of one mole of air would be (0.79 × 28) + (0.21 × 32) = 28.84 gm/mole.

= S Molecular mass of a gas × mole fraction of a gas

xi (mole fraction of a gas) =



Class exercise 1
Class exercise 1

A gas at a pressure of 5 atm is heated from 0° C to 546° C and compressed to one-third of its original volume. Hence final pressure is(a) 10 atm (b) 30 atm(c) 45 atm (d) 5 atm

Solution:

Hence, the answer is (c).


Class exercise 2
Class exercise 2

What weight of CO2 at STP could be contained in a vessel that holds 4.8 g of O2 at STP(a) 5.5 g (b) 6.6 g(c) 7.7 g (d) 3.3 g


Solution4
Solution

According to ideal gas equation

Comparing both equations

Hence, the answer is (b).


Class exercise 3
Class exercise 3

The volume of helium is 44.8 L at

(a) 100° C and 1 atm (b) 0° C and 1 atm

(c) 0° C and 0.5 atm d) 100° C and 0.5 atm


Solution5
Solution

At STP, volume of 1 mole of helium = 22.4 L

New volume = 2 × 22.4 = 44.8 L

(Keeping pressure constant)

T2 = 546 K = 273° C

(Keeping temperature constant)

Hence, the answer is (c).

P2 = 0.5 atm


Class exercise 4
Class exercise 4

The density of O2 at NTP will be

(a) 1.43 g/L (b) 1.45 g/L (c) 1.55 g/L (d) 1.59 g/L

Solution:

PV = nRT

Hence, the answer is (a).


Class exercise 5
Class exercise 5

Which of the following gases has the vapour density 14? (a) O2 (b) CO2(c) CO (d) NO

Solution:

  • If vapour density = 14

Hence, the answer is (c).


Class exercise 6
Class exercise 6

For a particular gas at NTP, the pressure will be _______ at 100° C.

(a) 1.2 atm (b) 1.34 atm(c) 1.4 atm(d) 1.9 atm

Solution:

Hence, the answer is (b).


Class exercise 7

Let the vapour pressure of water = p mm

Pressure of dry gas, P1 = (750 – p) mm

V1 = 100 ml P2 = 760 mm

T1 = 288 K V2 = 91.9 ml

T2 = 273 K

Class exercise 7

100 ml of an ideal gas was collected over water at 15° C and 750 mm of Hg pressure. The volume of the dry gas is 91.9 ml at NTP. Find the aqueous vapour pressure at 15° C.

Solution:


Solution6
Solution

p = 13.2 mm


Class exercise 8
Class exercise 8

The density of a mixture of CO and CO2 is 1.5 g/L at 30° C and 730 torr. What is the composition of the mixture.

Solution:

We know, PV = nRT

= 38.8 g

Mmix = 28x + (1 – x)44

38.8 = 28x + (1 – x)44


Solution7
Solution

% of CO = 32.5

% of CO2 = 67.5