Homework: 1, 2, 3, 5, 8 (page 556)

Final Exam: Chapter 18, 19, and 20 (not including “Engines”)

1.  Suppose 4.0 mol of an ideal gas undergoes a reversible isothermal expansion from volume V1 to volume V2=2.0V1 at temperature T=400K. Find (a) the work done by the gas and (b) the entropy change of the gas. (c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?

1

2lnVVnRTW =

(J) 92160.2ln40031.80.41

1 =×××=VVW

(a)

(b) ∫=Δf

i TdQS

For an isothermal process: T=constant

WQWQETQS =⇒=−=Δ=Δ 0; int

(J/K) 0.23400

9216===Δ

TWS

(c) 00 =Δ⇒= SQ

2. An ideal gas undergoes a reversible isothermal expansion at 77.00C, increasing its volume from 1.30 L to 3.90 L. The entropy change of the gas is 22.0 J/K. How many moles of gas are present?

i

fV

i

fif T

TnC

VV

nRSSS lnln +=−=Δ

3.19.3ln31.8

0.22

lnln

×=

Δ=⇒=Δ

i

fi

f

VV

R

SnVV

nRS

(mol) 41.2=n

3. A 2.5 mol sample of an ideal gas expands reversibly and isothermally at 360 K until its volume is doubled. What is the increase in entropy of the gas?

i

fV

i

fif T

TnC

VV

nRSSS lnln +=−=Δ

For an isothermal process, T=constant:

i

fif V

VnRSSS ln=−=Δ

(J/K) 4.142ln31.85.2 =××=ΔS

5. Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.0 kg block of copper whose temperature is increased reversibly from 25.00C to 1000C. The specific heat of copper is 386 J.kg-1.K-1. (a) Energy absorbed as heat to increase the copper temperature:

TcmQ Δ=(J) 57900752386 =××=Q

(b) The change in entropy:

∫∫ ===Δ2

1

2

1 1

2lnT

T

T

T TTcm

TcmdT

TdQS

ΔS = 386×2× ln 373.15298.15

=173.2 (J/K)

T1 = 25+ 273.15= 298.150K; T2 =100+ 273.15= 373.150K

8. At very low temperatures, the molar specific heat CV of many solids is approximately CV=AT3, where A depends of the particular substance. For aluminum, A = 3.15 x 10-5 J mol-1 K-4. Find the entropy change for 4.0 mol of aluminum when its temperature is raised from 5.0 K to 10.0 K.

∫=Δf

i TdQS

dTnCdQ V=We assume that the volume change is negligible:

( ) (J/K) 037.05101015.30.431

31 3350.10

0.53

0.10

0.5

2

=−×××==

===Δ

−

∫∫

nAT

dTTnATdTnCS

f

iV

Review

• Work done by the gas: ∫=f

i

V

V

pdVW

+3 special cases:

p = constant (isobaric):

)()( ifif TTnRVVpVpW −=−=Δ=V = constant (isochoric):

0=WT = constant (isothermal):

i

fVV

nRTW ln=

• The First Law of Thermodynamics:

WQE −=Δ int

Four Special Cases

Process Restriction Consequence Adiabatic Q = 0 ΔEint = -W Constant volume W = 0 ΔEint = Q Closed cycle ΔEint = 0 Q = W Free expansion Q = W = 0 ΔEint = 0

• Equation of State: nRTpV =

f

ff

i

ii

TVp

TVp=

• For a closed cycle: 0int =ΔE p

V

• Work done by the gas:

-Expansion: 0>W

-Compression: 0<W• Energy transferred as heat Q:

-Heat transferred to the gas (receiving energy as heat):

0>Q-Heat transferred from the gas (releasing energy as heat):

0<Q

A

B C

p

V

A

B C

0>W

0<W

ABCareaW =||

ABCareaW =||

• Isothermal process:

i

fVV

nRTQW ln==

• RMS Speed:

MRTvrms3

=

• Translational Kinetic Energy per Molecule:

kTK23

=

• Total Translational Kinetic Energy (n moles):

nRTKtotal 23

=

0int =ΔE

MRTvv avg π8

==

• Mean Free Path:

pdkT

VNd 22 2/21

ππλ ==

• Adiabatic Process (Q = 0): constant;=γpV

RCCRfCCC

VpVV

p +=== ;2

;γ

monatomic: f=3; diatomic: f=5; polyatomic: f=6

• Molar Specific Heats of an Ideal Gas:

• The Change in Internal Energy: TnCE VΔ=Δ int

V = constant: TnCQ VΔ=

p = constant: TnCQ pΔ=

RCC Vp +=with

constant1 =−γTV

• Change in entropy:

i

fV

i

fif T

TnC

VV

nRSSS lnln +=−=Δ

∫∫ ===Δ2

1

2

1 1

2lnT

T

T

T TTcm

TcmdT

TdQS

1) Ideal gas:

2) Liquid, solid:

∫=Δf

i TdQS

+Some special cases: + T = constant:

i

fVV

nRS ln=Δ

+ V = constant: i

fV T

TnCS ln=Δ

TW

TQS ==Δor

+Phase change:

+Cooling or heating:

TLmS =Δ

L is heat of vaporization or heat of fusion