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Chemistry. States of Matter - 3. Order of diffusion would be. CO 2 > SO 2 > SO 3 > PCl 3. Session Opener. The rates of diffusion of SO 2 , CO 2 , PCl 3 and SO 3 are in the following order. Solution:. Rate of diffusion . Hence, answer is (d).

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Presentation Transcript
session opener

Order of diffusion would be

CO2 > SO2 > SO3 > PCl3

Session Opener

The rates of diffusion of SO2, CO2, PCl3

and SO3 are in the following order

Solution:

Rate of diffusion

Hence, answer is (d)

illustrative example 1

The root mean square velocity of an ideal gas at constant pressure varies with density (d) as

(a) d2 (b) d

\ Vrms

Illustrative example - 1

Solution:

Hence, answer is (d)

illustrative example 2
Illustrative example - 2

Calculate the temperature at which root mean square velocity of SO2 molecules is same as that of O2 molecules at 27° C.

Solution:

For O2 at 27° C,

For SO2 at t° C,

solution
Solution

Since both these velocities are equal,

  • or 600 = 273 + t
  • or t = 600 – 273 = 327° C
session objectives
Session Objectives
  • concept of real gas
  • Explanation for deviation from idealbehaviour.
  • van der Waals’ equation for real gases
  • Liquefaction of gases and critical phenomena
  • The liquid state and its characteristics:Compressibility, diffusion, evaporation.
deviation ideal behaviour

These gases are real gases

Deviation: Ideal behaviour

Gases behave ideally at low pressures and high

temperatures.

  • For an ideal gas, PV = ZnRT and Z = 1, Z is compressibility factor, measures extent of ideal nature
reason for deviation
Reason for deviation
  • Near the liquefaction pointthere is greater deviation from ideal behaviour.

1. The volume of the gas molecule is negligible as compared to the total volume of the gas.

2. The forces of attraction between gas molecules are negligible.

van der waals equation14
van der Waals’ equation

p = Correction of pressure

Intermolecular forces of attraction proportional to

(i) the number of molecules per unit volume in the bulk of the gas.

(ii) the number of molecules colliding per unit volume.

a is van der Waals’ constant

van der waals equation15

Excluded volume for a pair of molecules =

Excluded volume per molecule =

van der Waals’ equation

Vm = Actual volume of a gas molecule.

Excluded volume for one mole b = 4 × Vm × NA.

observed volume = (V – nb) (For n mole of gas)

Van der waals’ equation for real gas

illustrative example 3
Illustrative example - 3
  • Calculate the pressure exerted by 5 mol of CO2 in one litre vessel at 47°C using van der Waals equation.Also report the pressure of gas if it behaves ideally in nature. Given that a = 3.592 atm litre2 mol–2, b = 0.0427 litre mol–1.
solution18
Solution
  • Given n = 5, v = 1 litre, T = 320 K
  • a = 3.592, b = 0.0427
  • Using van der Waals equation
  • P = 77.218 atm
  • Also if gas behaves ideally, then
  • PV = nRT
  • P × 1 = 5 × .0821 × 320
  • P = 131.36 atm
illustrative example 4
Illustrative example - 4

Using van der Waals equation, calculate the constant ‘a’ when two mol of a gas confined in a four litre flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.05 litre/mol.

Solution:

van der Waals equation

a = 6.48 atm litre2 mol–2

ask yourself
Ask yourself

What is the significance of a and b?

Larger value of ‘a’ indicatehigher intermolecular forcesof attraction,i.e. liquefaction of gases become easier.

While value of ‘b’ indicates theeffective size of the gas molecules.

application

At very low pressure, volume will be too high. and b can be neglected.

Application

For extremely low pressure

Hence, PV = RT (Ideal gas equation)

application22

Hence negligible value of

and b.

Application

For very high temperature

The volume will be extremely high

\ PV = RT

application23

PV = RT + Pb –

Small size and less molecular mass makesnegligible

i.e. Z =

Application

For hydrogen gas

PV = RT + Pb

critical states
Critical states

Critical temperature

Critical pressure

VC = 3b

Critical volume

Boyle temperature(TB)

At TB Z=1 I.e., gases behave ideally

Every real gas has a characteristic value of TB.

ask yourself25
Ask yourself

What is triple point?

Triple point is the point at which solid, liquid and vapour are in equilibrium with eachother.

liquid state characteristic properties
Liquid State – Characteristic Properties
  • The liquid state of matter hasintermediate properties.
  • Less orderly thansolid state but more orderly than thegaseous state.
liquid state
Liquid State

Liquid state is explained on the basisof kinetic theory model as follows:

  • Liquids are made up of molecules.
  • Molecules in a liquid are quite close to each other.
  • Force of attraction between the molecules in a liquid is quite large.
  • The molecules in a liquid are in a state of constant random motion.
  • The average kinetic energy of the molecules in a liquid is to their absolutetemperature.
properties of liquid
Properties of Liquid

Diffusion

  • There is diffusion in liquids but it is slower than in gases.
  • Diffusion involves movement of molecules from higher concentration to lower concentration..

Compressibility

Liquids are relatively less compressible than gases.

properties of liquid29
Properties of Liquid

Evaporation

  • The process of conversion of a liquid into its vapours at room temperature is known as evaporation.
  • Evaporation can be easily explained on the basis of kinetic theory.
  • At a given temperature, the average kinetic energy of the liquid molecules is constant but all the molecules do not have the same kinetic energy.
properties of liquid30
Properties of Liquid
  • Consequently, a certain fraction ofmolecules will have kinetic energies large enough to overcome the attractive forces of their neighbours and to escape into space above the liquid surface.
  • Evaporation causes cooling.
  • Rate of evaporation is influenced by the surface area of the liquid, the temperature and the strength ofintermolecular forces of attraction.
class exercise 1

When an ideal gas undergoes unrestricted expansion, no cooling occurs because the molecules

  • exert no attractive forces on each other
  • (b) do not work
  • (c) collide without loss of energy
  • (d) are above the inversion temperature
Class exercise - 1
solution33
Solution

When a gas undergoes unrestricted expansion, there will be no intermolecular forces of attraction. Hence, no cooling is possible.

Hence, the answer is (a)

class exercise 2

The speed possessed by most of

the gas molecules is

(a) most probable speed

(b) average speed

(c) root mean square speed

(d) None of these

Class exercise - 2
solution35
Solution

Most probable velocity is the velocity possessed by most of the gas molecules.

Hence, the answer is (a)

class exercise 3
Class exercise - 3
  • The value of critical coefficient is
  • 2.66 (b) 3.55
  • (c) 1.55 (d) 4.33

Solution:

Critical coefficient

Hence, the answer is (a)

class exercise 4
Class exercise - 4

The value of van der Waals’ constant ‘a’ for the gases O2, N2, NH3, CH4 are 1.36, 1.39, 4.17 and 2.25 litre2 atm mol–2. The gas which can be most easily liquefied is

(a) O2 (b) N2

(c) NH3 (d) CH4

Solution:

The gas with the highest value of ‘a’ is having highest intermolecular forces of attraction. Hence, it can be liquefied most easily.

Hence, the answer is (c).

class exercise 5
Class exercise - 5

At the same temperature and pressure, which of the following gases will have the highest rate of diffusion?

(a) Hydrogen

(b) Oxygen

(c) Methane

(d) All will have the same rate of diffusion

Solution:

Hence, the answer is (a).

solution40
Solution

Hence, the answer is (c).

class exercise 7
Class exercise - 7

One litre of one mole of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors are 1.072 and 1.375 respectively at the initial and final steps. Find the volume.

Solution:

  • We know that P1V1 = Z1nRT1 ... (i)
      • P2V2 = Z2nRT2 ... (ii)
solution42
Solution

Dividing (ii) by (i)

= 370.15 ml

Ans. 370.15

class exercise 8
Class exercise - 8

Three moles of CO2 (a van der Waals’ gas) occupy 10 L at 15 atm. Find out the temperature.

(Given a = 3.59 L2 atm mol–2, b = 0.043 L mol–1)

solution44
Solution

According to van der Waals’ gas equation,

  • T = 614.1 K

Ans. T = 614.1 K

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