ThermodynamicsHonors Physics

Biblical ReferenceIt makes the depths churn like a boiling caldron and stirs up the sea like a pot of ointment.

Job 41:31

Work done by a gas

If a piston is filled with a specific amount of gas, as you add heat, the temperature rises and the volume of the gas expands.

The gas applies a force on the piston wall pushing it a specific displacement.

Thus, a gas can do work (Force x Distance).

Work is the AREA of a P vs. V graph

The Sign in the Work Equation

negativeW

V

VPW

by

gas theBY done is Work

Since work done by a gas has a positive volume change, the gas is using up energy (or losing energy) thus the negative sign.

When work is done on a gas the change in volume is negative. This cancels out the negative sign in the equation, because some

external agent is adding energy to the gas.

positiveW

V

VPW

on

gas theON done isWork

If Objects A and B are independently in thermal equilibrium with Object C, then Objects A and B are in thermal equilibrium with each other.

Why “Zeroth”? It was formed after 1st, 2nd, and 3rd, but was more fundamental.

Zeroth Law of Thermodynamics

First Law of Thermodynamics

“The internal energy of a system tends to increase when HEAT is added and work is done ON the system.”

If you ADD heat then transfer work TO the gas, the internal energy must go up, because you have MORE than what you started with.

ExampleSketch a PV diagram and find the work done by the gas during the following stages.

(a) A gas is expanded from a volume of 1.0 L to 3.0 L at a constant pressure of 3.0 atm.

(b) The gas is then cooled at a constant volume until the pressure falls to 2.0 atm

)001.0003.0(103 5xVPWBY 600 J

0 since

0

V

VPW

Example continued

c) The gas is then compressed at a constant pressure of 2.0 atm from a volume of 3.0 L to 1.0 L.

d) The gas is then heated until its pressure increases from 2.0 atm to 3.0 atm at a constant volume.

)003.001(.102 5xVPWON -400 J

0 since

0

V

VPW

Example continued

What is the NET WORK?

NET work isthe area inside the shape.

600 J + -400 J = 200 J

Rule of thumb: If the system rotates CW, the net work is positive.

If the system rotates CCW, the net work is negative.

Thermodynamic Processes - Isothermal To keep the temperature

constant both the pressure and volume change to compensate.

Volume goes up, pressure goes down

“BOYLES’ LAW”

Boyle’s Law

Boyle’s law states that when temperature is constant:◦ Pressure of a gas increases if the volume decreases.◦ Pressure of a gas decreases if the volume increases.

2211 VPVP

Pressure & Volume

Pressure and Volume are inversely proportional.

P2

Boyle’s Law – Solve for V2

If you add 15 L of air to a balloon at sea level (101 kPa) and then take the balloon to Denver, where the air pressure is 86 kPa, what will be the new volume of the balloon?

2211 VPVP P2

LkPa

LkPa

P

VPV 6.17

86

)15)(101(

2

112

V2

Boyle’s Law – Solve for P2

If you have 5.6 liters of gas in a piston at a pressure of 152 kPa and compress the gas until its volume is 4.8 L, what will be the new pressure inside the piston?

2211 VPVP V2

kPaL

LkPa

V

VPP 177

8.4

)6.5)(152(

2

112

Calculate the pressure in atmospheres in a motorcycle engine at the end of the compression stroke. Assume that at the start of the stroke, the pressure of the mixture of gasoline and air in the cylinder is 745.8 mm Hg and the volume of each cylinder is 246.8 mL. Assume that the volume of the cylinder is 24.2 mL at the end of the compression stroke.

Boyle’s Law – P1V1 = P2V2

2211 VPVP

mmHgmL

mLmmHg

V

VPP 7606

2.24

)8.246)(8.745(

2

112

Thermodynamic Processes - Isovolumetric

• Gas undergoes a temperature change but no change in volume.

• No work is done on or by the system.

• Gay-Lussac’s Law

A container is designed to hold a pressure of 2.5 atm. It is filled with air at room temperature (20°C – 293 K) and normal atmospheric pressure. Would it be safe to throw the container into a fire where temperatures of 600°C (873 K) would be reached?

Gay-Lussac’s Law - 2

2

1

1

T

P

T

P

2

2

1

1

T

P

T

P

atmK

Katm

T

TPP 98.2

293

)873)(0.1(

1

212

Thermodynamic Processes - Isobaric Heat is added to the gas

which increases the Internal Energy (U).

Work is done by the gas as it changes in volume.

The path of an isobaric process is a horizontal line called an isobar.

∆U = Q - W can be used since the work is positive.

Charles’ Law

Charles’ Law

Charles’s law states that when the pressure is constant:◦ The volume of a gas increases if the temperature

increases.◦ The volume of a gas decreases if the temperature

decreases.

2

2

1

1

T

V

T

V

Temperature and Volume

Volume and Temperature are directly proportional.

Charles’ Law – Solve for V2

If you have 45 L of helium in a balloon at 25 C (298 K) and increase the temperature of the balloon to 55 C (328 K), what will be the new volume of the balloon?

2

2

1

1

T

V

T

V 2T2T

LK

LK

T

VTV 5.49

298

)45)(328(

1

122

Charles’ Law – Solve for T2

If you have 130 L of gas in a piston at 250 C (523 K), and cool the gas until the volume decreases to 85 L, what will be the temperature of the gas?

2

2

1

1

V

T

V

T 2V2V

)69(342130

)523)(85(

1

122 CK

L

KL

V

TVT

Charles’ Law - 2

1

2

1

T

T

V

V

Assume that the volume of a balloon filled with H2 is 1.00 L at 25°C. Calculate the volume of the balloon when it is cooled to -78°C in a low-temperature bath made by adding dry ice to acetone.

2

1

2

1

T

T

V

V

LK

KL

T

TVV 65.0

298

)195)(00.1(

1

212

Thermodynamic Processes - Adiabatic Greek - adiabatos-

"impassable"

No heat can leave or enter the system.

Work is done and internal energy changes but heat does not change.

A sample of gas has a volume of 283 mL at 25C (298 K) and 0.500 atm. What is the volume at 100C (373 K) and 1.00 atm?

Combined Gas Law - 2

22

1

11

T

VP

T

VP

2

22

1

11

T

VP

T

VP

21

2112 PT

TVPV

mLatmK

KmLatmV 177

)00.1)(298(

)373)(283)(5.0(2

In Summary

Gay-Lussac’s Law

Charles’s Law

ExampleA gas is enclosed in a container fitted with a piston of cross sectional area .10 m2. The pressure of the gas is maintained at 8000 Pa while the heat is slowly added, pushing the piston up a distance of .04 m. If 42 J of heat is added to the system during the expansion, what is the change in internal energy?

VPQU

JJJ

mmPaU

103242

)04.0)(1.0)(8000(42 2

Second Law of Thermodynamics

“Heat will not flow spontaneously from a colder body to a warmer body AND heat energy cannot be transformed completely into mechanical work.”

The bottom line:1) Heat always flows from a hot body to a cold body2) Nothing is 100% efficient

Heat Engines

• Heat flows from a Hot reservoir to a Cold reservoir

CHoutput

CH

QQW

QWQ

QH = remove from, absorbs = hotQC= exhausts to, expels = cold

• Convert thermal energy into mechanical energy.

Engine Efficiency

In order to determine the thermal efficiency of an engine you have to look at how much energy you get out based on how much energy you take in.

H

C

H

CH

hotthermal Q

Q

Q

QQ

Q

We

1

heat as addedenergy

heat as removed 1

energy

Example – Engine Efficiency

An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?

H

C

H

CH

hotthermal Q

Q

Q

QQ

Q

We

1

Rates of Energy UsageSometimes it is useful to express the energy usage of an engine as a RATE.

POWERt

W

t

Q

t

Q

C

H

For example:

The RATE at which heat is absorbed.

The RATE at which heat is expelled.

The RATE at which WORK is DONE.

Is there an IDEAL engine model?• How efficient can an Engine be?• What’s the most work we can

possibly get for a given amount of fuel?

• The efficiency question was first posed - and solved - by Sadi Carnot in 1820, not long after steam engines had become efficient enough to begin replacing water wheels, the main power sources for industry. 

•

• Carnot visualized the heat engine as a kind of water wheel in which heat (the “fluid”) dropped from a high temperature to a low temperature, losing “potential energy” which the engine turned into work done, just like a water wheel. 

Entropy – the Measure of DisorderGreater disorder means there is less energy to do work.

Example – Tropical Ocean as a Heat EngineWater near the surface of a tropical ocean has a temperature of 298.2 K (25.0 ºC), whereas water 700 m beneath the surface has a temperature of 280.2 K (7.0 ºC). It has been proposed that the warm water may be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. What would be the efficiency for such an engine.

TH = 298.2 K and TC = 280.2 K

Cooling and Heating Systems