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