Physics 52 - Heat and Optics

Dr. Joseph F. BeckerPhysics Department

San Jose State University© 2005 J. F. Becker

Chapter 19

The First Law of Thermodynamics

Ch. 19 First Law of Thermodynamics

1. Thermodynamic systems2. Work done during volume changes3. Paths between thermodyn. states4. Internal energy and the First Law 5. Kinds of thermodynamic processes6. Internal energy of an ideal gas7. Heat capacities of an ideal gas8. Adiabatic processes for ideal gas

Sign conventions:a) Q is positiveb) Q is negativec) W is positived) W is negative

e) In a steam engine the heat in and the work out (done) are both defined to be POSITIVE.

(a) Molecule hits a piston moving away from it; speed and KE of molecule decrease. (expan. -> cooling)

(b) Molecule hits a piston moving toward it; speed and KE of molecule increase. (comp. -> heating)

When a molecule hits a wall moving away from it, the molecule does work on the wall; the molecule’s speed and kinetic

energy decrease. The gas does positive work (on the piston);

and the internal energy of the gas decreases.Molecule hits a wall moving toward it, the

wall does work on the molecule; the molecule’s speed and kinetic energy

increase. The gas does negative work (on the

piston); and the internal energy of the gas

increases.

WORK-ENERGY THEOREM

The infinitesimal work done by the system (gas) during the small expansion dx is dW = p A dx = F dx so W = p dV

The work done equals the area under the curve on a p-V diagram. (a) Volume increases: work and area are positive. (b) Volume decreases: work and area are negative. (c) A constant-pressure: Volume increases, W > 0.

(a) Three different paths between state 1 and state 2.

(b)-(d) The work done by the system during a transition between two states depends on

the PATH (or PROCESSES) chosen.

The final states are the same. The intermediate states (p and V) during the transition from state 1 to 2 are entirely different. Heat, like work, depends on the initial and final states AND the path between the states. (a) Heat is added slowly so as to keep the temperature constant. (b) Rapid, free expansion does no work and there is no heat transfer.

HEAT ADDED

In a thermodynamic process, the internal energy of a system may increase, decrease, or remain the same:(a) U > 0(b) U < 0 (c) U = 0

The First Law of Thermodynamics

U = Q - W

Daily thermodynamic process of your body (a thermodynamic system)

The net work done by the system in

the CYCLIC PROCESS aba is

–500 J.

A p-V diagram

of a cyclic

process.

Q = ?

<

KINDS OF THERMODYNAMIC PROCESSES

ADIABATIC – NO HEAT TRANSFER; Q=0; U=-W

ISOCHORIC – CONSTANT VOLUME; W=0; U = Q

ISOBARIC – CONSTANT PRESSURE; W=p (V2 – V1)

ISOTHERMAL – CONSTANT TEMPERATURE; only for ideal gas U = 0, Q =

W

The First Law of Thermodynamics

U = Q - W or Q = U + W

Processes for an ideal gas:

adiabatic Q=0isochoric W=0

isothermal U=0

Q = 0

W = 0U = 0

The internal energy of an ideal gas depends only on its temperature,

not on its pressure or volume.

Raising the temperature of an ideal gas by various

processes:Q = U + W and U = f(T)Isochoric W = 0 so Q = UIsobaric Q = U + p(V2-V1)

A p-V diagram of an adiabatic process

for an ideal gas from state a to state b.

pV = constant (isotherm)

pV = constant (adiabat)

ADIABATIC PROCESSES for IDEAL GAS

Q = 0 so U = Q – W = 0 – p V and U = n Cv dT for an ideal gas

U = n Cv dT = – p V = -(n R T/V) dV dT/T + (R/ Cv) dV/V = 0dT/T + ( - 1) dV/V = 0 where – 1 = (Cp/Cv

) –1 dT/T + ( - 1) dV/V = 0 ; lnT+(-1) lnV = const

ln (TV -1 ) = constant and T1V1 -1 = T2V2 -1

And since pV = nRT, (pV/nR) V -1 = constant’

(p/nR) V = constant’ p1V1 = p2V2

Adiabatic compression of air in a cylinder of a diesel engine.

pCarnot Cycle for an

ideal gas

HOT

COLD

The Carnot

cycle for an ideal

gas.

Isotherms in light-blue. Adiabats in dark-blue.

p

Along which path is a) work done and b) heat transferred by the system greatest?

Q = U + W

The absolute value of heat transfer during one cycle is 7200

J.Assigned for HW

A quantity of air is taken from state a to state b along the

straight line.

Is Tb > Ta?

Given values for all p’s and V’s,

calculate the work.

Q = 90 J W = 60 J

Q = ? W = 15 J

Q = ? |W| = 35 J

Wadc =120 J

Wabc = 450 J

Ua = 150 J.

Ub = 240 J. Uc = 680 J.

Ud = 330 J.

Qab = ?

Qbc = ?

Qdc = ?

Qad = ?

Review