Chapter 10: Thermodynamics

10-1 Relationship Between Heat and Work

• In a closed system there’s a direct relationship between heat and work. Heat and work both transfer energy to or from a system.

• Key idea: A system never has “heat” or “work”, it has internal energy which is affected by heat in/out or work done on/by the system

Transfer of heat and work• System: a set of particles or interacting

particles considered to be a distinct physical entity

• Environment: the combination of conditions and influences outside a system that affects the behavior of the system

• Ex of closed systems: a gas confined in a cylinder by a piston, a calorimeter, a thermos…

Work Done on or By a Gas

• Is represented in the equation:

W = PΔV

W - in Joules (J) P = pressure in Pascal (Pa) 1 Pa = 1 N/m2

ΔV= volume change in (m3)

Sample Problem

• Gas in a container is at a pressure of 1.6x105 Pa and a volume of 4.0 m3. What is the work by the gas if it expands at a constant pressure to twice its initial volume?

Solution

P= 1.6x105 Pa =1.6x105 N/m2

∆V=Vf – Vi = 8.0 m3 - 4.0 m3 =4.0 m3

W= P ∆ V

W = (1.6x105 N/m2)(4.0 m3)

W= 6.4x105 J

Thermodynamic Processes• Isovolumetric process: a

thermodynamic process that takes place at a constant volume so that no work is done on or by the system, ex: a car with closed windows parked in a hot garage.

• Isothermal process: a thermodynamic process that takes place at a constant temperature, ex usually a slow process like a balloon expanding slowly during the day.

• Adiabatic process: a thermodynamic process during which heat energy is transferred to or from the system. ex: usually a fast process like filling a tank

• Isobaric process: a process that takes place at a constant pressure. ex: heating an open pot of water

10.2 The First Law of Thermodynamics• The first law is a statement of conservation of

energy that takes into account a system’s internal energy (U) as well as the energy transfer to/from the system by work and heat.

• It is expressed as:

Signs of Q and W For a System

ΔQ = positive if heat is added to a systemΔQ = negative if heat is released from a systemΔW = positive if work is done by the systemΔW = negative if work is done on the system

First Law – Isovolumetric Process

Δ U = Q – WΔV = 0Since W = P ΔV, W = 0therefore, Δ U = Q

First Law – Isothermal Process

ΔU = Q – Wsince ΔT = 0 , ΔU = 0therefore Q = W

First Law – Adiabatic Process

Δ U = Q – WQ = 0, thereforeΔ U = – W

First Law – Isobaric Process

Δ U = Q – Wsince W = PΔVΔ U = Q – PΔV

First Law – Isolated System

Δ U = Q – Wsince Q = W = 0Δ U = 0

Sample Problem

• A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat?

Solution

W= -135 J (work done on the system is -)

∆ U= 114 J

∆ U = Q - W

Q =  ∆ U  + W

Q= 114 J + (-135 J)= -21 J

Q= -21 J

In this problem, energy is removed from the gas as heat, which is indicated by the negative sign on the Q value ( Q < 0 ).

Cyclic Processes

• A thermodynamic process in which a system returns to the same conditions under which it started (no change in system’s energy)

ΔUnet = 0 and Qnet = Wnet • Resembles an isothermal process in that all

energy is transferred as work and heat.

The Heat Engine

• Any device that exploits a temperature difference to do mechanical work

• The net work done is equal to the difference in energy taken in as heat from a high-temp. reservoir (Qh) and the energy expelled as heat to the low temp. reservoir(Qc).

• Wnet = Qnet = Qh − Qc

10.3 The Second Law of Thermodynamics

• States that no cyclic process that converts heat entirely into work is possible.

• Includes the requirement that a heat engine give up some energy at a lower temperature in order to do work.

• So a heat engine cannot transfer all energy as heat to do work.

Second Law of Thermodynamics

• No cyclic process that converts heat entirely into work is possible

chnetnet QQQW

Efficiency of a Heat Engine

• The smaller the fraction of usable energy that an engine can provide, the lower its efficiency is.

Qh

Qc

Qh

QcQh

Qh

Wneteff

1

heatasaddedenergy

enginebydoneworknetefficiency

Sample Problem

• Find the efficiency of a gasoline engine that, during one cycle, receives 204 J of energy from combustion and loses 153 J as heat into the exhaust.

Solution

Qh= 204 J

Qc= 153 J

eff= 1- Qc

Qh

eff= 1- 153 J 204 Jeff= 0.250 J

Entropy

• Entropy: a measure of the randomness or disorder of a system

• A greater disorder means there is less energy to do work

• The motion of the particles of a system is not well ordered and therefore is less useful for doing work

• Once a system has reached a state of the greatest disorder, it will tend to remain in that state and have maximum entropy.

• The second law of thermodynamics states that the entropy of the universe increases in all natural processes.