Chapter 2 first law of TMD

As the weights fall, they give up potential energy and warm the water accordingly.. If friction in mechanism is negligible, the work done by the paddle wheel on the water equal the chang

Dr Ngo Thanh An

PHYSICAL CHEMISTRY 1

Chapter 2 – First law of thermodynamics

1

As the weights fall, they give up potential energy and warm the water accordingly This was first demonstrated by James Joule, for whom the unit of energy is named.

Schematic diagram for Joule´s experiment Insulating walls to prevent heat transfer enclose water

As the weights fall at constant speed, they turn a paddle wheel, which does work on water

If friction in mechanism is negligible, the work done by the paddle wheel on the water equal the change of potential energy of the weights

1 Joule’s experiment

Chapter 2 – First law of thermodynamics

Internal energy involves energy on the microscopic scale

For thermochemistry  the internal energy

is the sum of the kinetic energy of motion of the molecules, and the potential energy represented by the chemical bonds between the atoms and any other intermolecular forces that may be operative.

2 Internal energy

Chapter 2 – First law of thermodynamics

∆ U = Q - A

3 Statement of First law of thermodynamics

Chapter 2 – First law of thermodynamics

∆ U = Q1 – A1 = Q2 – A2 = Q3 – A3 1

“d” used for describing state function

“ δ ” used for describing path function

U: state function

5

Chapter 2 – First law of thermodynamics

3 Statement of First law of thermodynamics

Chapter 2 – First Law of Thermodynamics

Meaning of exact differential

3 Statement of First law of thermodynamics

Thermodynamic definition of work: It is a kind of interaction that would occur at the

system boundaries It can be positive or negative.

Heat vs Work

Chapter 2 – First law of thermodynamics

Heat vs Work

Chapter 2 – First law of thermodynamics

Heat vs Work

Chapter 2 – First law of thermodynamics

Heat vs Work

Chapter 2 – First law of thermodynamics

Work (A) Heat (Q)

release > 0 < 0 absorb < 0 > 0

4 Sign convention

Chapter 2 – First law of thermodynamics

In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston What is the change in internal energy of the system?

∆ Q is positive: +400 J (Heat IN)

∆ W is negative: -120 J (Work OUT)

Example

Chapter 2 – First law of thermodynamics

Distinguish between heat, temperature, and thermal energy

• Thermal energy is an energy of the system due to the motion

of its atoms and molecules Thermal energy is a state variable,

it may change during a process The system’s thermal energy continues to exist even if the system is isolated and not interacting thermally with its environment

• Heat is energy transferred between the system and the environment as they interact Heat is not a particular form of energy, nor is it a state variable Heat may cause the system’s thermal energy to change, but that does not mean that heat and thermal energy are the same thing.

• Temperature is a state variable, it is related to the thermal energy per molecule But not the same thing.

Chapter 2 – First law of thermodynamics

Def The specific heat capacity (c) of a substance is the amount of energy needed to raise the temperature of 1kg of the substance by 1K (or 1 oC)

5 Specific heat capacity

Chapter 2 – First law of thermodynamics

5 Specific heat capacity

Effect of temperature on specific heat capacity:

For 01 mole of ideal gas: CP – CV = R

2

1

T T

Q = ∫ CdT

Chapter 2 – First law of thermodynamics

5 Specific heat capacity

• For ideal gas:

Cp = Cv +

R

Chapter 2 – First law of thermodynamics

5 Specific heat capacity

Chapter 2 – First law of thermodynamics

5 Specific heat capacity

• pressure specific

Constant-• heats for some gases

5 Specific heat capacity

Chapter 2 – First law of thermodynamics

V V

a Isochoric process (V = const or dV = 0)

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b Isobaric process (p = const or dp = 0)

Enthalpy is the sum of internal energy and the product of pV

Chapter 2 – First law of thermodynamics

6 Application of 1st law of TMD

Ap = p ∆ V = ∆ (pV) = ∆ (nRT) = nR ∆ T

⇒

R: ideal gas constant, R = 1,987 cal/mol.K

= 8,314 J/mol.K = 0,082 l.atm/mol.K

Joule’s law: (for ideal gas)

Internal energy of ideal gas just depends on

e Polytropic process

Chapter 2 – First law of thermodynamics

6 Application of 1st law of TMD

e Polytropic process

Chapter 2 – First law of thermodynamics

6 Application of 1st law of TMD

A biatomic ideal gas undergoes a cycle starting at point A (2 atm, 1L) Process from A to B is an expansion at constant pressure until the volume is 2.5 L, after which is cooled at constant volume until its pressure is 1 atm It is then compressed at constant pressure until the volume is again 1L, after which it is heated at constant volume until

it is back in its original state Find (a) the work, heat and change of internal energy in each process (b) the total work done on the gas and the total heat added to it during the cycle

A system consisting of 0.32 mol of a monoatomic ideal gas occupies a volume of 2.2 L, at a pressure of 2.4 atm The system is carried through a cycle consisting:

1 The gas is heated at constant pressure until its volume

7 Example

Chapter 2 – First law of thermodynamics

7 Example

Chapter 2 – First law of thermodynamics

7 Example

Chapter 2 – First law of thermodynamics

7 Example

Chapter 2 – First law of thermodynamics

7 Example

Chapter 2 – First law of thermodynamics

7 Example

Chapter 2 – First law of thermodynamics