First Law of Thermodynamics

Essential UNIVERSITY PHYSICS, Volume 1 Richard Wolfson

18.1 The First Law of Thermodynamics

First Law of Thermodynamics The change in the internal energy of asystem depends only on the net heat transferred to the system and thenet work done by the system, independent of the particular processesinvolved.

Mathematically, the first law is

U = Q - W

Change in Intenalenergy

Heat addedto thesystem

Work done by thesystem


Heat addedto thesystem

Work done bythe system

U U

U = (U-U) Change in internal energy

Q

Example: Heat a gas, it expands against a weight. Force (pressure times area) is applied over a distance, work is done.

U = Q - W


We are frequently concerned with rates of energy flow.

Differentiating the first law with respect to time gives a

statement about rates:

dU/dt is the rate of change of a systems internal energy.

dQ/dt is the rate of heat transfer to the system.

dW/dt is the rate at which the system does work.

The First Law of Thermodynamics: Thermal PollutionEXAMPLE 18.1

The reactor in a nuclear power plant supplies energy at the rate of3.0 GW, boiling water to produce steam that turns a turbine-generator.The spent steam is then condensed through thermal contact with watertaken from a river. If the power plant produces electrical energy atthe rate of 1.0 GW, at what rate is heat transferred to the river?

The entire power plant is our system, comprising the nuclear reactor,including its fuel, and the turbine-generator.We identify U as the internal energy stored in the fuel, W as themechanical work that ends up as electrical energy, and Q as the heattransferred to the river.

We know that: dU/dt = dQ/dt dW/dt. The reactor extracts internalenergy from its fuel, so the rate dU/dt is negative; the power plantdelivers electrical energy to the outside world, so dW/dt is positive.We need to solve for dQ/dt.

GIVEN: dU/dt = 3.0GW and dW/dt = 1.0 GW

-3.0 GW 1.0 GW

-3.0 GW + 1.0 GW

-2.0 GW

The First Law of Thermodynamics: Thermal PollutionEXAMPLE 18.1

18.2 Thermodynamic Processes

Although the 1st law applies to any system, its easiest to understandwhen applied to an ideal gas. The ideal gas law relates thetemperature, pressure, and volume of a given gas sample:

pV = nRt

The thermodynamic state is completely determined by any two of thisquantities p, V, or T.

p,V,T

p,V,T

V

p


Reversible and Irreversible Processes

Water Gas

T T

Temperature control

These temperatures stay the same as the water temperature increasesslowly

If we raise thereservoirtemperature veryslowly, both thewater and gas temperatures willrise in unison, and the gas willremain in

equilibrium.

Quasi-staticprocessA system is alwaysin thermodynamicequilibrium.


Reversible and Irreversible Processes

Water Gas

T T

Temperature control

These temperatures stay the same as the water temperature increasesslowly

Reversible Process:

We could reverse this heatingprocess by slowlylowering thereservoirtemperatur; the gas would cool, reversing its pathin the pV diagram.

Irreversible Process:If we suddenlyplunging a cool gas sample into hotwater, then itbecomes and irreversible process.The system is notin equilibrium.


Work and Volume Changes

The work done by the system is related to the changes in pressure and volume:

(work done during volume change)


Isothermal Processes

An isothermal process occurs at constant temperature.

To find work, we relate pressure and volume through the ideal gas law: p = (nRT)/V. Then it becomes:

For an isothermal process, the temperature T is constant, giving


Isothermal Processes

The first law of thermodynamics then gives U = 0 = Q W, so

T = constant

Q = W

W = nRT*ln(V/V)

pV = constant


Constant-Volume Processes and Specific Heat

A constant-volume process occurs in a rigid closed container whosevolume cant change.To express U = Q in terms of a temperature change T, we introduce the molar specific heat at constant volume Cv defined by:

V

p

V = constant

Q = U

W = 0

Q = nCvT


Isobaric Processes and Specific HeatIsobaric means constant pressure. Processes occurring in systemsexposed to the atmosphere are essentially isoberic. The work done asthe volume changes from V to V, is the area under the isobar, or

W = p(V-V) = pV

(isobaric process)

Cp is the molar specific heat at constant pressure.

(molar specific heats)


Isobaric Processes and Specific Heat

p = constant

Q = U + W

W = p(V-V)

Q = nCpT

Cp = Cv + R

V

p

W

T

T

isobar

VV


Adiabatic Processes

In an adiabatic process, no heat flows between a system and itsenvironment. The way to achieve this is to surround the system withperfect thermal insulation.

Since the heat Q is zero in an adiabatic process, the 1st law becomessimply

U = -W (adiabatic process)


Adiabatic Processes

As a gas expands adiabatically, its volume increases while itsinternal energy and temperature decrease. The ideal-gas law,pV=nRT, then requires that the pressure decrease as well and bymore than it would in an isothermal process where T remainsconstant.

(adiabatic process)


Adiabatic Processes

Q = 0

U = -W

W = (pV-pV)/(-1)

pV = constant

TV(-1) = constant

An Adiabatic Process: Diesel PowerEXAMPLE 18.3

Fuel ignites in a diesel engine from the heat of compression as thepiston moves toward the top of the cylinder; theres no spark plug asin a gasoline engine. Compression is fast enough that the process isessentially adiabatic. If the ignition temperature is 500C, WHATCOMPRESSION RATIO Vmax/Vmin is needed? Airs specific-heat ratio is=1.4, and before compression the air is at 20C.

The problem involves temperature and volume, we apply the formulaTminVmin(-1) = TmaxVmax(-1).

Solving for the compression ratio Vmax/Vmin gives

18.3 Specific Heats of an Ideal Gas

The specific heats of an ideal gas follow from the degrees offreedom of each molecule:

Monatomic

3 degrees of freedom

Cv = (3/2)R

Diatomic

5 degrees of freedom

Cv = (5/2)R

First Law of Thermodynamics

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