Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 3 • The specific heat of aluminum is more than twice the specific heat of copper. A block of copper and a block of aluminum have the same mass and temperature (20ºC). The blocks are simultaneously dropped into a single calorimeter containing water at 40ºC. Which statement is true when thermal equilibrium is reached? (a) The aluminum block is at a higher temperature than the copper block. (b) The aluminum block has absorbed less energy than the copper block. (c) The aluminum block has absorbed more energy than the copper block. (d) Both (a) and (c) are correct statements. Picture the Problem We can use the relationship TmcQ Δ= to relate the amount of energy absorbed by the aluminum and copper blocks to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum block:

TcmQ Δ= AlAlAl

Express the energy absorbed by the copper block:

TcmQ Δ= CuCuCu

Divide the second of these equations by the first to obtain:

TcmTcm

QQ

ΔΔ

=AlAl

CuCu

Al

Cu

Because the block’s masses are the same and they experience the same change in temperature:

1Al

Cu

Al

Cu <=cc

QQ

or

AlCu QQ < and )( c is correct.

11 • A real gas cools during a free expansion, while an ideal gas does not cool during a free expansion. Explain the reason for this difference. Determine the Concept Particles that attract each other have more potential energy the farther apart they are. In a real gas the molecules exert weak attractive forces on each other. These forces increase the internal potential energy during an expansion. An increase in potential energy means a decrease in kinetic energy, and a decrease in kinetic energy means a decrease in translational kinetic energy. Thus, there is a decrease in temperature.

363

Chapter 18

364

21 •• An ideal gas in a cylinder is at pressure P and volume V. During a quasi-static adiabatic process, the gas is compressed until its volume has decreased to V/2. Then, in a quasi-static isothermal process, the gas is allowed to expand until its volume again has a value of V. What kind of process will return the system to its original state? Sketch the cycle on a graph. Determine the Concept During a reversible adiabatic process, is constant, where γ < 1 and during an isothermal processes is constant. Thus, the pressure rise during the compression is greater than the pressure drop during the expansion. The final process could be a constant volume process during which heat is absorbed from the system. A constant-volume cooling will decrease the pressure and return the gas to its original state.

γPV

PV

V

P

adiabatic

isothermal

iVi21

f VV =

constant volume

Estimation and Approximation 25 •• A ″typical″ microwave oven has a power consumption of about 1200 W. Estimate how long it should take to boil a cup of water in the microwave assuming that 50% of the electrical power consumption goes into heating the water. How does this estimate correspond to everyday experience? Picture the Problem Assume that the water is initially at 30°C and that the cup contains 200 g of water. We can use the definition of power to express the required time to bring the water to a boil in terms of its mass, heat capacity, change in temperature, and the rate at which energy is supplied to the water by the microwave oven. Use the definition of power to relate the energy needed to warm the water to the elapsed time:

tTmc

tWP

ΔΔ

=ΔΔ

= ⇒P

Tmct Δ=Δ

Substitute numerical values and evaluate Δt:

( ) ( )min1.6s63.97

W600

C30C100Kkg

kJ184.4kg200.0Δ ≈=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t ,

an elapsed time that seems to be consistent with experience.

Heat and the First Law of Thermodynamics

365

Heat Capacity, Specific Heat, Latent Heat 29 • How much heat must be absorbed by 60.0 g of ice at –10.0ºC to transform it into 60.0 g of water at 40.0ºC? Picture the Problem We can find the amount of heat that must be absorbed by adding the heat required to warm the ice from −10.0°C to 0°C, the heat required to melt the ice, and the heat required to warm the water formed from the ice to 40.0°C. Express the total heat required: waterwarmicemelticewarm QQQQ ++=

Substitute for each term to obtain: ( )waterwaterficeice

waterwaterficeice

TcLTcmTmcmLTmcQ

Δ++Δ=Δ++Δ=

Substitute numerical values (See Tables 18-1 and 18-2) and evaluate Q:

( ) ( )( )

( )

kJ3.31

C0C0.04Kkg

kJ184.4

kgkJ5.333C0.01C0

KkgkJ05.2kg0.0600

=

⎥⎦

⎤°−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+

+⎢⎣

⎡°−−°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

=Q

Calorimetry 33 • While spending the summer on your uncle’s horse farm, you spend a week apprenticing with his farrier (a person who makes and fits horseshoes). You observe the way he cools a shoe after pounding the hot, pliable shoe into the correct size and shape. Suppose a 750-g iron horseshoe is taken from the farrier’s fire, shaped, and at a temperature of 650°C, dropped into a 25.0-L bucket of water at 10.0°C. What is the final temperature of the water after the horseshoe and water arrive at equilibrium? Neglect any heating of the bucket and assume the specific heat of iron is K)J/(kg 460 ⋅ . Picture the Problem During this process the water will gain energy at the expense of the horseshoe. We can use conservation of energy to find the equilibrium temperature. See Table 18-1 for the specific heat of water.

Chapter 18

366

Apply conservation of energy to obtain:

0horseshoe thecool water thewarmi

i =+=∑ QQQ

or ( ) ( ) 0C650C0.10 fFeFefwaterwater =°−+°− tcmtcm

Solve for tf to obtain:

( ) ( )FeFewaterwater

FeFewaterwaterf

C650C0.10cmcmcmcmt

+°+°

=

Substitute numerical values and evaluate tf:

( ) ( ) ( ) ( )

( ) ( )

C1.12

KkgkJ 460.0kg 750.0

KkgkJ 184.4kg 0.25

C650Kkg

kJ 460.0kg 750.0C0.10Kkg

kJ 184.4kg 0.25

f

°=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=t

37 •• A 200-g piece of ice at 0ºC is placed in 500 g of water at 20ºC. This system is in a container of negligible heat capacity and is insulated from its surroundings. (a) What is the final equilibrium temperature of the system? (b) How much of the ice melts? Picture the Problem Because we can not tell, without performing a couple of calculations, whether there is enough heat available in the 500 g of water to melt all of the ice, we’ll need to resolve this question first. See Tables 18-1 and 18-2 for specific heats and the latent heat of fusion of water. (a) Determine the energy required to melt 200 g of ice: ( )

kJ70.66kgkJ5.333kg0.200ficeicemelt

=

⎟⎟⎠

⎞⎜⎜⎝

⎛== LmQ

The energy available from 500 g of water at 20ºC is:

( ) ( )

kJ84.14

C02C0Kkg

kJ184.4kg0.500Δ waterwaterwatermax available,

−=

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

== TcmQ

Because max available,Q < : icemeltQ C.0 is re temperatufinal The °

Heat and the First Law of Thermodynamics

367

(b) Equate the energy available from the water max available,Q to miceLf and

solve for mice to obtain:

f

max available,ice L

Qm =

Substitute numerical values and evaluate mice:

g125

kgkJ5.333

kJ84.14ice ==m

43 •• A 100-g piece of copper is heated in a furnace to a temperature tC. The copper is then inserted into a 150-g copper calorimeter containing 200 g of water. The initial temperature of the water and calorimeter is 16.0ºC, and the temperature after equilibrium is established is 38.0ºC. When the calorimeter and its contents are weighed, 1.20 g of water are found to have evaporated. What was the temperature tC? Picture the Problem We can find the temperature t by applying conservation of energy to this calorimetry problem. See Tables 18-1 and 18-2 for specific heats and the heat of vaporization of water. Use conservation of energy to obtain:

0sampleCu

thecoolrcalorimete

thewarm waterthe

warmwater

vaporizei

i =+++=∑ QQQQQ

or 0ΔΔΔ CuCuCuwcalcalOHOHOHwf,vaporizedO,H 2222

=+++ TcmTcmTcmLm

Substituting numerical values yields:

( ) ( ) ( )

( ) ( ) ( ) ( ) 0C0.38Kkg

kJ386.0g100C0.16C0.38Kkg

kJ386.0g150

C0.16C0.38Kkg

kJ184.4g200Kkg

kJ2257g1.20

C =−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+

°−°⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

t

Solving for tC yields: C618C °=t Work and the PV Diagram for a Gas 51 • The gas is first cooled at constant volume until it reaches its final pressure. It is then allowed to expand at constant pressure until it reaches its final volume. (a) Illustrate this process on a PV diagram and calculate the work done by the gas. (b) Find the heat absorbed by the gas during this process.

Chapter 18

368

Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat absorbed by the gas during this process (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram. 2

0

1

0

P, atm

3.00

2.00

1.00

1.00 2.00 3.00V, L

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

( )( )

J054

Lm10L2.00

atmkPa101.325atm2.00L2.00atm2.00Δ

33

gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×===

−

VPW

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain:

( ) ( )( ) gasby int,1int,2

gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−Δ=

Substitute numerical values and evaluate Qin:

( ) J618J054J456J912in =+−=Q

57 •• An ideal gas initially at 20ºC and 200 kPa has a volume of 4.00 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 100 kPa. Find (a) the work done by the gas, and (b) the heat absorbed by the gas during the expansion.

Heat and the First Law of Thermodynamics

369

Picture the Problem The PV diagram shows the isothermal expansion of the ideal gas from its initial state 1 to its final state 2. We can use the ideal-gas law for a fixed amount of gas to find V2 and then evaluate for an

isothermal process to find the work done by the gas. In Part (b) of the problem we can apply the first law of thermodynamics to find the heat added to the gas during the expansion.

∫ PdV

200

100 2

1

V2

P, kPa

293 K

4.00V, L

(a) Express the work done by a gas during an isothermal process:

∫∫∫ ===2

1

2

1

2

1

11gasby

V

V

V

V

V

V VdVVP

VdVnRTdVPW

Apply the ideal-gas law for a fixed amount of gas undergoing an isothermal process:

2211 VPVP = ⇒ 12

12 V

PPV =

Substitute numerical values and evaluate V2:

( ) L00.8L4.00kPa100kPa200

2 ==V

Substitute numerical values and evaluate W:

( )( ) ( )[ ]

( )

J 555

Lm 10LkPa 5.554

L 4.00L 00.8lnLkPa 800

lnLkPa 800L 00.4kPa 200

33

L 8.00L 00.4

L 8.00

L 00.4gasby

=

×⋅=⎟⎠⎞

⎜⎝⎛⋅=

⋅==

−

∫ VVdVW

(b) Apply the first law of thermodynamics to the system to obtain:

onintin WEQ −Δ= or, because ΔEint = 0 for an isothermal process,

onin WQ −=

Because the work done by the gas is the negative of the work done on the gas:

( ) J555gasby gasby in ==−−= WWQ

Chapter 18

370

Remarks: in an isothermal expansion the heat added to the gas is always equal to the work done by the gas (ΔEint = 0). Heat Capacities of Gases and the Equipartition Theorem 59 •• The heat capacity at constant pressure of a certain amount of a diatomic gas is 14.4 J/K. (a) Find the number of moles of the gas. (b) What is the internal energy of the gas at T = 300 K? (c) What is the molar heat capacity of this gas at constant volume? (d) What is the heat capacity of this gas at constant volume? Picture the Problem (a) The number of moles of the gas is related to its heat capacity at constant pressure and its molar heat capacity at constant pressure according to . For a diatomic gas, the molar heat capacity at constant pressure is

PP nc'C =Rc' 2

7P = . (b) The internal energy of a gas depends on its number of

degrees of freedom and, for a diatomic gas, is given by nRTE 25

int = . (c) The molar heat capacity of this gas at constant volume is related to its molar heat capacity at constant pressure according to Rc'c' −= PV . (d) The heat capacity of this gas at constant volume is the product of the number of moles in the gas and its molar heat capacity at constant volume. (a) The number of moles of the gas is the ratio of its heat capacity at constant pressure to its molar heat capacity at constant pressure:

P

P

c'Cn =

For a diatomic gas, the molar heat capacity is given by: Kmol

J1.2927

P ⋅== Rc'

Substitute numerical values and evaluate n:

mol 495.0

mol 4948.0

KmolJ1.29

KJ4.14

=

=

⋅

=n

(b) With 5 degrees of freedom at this temperature:

nRTE 25

int =

Heat and the First Law of Thermodynamics

371

Substitute numerical values and evaluate Eint:

( ) ( ) kJ 09.3K 300Kmol

J314.8mol 4948.025

int =⎟⎠⎞

⎜⎝⎛

⋅=E

(c) The molar heat capacity of this gas at constant volume is the difference between the molar heat capacity at constant pressure and the gas constant R:

Rc'c' −= PV

Because Rc' 27

P = for a diatomic gas:

RRRc' 25

27

V =−=

Substitute the numerical value of R to obtain:

KmolJ8.20

KmolJ79.20

KmolJ314.82

5V

⋅=

⋅=⎟

⎠⎞

⎜⎝⎛

⋅=c'

(d) The heat capacity of this gas at constant volume is given by:

VV nc'C' =

Substitute numerical values and evaluate : VC' ( )

KJ3.10

KmolJ79.20mol 4948.0V

=

⎟⎠⎞

⎜⎝⎛

⋅=C'

65 •• Carbon dioxide (CO2) at a pressure of 1.00 atm and a temperature of –78.5ºC sublimates directly from a solid to a gaseous state without going through a liquid phase. What is the change in the heat capacity at constant pressure per mole of CO2 when it undergoes sublimation? (Assume that the gas molecules can rotate but do not vibrate.) Is the change in the heat capacity positive or negative during sublimation? The CO2 molecule is pictured in Figure 18-22. Picture the Problem We can find the change in the heat capacity at constant pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in the solid and gaseous states. Express the change in the heat capacity (at constant pressure) per mole as the CO2 undergoes sublimation:

solidP,gasP,P CCC −=Δ (1)

Chapter 18

372

Express Cp,gas in terms of the number of degrees of freedom per molecule:

( ) NkNkfC 25

21

gasP, == because each molecule has three translational and two rotational degrees of freedom in the gaseous state.

We know, from the Dulong-Petit Law, that the molar specific heat of most solids is 3R = 3Nk. This result is essentially a per-atom result as it was obtained for a monatomic solid with six degrees of freedom. Use this result and the fact CO2 is triatomic to express CP,solid:

NkNkC 9atoms3atom3

solidP, =×=

Substitute in equation (1) to obtain: NkNkNkC 213

218

25

PΔ −=−= Quasi-Static Adiabatic Expansion of a Gas 69 •• A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is (a) isothermal and (b) adiabatic. Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In Part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In Part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =

Substitute numerical values and evaluate Vi:

( ) ( )33

i m103.118kPa400

K300Kmol

J8.314mol0.500−×=

⎟⎠⎞

⎜⎝⎛

⋅=V

(a) Because the process is isothermal: K300if == TT

Heat and the First Law of Thermodynamics

373

Use the ideal-gas law for a fixed amount of gas with T constant to express Vf:

ffi VPVP = ⇒f

iif P

PVV =

Substitute numerical values and evaluate Vf:

( )

L7.80

L 795.7kPa160kPa400L3.118f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Express the work done by the gas during the isothermal expansion:

i

fgasby ln

VVnRTW =

Substitute numerical values and evaluate Wby gas:

( )

( )

kJ14.1

L3.118L7.795lnK300

KmolJ8.314mol0.500gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⎟⎠⎞

⎜⎝⎛

⋅=W

Noting that the work done by the gas during the process equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

( )kJ1.14

kJ1.140onintin

=

−−=−Δ= WEQ

(b) Using γ = 5/3 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate Vf: ( )

L5.40

L 403.5kPa160kPa400L118.3

53

f

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Apply the ideal-gas law to find the final temperature of the gas: nR

VPT fff =

Chapter 18

374

Substitute numerical values and evaluate Tf:

( )( )( )

K208

KmolJ8.314mol0.500

m105.403kPa160 33

f

=

⎟⎠⎞

⎜⎝⎛

⋅

×=

−

T

For an adiabatic process: 0in =Q

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic process:

TnRTCQEW Δ=−Δ=−Δ= 23

Vininton 0

Substitute numerical values and evaluate Won:

( )( )( )

J745K300K208

KJ/mol8.314mol0.50023

on

−=−×

⋅=W

Because the work done by the gas equals the negative of the work done on the gas:

( ) J745J574gasby =−−=W

Cyclic Processes 73 •• A 1.00-mol sample of an ideal diatomic gas is allowed to expand. This expansion is represented by the straight line from 1 to 2 in the PV diagram (Figure 18-23). The gas is then compressed isothermally. This compression is represented by the straight line from 2 to 1 in the PV diagram. Calculate the work per cycle done by the gas. Picture the Problem The total work done as the gas is taken through this cycle is the area bounded by the two processes. Because the process from 1→2 is linear, we can use the formula for the area of a trapezoid to find the work done during this expansion. We can use ( )ifprocess isothermal ln VVnRTW = to find the work done

on the gas during the process 2→1. The net work done during this cycle is then the sum of these two terms. Express the net work done per cycle:

1221

gas on thegas by thenet

→→ +=

+=

WW

WWW (1)

Heat and the First Law of Thermodynamics

375

Work is done by the gas during its expansion from 1 to 2 and hence is equal to the negative of the area of the trapezoid defined by this path and the vertical lines at V1 = 11.5 L and V2 = 23 L. Use the formula for the area of a trapezoid to express W1→2:

( )( )

atmL3.17atm1.0atm2.0

L11.5L2321

trap21

⋅−=+×

−−=

−=→ AW

Work is done on the gas during the isothermal compression from V2 to V1 and hence is equal to the area under the curve representing this process. Use the expression for the work done during an isothermal process to express W2→1:

⎟⎟⎠

⎞⎜⎜⎝

⎛=→

i

f12 ln

VVnRTW

Apply the ideal-gas law at point 1 to find the temperature along the isotherm 2→1:

( )( )( )( ) K280

Katm/molL10206.8mol1.00L5.11atm0.2

2 =⋅⋅×

== −nRPVT

Substitute numerical values and evaluate W2→1:

( )( )( ) atmL9.15L23L5.11lnK280Katm/molL10206.8mol00.1 2

12 ⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅×= −

→W

Substitute in equation (1) and evaluate Wnet:

kJ14.0atmL

J101.325atmL40.1

atmL15.9atmL3.17net

−=⋅

×⋅−=

⋅+⋅−=W

75 ••• At point D in Figure 18-24 the pressure and temperature of 2.00 mol of an ideal monatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point C. Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle.

Chapter 18

376

Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle:

DCCBBAADtotgas,by →→→→ +++= WWWWW

1. Use the ideal-gas law to find the volume of the gas at point D:

( ) ( )

( )

L29.54atmkPa101.325atm2.00

K360Kmol

J8.314mol2.00

D

DD

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⋅=

=P

nRTV

2. We’re given that the volume of the gas at point B is three times that at point D:

( )L62.88

L 54.2933 DCB

==== VVV

Use the ideal-gas law to find the pressure of the gas at point C:

( ) ( )atm6667.0

L62.88

K360Kmol

atmL10206.8mol2.00 2

C

CC =

⎟⎠⎞

⎜⎝⎛

⋅⋅

×==

−

VnRT

P

We’re given that the pressure at point B is twice that at point C:

( ) atm333.1atm6667.022 CB === PP

3. Because path DC represents an isothermal process:

K360CD == TT

Use the ideal-gas law to find the temperatures at points B and A: ( )( )

( )

K719.8Kmol

atmL108.206mol2.00

L88.62atm1.3332

BBBA

=

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

==

−

nRVPTT

Heat and the First Law of Thermodynamics

377

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that:

atm00.42 DA == PP

The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

Point P V T

(atm) (L) (K) A 4.00 29.5 720 B 1.33 88.6 720 C 0.667 88.6 360 D 2.00 29.5 360

4. For the path D→A, 0AD =→W and: ( )DA2

3

AD23

AD int,AD

TTnRTnREQ

−=

Δ=Δ= →→→

Substitute numerical values and evaluate QD→A:

( ) ( ) kJ979.8K360K720Kmol

J8.314mol2.0023

AD =−⎟⎠⎞

⎜⎝⎛

⋅=→Q

For the path A→B:

⎟⎟⎠

⎞⎜⎜⎝

⎛== →→

A

BBA,BABA ln

VVnRTQW

Substitute numerical values and evaluate WA→B:

( ) ( ) kJ15.13L29.54L88.62lnK720

KmolJ8.314mol2.00BA =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=→W

and, because process A→B is isothermal, 0BA int, =Δ →E

For the path B→C, , and: 0CB =→W

( )BC23

VCBCB ΔΔTTnR

TCUQ−=

== →→

Substitute numerical values and evaluate QB→C:

( )( )( ) kJ979.8K720K360KJ/mol8.314mol2.0023

CB −=−⋅=→Q

For the path C→D: ⎟⎟⎠

⎞⎜⎜⎝

⎛=→

C

DDC,DC ln

VVnRTW

Chapter 18

378

Substitute numerical values and evaluate WC→D:

( ) ( ) kJ576.6L62.88L54.92lnK603

KmolJ8.314mol2.00DC −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅=→W

Also, because process A→B is isothermal, 0BAint, =Δ →E , and kJ58.6DCDC −== →→ WQ

Qin, Won, and ΔEint are summarized for each of the processes in the table to the right.

Process Qin Won ΔEint

(kJ) (kJ) (kJ) D→A 98.8 0 8.98

A→B 2.13 −13.2 0

B→C 98.8− 0 −8.98

C→D 58.6− 6.58 0

Referring to the table, find the total work done by the gas per cycle:

kJ6.6

kJ6.580kJ13.20DCCBBAAD totgas,by

=

−++=

+++= →→→→ WWWWW

Remarks: Note that, as it should be, ΔEint is zero for the complete cycle. General Problems 79 • The PV diagram in Figure 18-25 represents 3.00 mol of an ideal monatomic gas. The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path AEC, find (a) the initial and final temperatures of the gas, (b) the work done by the gas, and (c) the heat absorbed by the gas. Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process EDC is isobaric, we can find the area under this line geometrically and then use the 1st law of thermodynamics to find QAEC.

Heat and the First Law of Thermodynamics

379

(a) Using the ideal-gas law, find the temperature at point A:

( )( )( )

K65K65.2

KmolatmL108.206mol3.00

L4.01atm4.02

AAA

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

Using the ideal-gas law, find the temperature at point C: ( )( )

( )

K81K81.2

KmolatmL108.206mol3.00

L0.02atm1.02

CCC

==

⎟⎠⎞

⎜⎝⎛

⋅⋅

×=

=

−

nRVPT

(b) Express the work done by the gas along the path AEC:

( )( )

kJ1.6kJ1.62atmL

J101.325atmL15.99

L4.01L20.0atm1.0Δ0 ECECECAEAEC

==

⋅×⋅=

−=+=+= VPWWW

(c) Apply the first law of thermodynamics to express QAEC: ( )ATTnRW

TnRWTCWEWQ

−+=

Δ+=

Δ+=Δ+=

C23

AEC

23

AEC

VAECintAECAEC

Substitute numerical values and evaluate QAEC:

( )( )( ) kJ2.2K65.2K81.2KJ/mol8.314mol3.00kJ1.62 23

AEC =−⋅+=Q

Remarks The difference between WAEC and QAEC is the change in the internal energy ΔEint,AEC during this process. 83 •• As part of a laboratory experiment, you test the calorie content of various foods. Assume that when you eat these foods, 100% of the energy released by the foods is absorbed by your body. Suppose you burn a 2.50-g potato chip, and the resulting flame warms a small aluminum can of water. After burning the potato chip, you measure its mass to be 2.20 g. The mass of the can is 25.0 g, and the volume of water contained in the can is 15.0 ml. If the temperature increase in the water is 12.5°C, how many kilocalories

Chapter 18

380

(1 kcal = 1 dietary calorie) per 150-g serving of these potato chips would you estimate there are? Assume the can of water captures 50.0 percent of the heat released during the burning of the potato chip. Note: Although the joule is the SI unit of choice in most thermodynamic situations, the food industry in the United States currently expresses the energy released during metabolism in terms of the ″dietary calorie,″ which is our kilocalorie. Picture the Problem The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it. The ratio of the energy in a 150-g serving to the energy in 0.30 g of potato chip is the same as the ratio of the masses of the serving and the amount of the chip burned while heating the aluminum can and the water in it:

500g 0.30g 150

g 0.30

serving g-150 ==Q

Q

or g 0.30serving g 150 500QQ =

Letting f represent the fraction of the heat captured by the can of water, express the energy transferred to the aluminum can and the water in it during the burning of the potato chip:

( ) Tcmcm

TcmTcm

QQfQ

Δ

ΔΔ

OHOHAlAl

OHOHAlAl

OH Alg 0.30

22

22

2

+=

+=

+=

where ΔT is the common temperature change of the aluminum cup and the water it contains.

Substituting for yields and solving

for yields: g 0.30Q

serving g-150Q

( )f

TcmcmQ

Δ500 OHOHAlAlserving g-150

22+

=

Substitute numerical values and evaluate : serving g-150Q

( ) ( ) ( )

kcal 652cal 10256J 4.184

cal 1J 1007.1

500.0

C 5.12Kkg

kJ184.4kg 0150.0Kkg

kJ900.0kg 0250.0500

36

serving g-150

≈×=××=

°⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

=Q

89 •• A thermally insulated system consists of 1.00 mol of a diatomic gas at 100 K and 2.00 mol of a solid at 200 K that are separated by a rigid insulating

Heat and the First Law of Thermodynamics

381

wall. Find the equilibrium temperature of the system after the insulating wall is removed, assuming that the gas obeys the ideal-gas law and that the solid obeys the Dulong–Petit law. Picture the Problem We can use conservation of energy to relate the equilibrium temperature to the heat capacities of the gas and the solid. We can apply the Dulong-Petit law to find the heat capacity of the solid at constant volume and use the fact that the gas is diatomic to find its heat capacity at constant volume. Apply conservation of energy to this process:

0Δ solidgas =+= QQQ

Use to substitute for QTCQ ΔV= gas and Qsolid:

( ) ( ) 0K200K100 equilsolidV,equilgasV, =−+− TCTC

Solving for Tequil yields: ( )( ) ( )( )

solidV,gasV,

solidV,gasV,equil

K200K100CC

CCT

+

+=

Using the Dulong-Petit law, determine the heat capacity of the solid at constant volume:

RnC solidsolidV, 3=

The heat capacity of the gas at constant volume is given by:

RnC gas25

gasV, =

Substitute for CV,solid and CV,gas and simplify to obtain:

( )( ) ( )( ) ( )( ) ( )( )solidgas2

5

solidgas25

solidgas25

solidgas25

equil 33K200K100

33K200K100

nnnn

RnRnRnRn

T+

+=

+

+=

Substitute numerical values for ngas and nsolid and evaluate Tequil:

( )( )( ) ( )( )( )( ) ( ) K 171

mol 00.23mol 00.1mol 00.23K200mol 00.1K100

25

25

equil =++

=T

95 ••• (a) Use the results of Problem 94 to show that in the limit that

ETT >> , the Einstein model gives the same expression for specific heat that the Dulong–Petit law does. (b) For diamond, TE is approximately 1060 K. Integrate numerically dEint = dT to find the increase in the internal energy if 1.00 mol of diamond is heated from 300 to 600 K.

v′c

Chapter 18

382

Picture the Problem (a) We can rewrite our expression for by dividing its numerator and denominator by

'cVTTe E and then using the power series for ex to

show that, for T > TE, . In Part (b), we can use the result of Problem 94 to obtain values for every 100 K between 300 K and 600 K and use this data to find ΔU numerically.

Rc' 3V ≈'cV

(a) From Problem 94 we have:

( )2

2E

V1

3E

E

−⎟⎠⎞

⎜⎝⎛=

TT

TT

ee

TTRc'

Divide the numerator and denominator by TTe E to obtain:

TTTT

TT

TTTT

eeTTR

eeeT

TRc'

EE

E

EE

213

1213

2E

2

2E

V

−+−⎟⎠⎞

⎜⎝⎛=

+−⎟⎠⎞

⎜⎝⎛=

Express the exponential terms in their power series to obtain:

E

2E

2EE

2EE

for

...2112...

2112 EE

TTTT

TT

TT

TT

TTee TTTT

>>⎟⎠⎞

⎜⎝⎛≈

+⎟⎠⎞

⎜⎝⎛+−+−+⎟

⎠⎞

⎜⎝⎛++=+− −

Substitute for TTTT ee EE 2 −+− to obtain:

R

TTT

TRc' 313 2E

2E

V =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛≈

(b) Use the result of Problem 94 to verify the following table:

T (K) 300 400 500 600 cV (J/mol⋅K) 9.65 14.33 17.38 19.35

Heat and the First Law of Thermodynamics

383

The following graph of specific heat as a function of temperature was plotted using a spreadsheet program:

5

7

9

11

13

15

17

19

21

300 350 400 450 500 550 600

T (K)

CV (J

/mol

-K)

Integrate numerically, using the formula for the area of a trapezoid, to obtain:

( )( )( )

( )( )( )

( )( )( )

kJ62.4Kmol

J35.1938.17K100mol 00.1

KmolJ38.1733.14K100mol 00.1

KmolJ33.1465.9K100mol 00.1Δ

21

21

21

=⋅

++

⋅++

⋅+=U

Chapter 18

384