Lecture 1 Laws of Thermodynamics

Thermodynamic state - equilibrium Thermodynamic processes Laws of thermodynamics Absolute Temperature Problems 2.5, 2.6, 2.8

Thermodynamic intensive coordinates are uniform across the whole system (T, P, ) or across each macroscopic phase (e.g., water and ice density density at the melting point.

All thermodynamic coordinates are time independent

Mechanical equilibrium, thermal equilibrium and chemical equilibrium

Thermodynamic state - equilibrium

Thermodynamic coordinates (T, P, ) define macroscopic state of equilibrium.

Microscopic state is defined by atomic positions, and momenta - many microscopic states are consistent with a macroscopic state

Statistical mechanics connects microscopic description and detail with macroscopic state via ensemble average

Macroscopic vs. microscopic state

Infinitesimal process infinitesimal change of coordinates, e.g., dT, dV, dP

Quasi static process always near equilibrium

Adiabatic process no heat

Reversible process can be restored to the initial state without charging surroundings

Thermodynamic process change of the thermodynamic state

Two or more systems in equilibrium do not exhibit heat flow among each other, they are at the same temperature

Later we will see that criterion of equilibrium for isolated system, i.e., const E, V, T is the maximum entropy state

dS(E, V, N) = 0

0th law of thermodynamics

E1, V1, N1 E2, V2, N2

dS = dS1+dS2 = 0

€

dS1 =∂S1

∂E1

dE1 +∂S1

∂V1

dV1 +∂S1

∂N1

dN1 =∂S1

∂E1

dE1

Allowing only energy exchange between

two isolated systems

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dS2 =∂S2

∂E21

dE2 +∂S2

∂V2

dV2 +∂S2

∂N2

dN2 =∂S2

∂E2

dE2€

dS1 + dS2 =∂S1

∂E1

dE1 −∂S2

∂E2

dE1

€

dE2 = −dE1

From conservation of energy €

∂S1

∂E1

=∂S2

∂E2

T =∂E

∂S

1st law of thermodynamics - conservation of energy

dE = dQ-dW

d indicates inexact differential - depends on the integration path

In a cycle, E = 0

net Q in = net W out

Work and heat are not state functions

Energy is a state function

dW = Fdx can be PdV, -dl, -it

2nd law of thermodynamics - entropy

For a reversible process

dQ= TdS

Where S is entropy which a state function, and T is an absolute temperature

The entropy can by calculated by integrating heat over a reversible path

€

S f − Si =dQ

TR∫

Absolute temperatureConsider the Carnot cycle

T

S

T2

T1

Q1

Q2

€

S1 =Q1

T1

ΔS2 =Q2

T2

Since entropy is the state function

€

Q1

T1

=Q2

T2€

S1 = −ΔS2

Using reference T3 = 273.16 K

€

T = 273.16Q

Q3

€

Q

Q3

Q > 0

Problem 2.5When a system is taken from a to b state along abc 80 joules of heat flows

into the system and the system does 30 joules of work.

a) How much heat flows into the system along path adb, if the work done by the systems is 10 joules.

b) When the system is returned from b to a along the curved path the work done on the system is 20 joules. Does the system absorb or liberate the heat? How much?

c) If Ea = 0 and Ed = 40 joules, find the heat absorbed in process ad and db.

P

V

Answers:

a) Qadb = 60 joules.

b) Qba = - 70 joules joules (liberate heat) of the heat

c) Qad= 50 J, Qdb= 10 J

b

da

c

Problem 2.6

A vessel of volume VB contains n moles of gas at high pressure. Connected to the vessel is a capillary tube trough which the gas may slowly leak out into the atmosphere, where P=P0. Surrounding the vessel and capillary is a water bath, in which is immersed an electric resistor. The gas is allowed to leak slowly trough the capillary into the atmosphere while, at the same time, electrical energy is dissipated in the resistor at such a rate that the temperature of the gas, the vessel, the capillary and the water is kept equal to that of the surrounding air. Show that, after as much gas is leaked as is possible during time , the change of internal energy is

where, v0 = molar volume of gas at P=P0, = potential on the resistor, and i is the current.

€

E = εiτ − P0(nv0 −VB )

Problem 2.8

The tension force in a wire is increased quasi statically and isothermally from 1 to 2. If the length, cross-sectional area and isothermal Young’s modulus (Y) remain practically constant, show that the work done by the wire is:

€

W = −L

2AY(τ 2

2 − τ 12)