�
"!#$&%'$)(+*,)-/.10324%4*5-/*624# 7/%8.39:<;=!?>A@CBBD$FEG!?>A@B
BD$IHKJ
%4%
$MLN.3LO*697/PLQ.1RG7/S+*7T*69U9U%FVXW#1242?:EG!?>A@B
BD$IHZY[!?>\@BBD$^]!?>A@CBBD$
_&SPUV`;=!?>A@CBBD$IHaJbc!
Y[!?>\@BBD$^]!?>A@CBBD$T$
;=!?>A@CBBD$IH !j j3s3tNuC0C$ defd1g1! dij3h s kjlkmoj1nqj1jp+b rV v j1jwLixVA$ H ge j y h kj lmon L
e
h g h
d
!oz$
!
W$`;!?>A@B
BD$IH|{!?>A@CBBD$X.39w7/SU*~}.39U(U%87/%4.39€7/SU# 7k:M{!?>A@BB?$^Y[!?>A@BB?$^]‚!?>A@CBBD$MHaJ
Y[!?>A@B
BD$IH {!?>A@CBBD$^J ]!?>A@BB?$ H !j j siLƒ$ defd1g1! dƒj3h s kj lkj1mon j1jp+b rvV j1j„L„xV\$ H… e t † h kj lmon uC0 ! g $
e d
h g h
d
Mj$`‡)R^Iˆ`*6#1(U%49+0N‰<9+01*62?ŠUˆ`*6%4(‹ fv
%'g $ŒŽ*&LO.c(+*627/S+*LO.328*6}kPU24*6VG%F9~7/S+*d }.39+*&}*62F24V=#1V%F9+‘9U%87T*<W‚.6’~.1R“`%4(+7/S@_&SU*)}.1-/-”*6VT•‚.39U(U%49+0
*69+*k-/03%4*6V€!
*6%401*69– #124P+*6V.1R`7/SU*w—c}\S+-™.3˜ (U%F9+01*k-q*6šPU# 7/%8.39$# -/*w›`œHž!?JŸ Ÿ6$Tbc!?¡Y@CŸ™$i!
“`S+*k-”*iY¢H
† e 1 h kjlmA£`uC0N%FV7/S+*5*628*6}7T-/.39¤LN#1V/V\$\
¥ V/V/PULƒ%49+0¦7/SU# 7M24%403S7I# WV/.1-/•U7/%8.39N}.1-”-/*6VT•‚.39U(UV7T.#57T-”#19UV/%47/%8.39§R'-”.3L¨7/S+*&01-/.3P9U(NVT7/# 7T*§!9H§z$
7T.N7/S+*~‘U-”VT7X*’+}k%87T*6(VT7/# 7T*€!9H g $\ŠU7/SU*5*69+*k-/01©w# WVT.1-/W‚*6(w}k#19W*¦“&-\%87T7T*69#1Vk:
ª ›¨H ¡YJŸ @ !
y Ÿ&« Ÿ $IH ¡v1YJU@ Ÿ
!?v3$
Ÿ
Ÿ
%4%
$M_`S+*5*69+*k-/01©w# WVT.1-”W*6(w%FVX-/*624# 7T*6(7T.ƒ7/S+*5“)#z–1*628*69+017/SŽ;!B
¬?­cJ+®/$X7/S+-/.3PU03S:
ª ›¨HaJ¯!B'¬D­J+®/$MHaJ°6b1;!B
¬?­JU®T$
!
yC$
!°6H¦VT•‚*k*6(w.1R24%803SC7\H5v e j1j h kj1±~LixV\$
‡).39VT*6šCPU*697/28©1Š
v1JŸ
¡Y@ Ÿ
² v3;=!¡B
¬?Y­JU° ®T$oJ
J ;=!B
¬?° ­J+®/$ H
@ H
@¹Tº¼»"HKt6v§•L
·
_&S+*57/SU-/*k*LO.328*6}kPU2F# -`V/%8³k*6V&# -”*1:µ´¶ @ ½ ¹oº¼º œ§H d †yN•L
¶¸ C@ ¾?¿fÀzº)H d 6¡O•L
!?s3$
!d $
D! t1$
v )‡XR^IÁ*6}7/P+-/*ƒƒ!V”24%4(+*wÂoÃo97T*k-/R
*k-/*69U}*q%49 ¥ }7/%4.39UÄ$
¥ 9q%49C7T*k-/R#1}*MW‚*k7œ)*k*697œ).5(U%ÇƂ*k-/*697=24%803SC7oÈD7T-”#19UV/LN%47T7/%49+0&LO*6(U%F#&-/*kÉU*6}7/VG#"R'-”#1}7/%4.39~.1R+7/S+*)%49U}k%4(U*697
“X#6–1*1`‡).39UVT*6šP+*69C7/28©1ŠVT•-/*6#1(U%49+0N#19.1-/03#19U%4}57/S%49w‘24LÊ.397/S+*~V/PU-/R
#1}*~.1R#ƒ0324#1V/V-”*6V/PU287/V&%F9w7œ).
-/*kÉU*6}7T*6(˓X#6–1*6Vi!
.39+*O%4V"(P+*~7T.€7/S+*O#1%8-Ax™‘24LÌ%49C7T*k-/R#1}*1͍7/S+*.17/S+*k-¦7T.€7/S+*§‘24Lix™0324#1V”V"%49C7T*k-/R#1}*™$\
,)©Ë*69+03%49+*k*k-”%F9+0€7/S+*N7/SU%4}”u9U*6V/V#19U(Ë7/S+*i}.3LO•‚.3V/%87/%8.39Ë.1RX7/S+*O‘2FLwŠ%87%4V¦•‚.3V/V”%8W28*7T.01*69+*k-\# 7T*i#
(+*6VT7T-\PU}7/%8–1*~%49C7T*k-/R
*k-/*69U}*W‚*k7^“<*k*69¤7/S+*-/*kÉ*6}7T*6(“X#6–1*6V§!
-/*kR
*k-`7T.ƒ7/SU*~28*6}7/P+-/*qV/24%4(+*5V/SU.z“`%49U0N7œ).
-/*kÉU*6}7T*6(“)#z–1*6V`.3PU7oÈD.1RFÈD•S#1VT*“`%87/S€-”*6VT•‚*6}7&7T.N.39+*#19+.17/SU*k-A$\XΦ2F# -/*%4VX7/SPUV&}kP+7k
ÒÛ« Ú
y+‹ŒŽ*}k#19ϓ`-”%87T*„7/S+*€•24#19+*i“)#z–1*#1V§ÐÑÒÛ« Ú ! Òf« Ó~Ô ®/$OHÖÕ¦×kØ
Ù Ò« Ó Ò« l+Ú ÜÝFÞ ‹_`S+*„V”šCPU# -”ÒÛ*€
LN.(UP24PUV.1R
Ò« Ó «ãâ ®T$T$!}.3V! « Ú Ò« Ó «ãâ ®/$Iä
â
«
7/S+*w“X« #6–1*ß%4V§Ð)à1! Ò« ÓÔ ®T$^Ц! Ò« Ó~Ô ®/$ƒ« HáÕ¦àÕ§!}.3V™! Ò« Ó «µ
/
®
$
¬
/
V
4
%
9
!
¬V/%F9! ÒÛÚ Ò« Ó «Žâ ®/$T$IHæå ÕOå Ÿ3!}.3V”Ÿ3! Ò Ú Ò« Ó «Žâ ®T$äçV/%49 Ÿ ! ÒÛ« Ú Ò« Ó «Žâ ®/$T$IHæå ÕOå Ÿ`HZ°kèz®\
_&SU*V/šCP# -/*LO.c(UPU24PUVX%4V#ƒ}.39UVT7/#19C7kŠ%87/V•‚*k-”%4.(U%F}k%87Å©N%4V&7/SCPUV&³k*k-/. s"!#$"ä…}k#1VT*
_&SU*501*69+*k-”#12VT.32FP+7/%8.39€.1R{Ÿké!
®/$Tb{3®oŸ"HKä ⠟Aé!
®T$X}k#19W*5“`-”%87T7T*69#1Vk:
é!
®T$IHaÕÑ× ÜÝ äÏêO× l+ÜCÝ
!?¡3$
!
W$ « }k#1VT*
_&SU*501*69+*k-”#12VT.32FP+7/%8.39€.1R{ Ÿ é!
®/$Tb{3® Ÿ H «&⠟ é!
®T$X}k#19W*5“`-”%87T7T*69#1Vk:
é!
®T$MHZÕÑ× Ø ÜÝ äÏêO× l Ø ÜÝ
!?†3$
.1-"*6šCPU%4–#128*69C7/28©„#1Vk:
é!
®T$MHZÕ¤}.3V! â ®T$äëêëV”%49! â ®T$
!okj$
dŒZ)S+‡X*6RG9ˆ`›ì*6#1%4(UV&%4%49+9€07/ SUd *4 *6 9+*k-/01©€-”#19+01* «"íUî5ï › ï jƒ#19U(%4V&9U.17`#19w*6%801*69C–#12FP+*1ŠU“)*.1WVT*k-/–1*~7/SU# 7
7/S+*NVT.32FP+7/%8.39ß.1R<7/SU*i—ð—r‰&%4V59U.175“<*62428ÈDW*6S#6–1*6(:Ñ7/S+*O“X#6–1*kRPU9U}7/%8.39[ÐÑ!
ñ‚$"01.*6V7T.¤ä~òó.39Ž7/S+*
-”%803SC7•# -/7=.1R7/S+*<01-\# •Sw!
7/SU# 7%4VkŠñ Ò ä~òµ$\G_`SU%4VW‚*6SU#z–%4.3P+-=%4VG9+.17•SC©V”%4}k#12?:7/S+*)•U-/.1W# W%424%87^©Ñ.1R
‘9U(%49+0~#19i*628*6}7T-/.39w# 7)ñ Ò äòô!
“"SU%4}”S„%4V<03%8–1*69„WC©i24%4Lõzö÷øÐ à !
ñ$^ÐÑ!
ñ‚$T$)}k#19€9+.17XW‚*"%F9+‘9U%87T*1
_&S+*6V/*VT.324P+7/%8.39V# -/*~9+.17*6%801*69VT7/# 7T*6V`.1RG7/S+*—‚ð—r‰&
¥ VT.324PU7/%8.39w%4V#1(ULN%FV/V/%8W24*`.397/SU*}.39U(U%87/%8.39€7/S# 7`›ì%FV#19w*6%801*69C–#124PU*.1R=7/SU*—ð—r‰` :7/S+*I“X#6–1*È
RPU9U}7/%8.39€01.*6V&7T.N³k*k-”.ƒ# 7&7/S+*W‚.3PU9(U# -”%8*6V)“`S+*69€7/S%4V&}.39U(U%47/%8.39w%4VV/# 7/%4V/‘U*6(
Ct M%
$)}k24#1V/V/%4}k#12šCPU#19C7/%87^©:wmoù6û ú
%4%
$M}.1-”-/*6VT•‚.39U(+*69U}*5•-”%49U}k%8•28*1:UE Ò « ¬TJü » » õ
%4%F%'$<šCPU#19C7/PULÊ.1•‚*k-”# 7T.1£ þ « ¬TJü » » õ§ÿ « ¬TJü » » õ� « ¬TJü » » õ ÐÑ!
ñ‚$"H¨! « ¬TJ‚ü $¼m » » õ ú ÐÑ!
ñ$IH|¬TJü m » » õ ú ÐÑ!
ñ$
ùmo6ý û ú ÐÑ!
ñ$IH moûÊ
ú
ú
%49U#12F28©
{Cm
Em
vY H|¬/Jü m {3ñ m
g
!o1z$
¡ß_&S+*ƒ*6%401*69+R
P9U}7/%8.39UV#19([
*6%801*69C– #124P+ *6VqR'.1-§#•# -/7/%4}k28*ƒ%F9[#¤7œ). ÈÅ(U%4LO*69UV”%8.39U#12IW‚.z’‹.1R&V/%8³k*
¿
õ
Ô
@ h @€# -/*1:IА¿ û !
ñ $MHZÕ¤V”%49! $V/%49! û $)#19U(›"¿
û H ±oû B Ÿ äëY Ÿ ‡).39VT*6šCPU*697/28©1Š+7/SU*7/S+-/*k*~24.z“)*6 VTõ 7oÈD*69+*k-/01©€*6%801*69+RPU9U}7/%4.39UV# -/* А¿ £ û £ !
ñ Ô $IHZÕ £ V/%49! $V/%49! $ !›`¿ £ û £ H nTû $
· ´¶ А¿ !
ñ Ô $IHZÕ V/%49! õ $V/%49! Ÿ $ !›`¿ H $ !VT*k*01-”# •SVX.399+*’c7•# 01*™$
£û Ÿ
Ÿ
õ $V/%49! $ !›`¿ £ û Ÿ H ±oû $
Ô
Ÿ

Ð
¿
!
ñ
I
$
Z
H
Õ
/
V
4
%

9
!
¶¸
m
Ÿ û £ ±oû Ÿû £
!#$.1-/7/S+.101.39U#124%47Å©iÈ<#19U#128©7/%4}k#12428©
_&SU*5.1-/7/S+.101.39U#124%47Å©w}.39U(U%87/%4.39€W*k7^“<*k*69А¿ £ û £ !
ñ Ô $&#19U(€Ð¿ £ û Ÿ !
ñ Ô $%4Vk:
õ Õ V/%49!!$ ñ $V”%49!!$ $% à Õ V”%49!!$ ñ $V/%F9! g $ $¼{ñ{ Haj
!o g $
Ÿ
î " £
õ î ! #
@
@
@
@
“`SU%F}”S€“)*~}k#19-/*k“`-”%87T*5#1Vk:
& õ V/%49!! $ ñ $V/%49!! $ ñ $¼{3(ñ ' & V”%49!! $ $V”%49! g $ $¼{ 'ëHZj
!o6v3$
! î
õî
@
@
@
@
) V/%49+0NLƒ# 7/S+*6LN# 7/%4}k#O.1-ÑLN# •28*1Šc*6#1}\S¤%49C7T*k01-”#12}k#19W‚*¦}k#12F}kPU24# 7T*6(
Ão9w•# -/7/%F}kPU24# -6Š+* î V/%49! $V/%F9! Ÿ $¼{ HZjc
_&SPUVkŠ+А¿ £ û £ !
ñ Ô $X#19U(wА¿ £ û Ÿ !
ñ Ô $X# -/*~.1-/7/S+.101.39U#12 ! W$X.1-/7/S+.101.39U#124%47Å©„ÈI01-”# •SUõ %F }k#12428© !
ŒŽ*5“X#197X7T.O•U-”.z–1*¦7/SU# 7 * õ î * î Ð ¿à £ û £ !
ñ Ô $^п £ û Ÿ !
ñ Ô $¼{ñ{ *6šPU#124V)³k*k-/.+)ÎÑ-\# •ScÈ
%4}k#12424©1Š+7/SU%4V%49C7T*k01-”#12%4VX7/SU*–1.324PULO*5W‚*k7^“<*k*697/S+*•24#19U*¦é§HZjO#19U(w7/SU*~V/P+-/R#1}*5-/*k•U-”*6VT*697/%F9+0O7/S+*
RPU9U}7/%8.39€Ð ¿à £ û £ !
ñ Ô $^А¿ £ û Ÿ !
ñ Ô $\<ŒŽ*~}k#19•28.177/SU%4VXRPU9U}7/%8.39:
Ð £ £ !
ñ Ô ^$ Ð £ Ÿ !
ñ Ô $
õ
-”.3LÌ7/S+*~01-”# •SŠ%47`}k#19W‚*~VT*k*697/SU# 7`Ð ¿à !
ñ Ô
VT•‚*6}77T.&7/S+*M•24#19+* HZ@cb g -,"*69U}*1Š1%49¦7/S+*I%49C7T*k£ 01û -”#12.£ * õ õ î v
$^А ¿ £ û Ÿ !
ñ Ô $%4V&V/©LNLN*k7T-”%4}5“`%47/Sw-/*È
* ! î ÐX¿à £ û £ ! ñ Ô $^А¿ £ û Ÿ !
ñ Ô $¼{3ñ{ o 7/S+*M9U*k03# 7/%8–1*I•# -/7*’c#1}7/28©5}k#19U}*62FV7/S+*I•‚.3V/%87/%4–1*G•# -/7/Äc=_&SU*I%497T*k01-\#123%4V‚7/SPUV³k*k-/.+:=п
#19U(wА¿ £ û Ÿ !
ñ Ô $# -/*5.1-”7/S+.101.39U#12 —c%FLN%424# -”24©1Š.39+*}k#19¤V/S+.z“ 7/SU# 7Ð £ £ !
ñ Ô $X%4VX.1-/7/S+.101.39#127T.NÐ Ÿ £ !
ñ Ô $ ŠU#19U(7/SU# 7Ð
.1-/7/S+.101.39#127T.NÐ Ÿ £ !
ñ Ô $ V/%49! õ $V/%49! Ÿ $
£ û £ !
ñ Ô $
£ Ÿ !
ñ Ô X$ %4V
V/%49! Ÿ õ $V/%49! $
õ
õ
/EF243653G798;:<>B@=@C4?HD A A
/10!243653798;:<>B@=@C4?D A A
V”%49! õ $V”%49! $
õ
/IH243653798;:< J C4? A D A
y
of them with its corresponding eigenfunction. Can you describe what is happening for E
lesser than zero ? Do we still have eigenstates ? What happens when E is not an eigenvalue
? Note that the applet starts its numerical integration from left to the right.
7. In a parallel universe, the kinetic energy of a particle is p3/(3m). What would be the
corresponding quantum mechanical operator ?
8. Sketch the graph of the three lowest energy eigenfunctions of a particle in a 2-dimensional
box. Show graphically and analytically that they are orthogonal (Use the method of
separation of variables to write the ground-state wavefunction of this 2-dimensional problem
using the 1-dimensional solutions). What are the three lowest-energy states ?
THERMODYNAMICS
9. Thermodynamic Variables. We have described two classes of thermodynamic variables:
intensive (such as temperature and pressure) and extensive (such as volume).
a. What class would the variable mass (let’s use the symbol M for mass) fall intointensive or extensive?
Mass is an extensive variable: it depends linearly on the size of the system: if I double the number
of moles of some material, I double the mass of that material.
b. What about variables that are derived from other variables? Consider density: We
define density as the mass of a material per unit volume:
"=
M
V
Can density be classified simply as an intensive or extensive variable? Explain your
answer. (Hint: Look carefully at the definitions!)
!
Density is a variable formed by the ratio of two other extensive variables. But is density extensive?
Suppose I have system A with mass MA and volume VA, and I combine it with system B with mass
MB and volume VB. Then the density of the composite system is, by definition:
" A +B =
M A +B M A + M B
VA
VB
=
=
"A +
"B
VA +B
VA + VB
VA + VB
VA + VB
Comparing this result to a linear addition of the densities of A and B:
"A + "B =
!
MA MB
+
VA
VB
Density is neither explicitly intensive nor extensive- the density of a composite system is the
weighted average of the densities of the individual subsystems.
!
10. Thinking about systems and boundaries. Read the descriptions below of some
thermodynamic systems (any macroscopic material can by a thermodynamic system!)
3.012 PS 1
2 of 6
9/22/05
undergoing various changes and identify the type of system (open, closed, etc.). Explain in
one sentence for each case why your choice is appropriate- state your assumptions to
defend your choice.
(a) The system is a liter of hot coffee placed inside a high-quality, sealed thermos that
has an evacuated space between an inner rigid cylinder where the liquid is placed and
the outside walls of the container.
Based on the given information, which is that the system (the coffee) is stored in a rigid, sealed
volume that restricts heat transfer, we could state that this is approximately an isolated system. Of
course, as with any real system, the isolation is not perfect… the coffee eventually gets cold as heat
slowly escapes the container.
(b) If the thermos described above had flexible rather than perfectly rigid walls, what kind
of system would we have?
If the walls of the container are flexible, then the volume of the system is not forced to be constantand a system that restricts heat transfer but can change volume (allowing work to be done on/by the
system) is an adiabatic system.
(c) The system is a puddle of water on the sidewalk.
Because water molecules can evaporate or condense on the puddle, and the puddle could be
heated or cooled depending on its surroundings, this is an open system.
(d)You have a glass jar of strawberry jam and the metal screw-top lid is stuck (the jam is
your system). You really want a peanut butter and jelly sandwich, so you place the jar
under a stream of hot water to loosen the lid.
You have a closed system (jam can’t get out of the jar, but heat is clearly going to pass through the
glass walls of the jar).
(e) You are the system. You’re on a ski trip at Killington and you are wearing a high-tech
insulated parka, hood, and snow pants to keep you warm. It’s -10°C on the ski slope,
but as you trudge up to the ski lift, you are getting warm.
Assuming your parka is a perfect (or at least very good) insulator, we could state that you are an
isolated system. (If you assume the parka isn’t a good insulator, then you are a closed system.)
The fact that you feel warmer after exerting yourself is due to the transformation of internal chemical
energy stored in your body into some waste heat during the process of walking to the ski lift. This
fact might further suggest that the parka is a good insulator and no heat is transferring out of you into
the surroundings (in practice of course, some heat is always leaking into the surroundings).
3.012 PS 1
3 of 6
9/22/05
11. Using the First Law. We introduced the theoretical concept of an ideal gas in lecture- ‘ideal’
because real gases only approach ideal behavior under select conditions. Recall that the two
key thermodynamic characteristics of an ideal gas are an equation of state PV = nRT and
that the internal energy is only a function of temperature (U = 1.5nRT). Suppose we have 1
mole of an ideal gas confined in a cylinder with a movable piston top, which allows the
volume of the gas to be controlled. The gas starts at an initial volume Vo = 10 L at a
pressure of 1 atm. We want to consider the effect of two different two-step processes, each
of which starts with the gas in the same initial state and ends with the gas in the same final
state:
Process A: (i) The gas is compressed to a volume of 1 L at constant pressure (P = 1
atm). (ii) The pressure is then slowly increased from 1 atm to 10 atm at constant
volume.
Process B: (i) The pressure is slowly increased from 1 atm to 10 atm at constant
volume (V = Vo). (ii) The gas is then compressed from 10 L to 1 L at constant
pressure (P = 10 atm).
Assume the changes occur slowly enough for the gas to remain in equilibrium at all times
(reversible processes).
a.
By which composite process, A or B, is more work done by the gas? Show why
using a P vs. V diagram and also quantitatively by calculation.
A P-V diagram of this process looks like:
Mathematically, the work in each composite process is:
V final
Process A:
(i) dw = -PextdV
w="
#P
ext
dV = "(1atm)(1L "10L) = 9L $ atm
Vo
Note that to put this result in more conventional units of energy (Joules), we
can use the following conversion factor (which is just a ratio of the gas constant R in two different
units- something readily memorizable!):
!
#
&
8.3144J
( = 911J
$ 0.082057L " atm '
(9L " atm)%
The work is positive: work is done on the system.
(ii) dw = 0
(dV = 0, no volume change so no hydrostatic work is performed).
3.012 PS 1
!
4 of 6
9/22/05
Total work: 911 J
Process B:
(i) dw = 0
no volume change in the first step
V final
(ii) w = "
#P
ext
Vo
%
(
8.3144J
dV = "(10atm)(1L "10L) = 90L $ atm'
* = 9,119J
& 0.082057L $ atm )
Total work: 9,119 J
Thus, more work is performed in process B.
!
b.
Is the change in internal energy of the gas the same for each composite process?
Why?
Yes. Internal energy is a state function, and thus changes in internal energy do not depend on the
path taken, only the identity of the initial and final states. Since the initial and final states of the
system undergoing each process are the same, the change in internal energy must be the same.
c.
With the given information, can you calculate the heat transferred into/out of the gas
during process A? Why or why not? If yes, show your calculation.
The internal energy change in process A can be determined, using the given expression for internal
energy:
"U process _ A = U final # U initial = 1.5nR(Tfinal # Tinitial )
We can determine the temperature at the initial and final states using the given information and the
ideal gas equation of state:
!
PV = nRT
PV
(1atm)(10L)
Tinitial =
=
= 121.9K
nR (0.082057 L " atm )
K " mole
PV
(10atm)(1L)
Tfinal =
=
= 121.9K
nR (0.082057 L " atm )
K " mole
We have calculated the work in this process above. Using these two calculations and the first law,
we can obtain the heat transfer:
!
"U process _ A = 1.5nR(Tfinal # Tinitial ) = 0 = q + w = q + 911J
$q = #911J
Thus, 911 J of heat leaves the system.
Suppose now!we carry out a four-step process: Starting in the initial state (V = Vo, P = 1 atm), we
perform composite process A, followed by composite process B in reverse.
d.
3.012 PS 1
What is the total internal energy change for the gas after all 4 steps?
5 of 6
9/22/05
The composite process described is a cycle- the initial and final states are the same; therefore, the
change in internal energy is zero.
e. Can you determine the total heat transferred into/out of the gas for this 4-step
composite process? If yes, show your calculation.
Yes: applying the first law as in part c:
"U process = w A + qA + w B + qB = 0
qtotal = qA + qB
"U process = 911# 9119 + qtotal = 0
$qtotal = 8,208J
!
3.012 PS 1
6 of 6
9/22/05