chapter 15. work, heat, and the first law of thermodynamics

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Chapter 15. Work, Heat, and the First Law of Thermodynamics


Heat and Work in Ideal-Gas Processes

• Consider a gas cylinder sealed at one end by a moveable piston.• Assume KE =0 (cylinder not moving) and PE = 0 (cylinder position at y = 0). • But since T > 0 kelvin, Uint ≠ 0.•For a monatomic gas:

U = 3/2 nRT• For non-monatomic gases, U is still proportional to T:

U α T


Heat and Work in Ideal-Gas Processes

•How can we change the amount of internal energy in our system of an ideal gas? • We have studied 2 energy transfer mechanisms, heat (Q) and work (W). Let’s look at both of these mechanisms.


Review: What is the best definition of heat?

A. the amount of thermal energy in an object. B. the energy that moves from a hotter object

to a colder object. C. how high the temperature of an object is.D. all of the above


What is the best definition of heat?

A. the amount of thermal energy in an object. B. the energy that moves from a hotter object

to a colder object. C. how high the temperature of an object is.D. all of the above


Heat, Temperature, and Thermal Energy• Thermal energy U is an energy of the system due to the motion of its atoms and molecules. Any system has a thermal energy even if it is isolated and not interacting with its environment. The units of U are Joules.

• Heat Q is energy transferred between the system and the environment as they interact due to a difference in temperature. The units of Q are Joules.

• Temperature T is a state variable that quantifies the “hotness” or “coldness” of a system. A temperature difference is required in order for heat to be transferred between the system and the environment. The units of T are degrees Celsius or Kelvin.


If heat is the only energy transfer mechanism

ΔU = Uf -Ui = Q

Q is positive when the system gains energy. This means that the environment has a higher temperature than the system.

Q is negative when the system loses energy. This means that the environment has a lower temperature than the system.


Work done by the system and on the system

W = |F| |Δx| cos θ• If the piston moves to the right:• Work done by the gas molecules on the piston is positive (force is to the right, piston moves to the right).• energy is added to the piston, energy is taken away from the gas and the gas expands.

•If the piston moves to the left: •Work done by the gas molecules on the piston is negative (force is to the right, gas molecules move left). •Energy is added to the gas and the gas compresses.


15.3 The First Law of Thermodynamics

THE FIRST LAW OF THERMODYNAMICS

The internal energy of a system changes due to heat and work:

byif WQUUU

Work is positive when it is done by the system and negative when it is done on the system.

Heat is positive when the system gains heat and negative when the system loses heat.


A gas cylinder and piston are covered with heavy

insulation so there can be no heat exchange with the

environment. The piston is pushed into the cylinder,

compressing the gas. According to the 1st Law of

Thermodynamics, the gas temperature:

A. decreases.B. increases.C. doesn’t change.D. There’s not sufficient

information to tell.


ΔU = Q – Wby • insulation implies no heat exchange with environment•Work done by gas is negative. Gas pushes left, piston compresses gas to the right:

ΔU = Q – Wby

ΔU = – – Wby ΔU is proportional to temperature increase.

Therefore, temperature increases.


EOC # 1The internal energy of a system changes because the

system gains 165 J due to heat transfer and does 312 J of work. In returning to its initial state *, the system loses 114 J of heat.

a. How much work is involved during the return process?

b. Is the work done by the system, or on the system?

* Initial state (for a gas) means same voloume, pressure, temperature, and internal energy.


EOC # 1-AnswerThe internal energy of a system changes because the

system gains 165 J due to heat transfer and does 312 J of work. In returning to its initial state *, the system loses 114 J of heat.

a. How much work is involved during the return process? 261 J involved

b. Is the work done by the system, or on the system? Work is done on the system.



Example 2 An Ideal Gas

The temperature of three moles of a monatomic ideal gas is reduced from 540K to 350K as 5500J of heat flows into the gas.

Find (a) the change in internal energy and (b) the work done by the gas.

nRTU 23WQUUU if



J 7100K 540K 350KmolJ 31.8mol 0.323

23

23

if nRTnRTU

J 12600J 7100J 5500 UQW

(a)

(b)


15.4 Thermal Processes

isobaric: constant pressure

isochoric: constant volume

isothermal: constant temperature

adiabatic: no transfer of heat

Ideal Gas Processes



An isobaric process is one that occurs atconstant pressure.

VPsAPsFWby

ifby VVPVPW

According to the 1st Law of Thermodynamics:ΔU = Q – Wby

During an isobaric process:ΔU = Q – P ΔV: energy is transferred by both work and heat.


if VVPVPW

The PV diagram for an work done during an isobaric process is a horizontal line


An isobaric process

• During an isobaric process, a system gains 1500 J of heat. The internal energy of the system increases b y 4500 J and the volume decreases by 0.01 m3 . Find the pressure of the system.


An isobaric process

• During an isobaric process, a system gains 1500 J of heat. The internal energy of the system increases b y 4500 J and the volume decreases by 0.01 m3 . Find the pressure of the system.

• Answer: 3.0 x 105 Pa


if VVPVPW

The work done by the gas equals the area under the PV curve



Example 4 Work and the Area Under a Pressure-Volume Graph

Determine the work for the process in which the pressure, volume, and temp-erature of a gas are changed along thestraight line in the figure.

The area under a pressure-volume graph isthe work for any kind of process.



Since the volume increases, the workis positive.

Estimate that there are 9colored squares in the drawing.

J 180

m100.1Pa100.29 345

W


Two process are shown that take an ideal gas from state 1 to state 3. Compare the work done by process A to the work done by process B.

A. WA > WB

B. WA < WB C. WA = WB = 0D. WA = WB but neither is zero



isochoric: constant volumeQWQU

For the isochoric process the area under the curve is equal to zero.

But W = PΔV = 0: during an isochoric process, energy is transferred by heat only


15.5 Thermal Processes Using and Ideal Gas

ISOTHERMAL EXPANSION OR COMPRESSION

Isothermalexpansion orcompression ofan ideal gas

i

f

V

VnRTW ln

ΔU = Q – Wby But ΔU α ΔTIf T does not change, ΔU = ΔT = 0!And Q = W


EXAMPLE The work of an isothermal compression

QUESTION:

A cylinder contains 7.0 g of N2 gas. The gas compresses to half its volume at a constant temperature of 80˚C. a.How much work must be done by the gas?b.By how much does the internal energy of the gas change?c.How much heat was added or taken away from the gas?


EXAMPLE The work of an isothermal compression

Answer:

A cylinder contains 7.0 g of N2 gas. The gas compresses to half its volume at a constant temperature of 80˚C. a.How much work must be done by the gas? -508 J (work was done on the gas to compress it) b.By how much does the internal energy of the gas change? 0Jc.How much heat was added or taken away from the gas? 508 J of heat was taken away from the gas.

Q – Wby = 0


15.5 Thermal Processes Using and Ideal Gas

ADIABATIC EXPANSION/COMPRESSION

fiby TTnRUW 23

The red curve shows an adiabatic expansion of an ideal gas.The blue curves are isotherms at Ti and Tf.Adiabatic curves can be approximated as linear.

According to the 1st Law of Thermodynamics: ΔU = Q – Wby but Q = 0, since walls are insulatedΔU = – Wby

For a monatomic ideal gas:


Adiabatic Processes without adiabatic (insulating) walls

• Rapid expansion or compression does not allow the gas and surroundings to come to equilibrium– bicycle pump– air being forced to rise

or sink due to atmospheric conditions

– loading/ unloading a pneumatic lift.


Mass Lifter Lab – thermodynamic process with 4 steps

1. mass is loaded onto movable piston of a cylinder of gas kept at a low temperature. This is a “very rapid” event. Energy transfer due to temperature difference between gas and surroundings assumed to be negligible.

2. gas of cylinder is put in contact with room temperature water and energy transfer allowed. The mass rises.

3. mass is unloaded at a “rapid” pace.4. gas of cylinder put in contact with ice water. The movable piston falls.


Molar specific heat capacity

• For a solid or liquid: Q = cm ∆T, where c is the specific heat capacity in Joules/kg kelvin.

• For a gas, we use number of moles (n) instead of mass: Q = Cn ∆T where C is the molar specific heat capacity in Joules/mol kelvin.


Molar specific heat capacity for a monatomic ideal gas

• At constant pressure Cp = 5/2R

• At constant volume Cv = 3/2R.

• Q = Cn ∆T ; so for the same amount of heat (Q) added to a gas, the gas at constant volume will show a greater temperature change because it is not losing energy by expanding.



Two containers hold equal masses of helium gas (monatomic) at equal temperature. You supply 10 J of heat to container A while not allowing the volume to change. You supply 10 J of heat to container B while not allowing the pressure to change. Is TfA

greater than, less than or equal to TfB? Explain,

using the 1st Law of Thermodynamics.



For an isochoric process:QA = Cvn ΔT = (3/2R)n ΔTA = 10J

For an isobaric process:

Q = Cpn ΔT = (5/2R)n ΔTB = 10J

ΔTA is greater than ΔTB


EOC # 33

Heat, Q, is added to a monatomic ideal gas at constant pressure. As a result, work, W is done by the gas. What is Q/W, the ratio of heat added to work done by the gas? This is a numerical value, with no variables.


EOC # 33

Heat, Q, is added to a monatomic ideal work at constant pressure. As a result, work, W is done by the gas. What is Q/W, the ratio of heat added to work done by the gas?

Q = Cp nR ∆T = 5/2 nR ∆T

W = P ∆V = PVf – PVi but PV = nR ∆T

so W = nR ∆T

Q/W = 5/2/1 = 2.5.


Which of the following processes involve heat (energy transfer due to ∆T)?

A. The brakes in your car get hot when you stop.

B. You push a rigid cylinder of gas across a frictionless surface.

C. A steel block is placed on top of a candle.D. You push a piston into a cylinder of gas,

increasing the temperature of the gas.E. All of the above


Which of the following processes involve heat?

A. The brakes in your car get hot when you stop.

B. You push a rigid cylinder of gas across a frictionless surface.

C. A steel block is placed on top of a candle.D. You push a piston into a cylinder of gas,

increasing the temperature of the gas.E. All of the above


EOC #26The drawing refers to 1 mole of a monatomic gas and a 4-step process.

Complete the following table: ∆U Wby Q

A to BB to CC to DD to A


EOC #26The drawing refers to 1 mole of a monatomic gas and a 4-step process.

Complete the following table: ∆U Wby Q

A to B 4990J 3320J 8310JB to C -4990J 0J -4990JC to D -2490J -1660J -4150JD to A 2490J 0J 2490J

chapter 15. work, heat, and the first law of thermodynamics

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