Chemistry. Chemical energetics-1 Session Objectives.

Chemistry

Chemical energetics-1

Session Objectives

Session Objective

1. Overview of thermodynamics

2. Types of systems

3. Types of thermodynamic processes

4. Zeroth law

5. Internal energy

6. Work

7.First law of thermodynamics

Definition - Thermodynamics

quantitative relationship between various forms of energies.

Importance of Thermodynamics

Predict the feasibility of a chemical reaction under a given set of conditions.

Predict the extent to which the reaction is carried out before attainment of equilibrium.

Basic Terms in Thermodynamics

Universe

system surroundings

Types of systems

Open system

Closed system

Isolated system

Homogeneous system

e.g. aqueous solution of NaCl, aqueous solution of sugar.

Heterogeneous system

e.g. ice in water, oil in water, etc.

Properties of a system

1. Macroscopic property

shown by large number of particles of a system.

2. Microscopic property

Property of individual atoms and molecules of a system at microlevel

volume, pressure, temperature,density.

atomic mass, molecular mass.

Macroscopic Properties of System

extensive properties intensive properties

mass, volume temperature, pressure, viscosity, density,

Macroscopic Properties

State function Thermodynamic properties which depend only upon the initial and final states of a system For example, internal energy, enthalpy, etc. mass, volume, etc.

State of a system

The macroscopic properties which causes a change in the state of a system are called state variables. For example, pressure, temperature, etc.

Types of Thermodynamic Process

(i) Isothermal process A process which is carried outat constant temperature.

(ii) Adiabatic process A process which is carried out in such a way that no heat flows from the system to surroundings and vice versa.

(iii) Isochoric process A process which is carried out at constant volume.

(iv) Isobaric process A process which is carried out at constant pressure.

Types of Thermodynamic Process(v) Reversible process

carried out infinitesimally slow so that all changes occurring can be reversed and the system remains almost in a state of equilibrium with the surroundings at every instant of time.

(vi)Irreversible process Occurs rapidly but the system cannot be reversed to its initial state immediately and the system does not remain in equilibrium during transition.

(vii)Cyclic process When a system undergoes different processes and finally returns to its initial state, is known as cyclic process.

Zeroth law (Law of thermal equilibrium)

Thermal equilibrium

Thermal equilibrium

Thermal equilibrium

(A) (B)

(B) (C)

(A) (C)

Internal energy

Energy stored within a substance of a system

E = Ee + Ev + Er

internal energy or intrinsic energy.

Internal energy

Actual (absolute) value can not be measured

change in internal energy (E) can be measured

It is an extensive property.

Internal energy is a state function

For a cyclic process,

E = 0

Work

Where,

W = work done

W = –ve, work done by the system (expansion)

W = +ve, work done on the system (contraction)

Work done = Force × Displacement

W = P(V2 – V1) (At constant pressure)

W = PV

Work

Work done in isothermal reversible expansion of an ideal gas

Pext– dP

Pg

dv

Pext

Pg

Work

Work done in this isothermal reversible process in each step is,

extP dV

Hence, the total work done in an isothermal expansion from volume V1 to V2 is

2

1

V

revv

W PdVSince PV = nRT

W = – (Pext – dP)dV

Work

2

1

V

revV

nRTW dV

V

2rev

1

VW nRT ln

V

Wrev = – 2.303 nRT log 2

1

V

V

1

Since PV

1

2

P

P Wrev = – 2.303 nRT log

Work done in irreversible isothermal expansion

Pext = 0, i.e. work done is zero when gasexpands in vacuum.

Pext < Pg. So, the work done when volume changes from V1 to V2 is Wirr = Pext(V2 – V1).

Intermediate expansion

Free expansion

First Law of Thermodynamics

E2 = E1 + q + w

E2 – E1 = q + w

Energy can neither be created nor destroyed

E = q + w

If work is done by the system, (in case of expansion)

w = - P v

E = q - P v

q = E + P v

First Law of Thermodynamics

c. In case, the process is carried out at constant volume

E = 0

q = P v

a. During isothermal expansion

b. In case of adiabatic process, q = 0

E = - P v = +w

E = +w

First Law of Thermodynamics

Quantity of heat supplied at constant volume, qv is equal to increase in the internal energy of the system.

V

V = 0

P V = 0 or w = 0

q = E

Class exercise

Class exercise 1

For an adiabatic process which of the following is correct?

(a) q = 0 (b) E = q(c) q = +w (d) pv = 0

Adiabatic process is carried out in a thermally insulated system.

Hence, q = 0.

Solution:

Hence, the answer is (a)

Class exercise 2

In an isothermal process

(a) temperature is kept constant

(b) no heat exchanges

(c) no work is done

(d) None of these

Solution:

In an isothermal process, temperature is kept constant.

Hence, the answer is (a)

Class exercise 3In case of an isothermal reversible process involving an ideal gas, work done is given by

(a) q = – w (b) q = E + w

(c) q = E (d) q = E – w

According to first law,

E = q + w

In case of isothermal reversible process

E = 0

q = – w

Solution:

Hence, the answer is (a)

Class exercise 4

What is meant by reversible process?

A reversible process is a process which is carried out infinitesimally slowly so that all changes can be exactly reversed and the system remains almost in a state of equilibrium with the surroundings at every stage of the process.

Solution:

Class exercise 5

What is the work done when an ideal gas expands from 10–3 m3 to 10–2 m3 at 300 K, against a constant pressure of 105 Nm–2?

5

5

Work done = -P( V)

= -10 (0.01 - 0.001)

= -10 (0.009)

= - 0.9 kJ

Solution:

Class exercise 6

Calculate the values of q, w and DE when one mole of a gas absorbs 500 J of heat at constant volume and temperature rises from 25° C to 35° C.

E = q + w

w = - PDV ( V = 0)

w = 0

E = q = 500 J

Solution:

Class exercise 7

Work done by the gas when the gas expands isothermally and reversibly is

(a) zero (b) minimum

(c) maximum (d) cannot be determined

E = q + w

E = 0 in case of an isothermal and reversible process.

q = - w

Work done is maximum.

Solution:

Hence, the answer is (c)

Class exercise 8A gaseous system changes from stateP(P1, V1, T1) to Q(P2, V2, T2), Q toR(P3, V3, T3) and finally from R to P. The whole process is called

(a) reversible process (b) cyclic process

(c) spontaneous process (d) isobaric process

When the initial and final states of a process are same then the process is said to be cyclic.

Solution:

Hence, the answer is (b)

Class exercise 9A gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of 1 atmosphere from a volume of 2L to a volume of 12 L. In doing so, it absorbs 800 J heat from surroundings. Determine the increase in internal energy of the process.

Given, q = 800 J

v = (12 - 2) = 10 L

w = - p v = -1 × 10 litre atm.

w = -1013.2 J

(1 lit atm = 101.32 J )

According to first law,

E = q + w = (800 - 1013.1) = - 213.2 J

Solution:

Class exercise 10

What work is to be done on 2 mol of a perfect gas at 27°C if it is compressed reversibly and isothermally from a pressure of 1.01 × 105 Nm–2 to 5.05 × 106 Nm–2 (R = 8.314 J mol–1 K–1).

Given that

n = 2 mol P1= 1.01 × 105 N/m2

T = 300 K R = 8.314 J /mol k

P2 = 5.05 × 106 N/m2

Solution:

Solution

W = - nRT ln = - 2.303 nRT log

W = - 2 × 2.303 × 8.314 × 300 × log

W = +19.518 kJ

Thank you