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Unformatted text preview: 7 a.va SP“? K97 {Leash F 00
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1)(30) a (3 pts)We developed an equation for the work done by a gas in an isothermal reversible process. It used the initial and ﬁnal volumes ( V1 and V2). Assuming the gas is ideal change this
equation to use the initial and ﬁnal pressures (P; and P2). Then go on and use it in part b m/ Teens’r. l/IE=V2.1?L $4: %=% V2
* —RTln — r:
w— in] ck RTACQL) ,_ “ﬁema, Gm Fawn If you cannot do part a raise your hand. We’ll give you the equation, but deduct the points. K mfg
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b (27 pts) One mole of @trlatonnc ideal gas a 100 IS compresse very slowly against a constantly changing pressure ﬁ'orn 0.0 bar to 5.0 bar. The compression is done
so that heat can ither enter or leave. Compute the following thermodynamic parameters
and place your answers in the boxes given. Show your work in the box for any partial credit. Use CT‘CST' 30$; 1.5 at
R = 8.31 .i/mole K and 1.00 L—atmos = 101 .1 Report your answers in Joules 5‘1"!" Nd PA 1LT m z. ) . ca 27‘
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r PM W than}! arm DO EITHER #2 or #3 Cross off the one you do not want graded. If no decision is made we
will grade only #3 .
2) (20 pts) When we derived the equation for work we started with w =  IP(ext)dV
Derive an expression for one mole reversible isothermal work using a modiﬁed vander Waal’s equation for gas behavior where the gas has volume but no attractive forces:
P(Vb) = nRT where b is a constant 776;: = gar 7?“ Weir/(V11) (V2 “53
. .. = .— m V .. . R £7
 fFotV RTE—mﬁb>  7" T (VF A)
515” 512* 539+ 01‘
3(20 pts) If U= U(T,V) write out the total diﬁ‘erential of U wit/ﬁe" e123” e?
U . At constant temperature = 0 for an ideal gas, but forsome real gases 6V T  6V T a) (51313) 2
an ........— V3 AOL
Again assuming constant T ; create an integrated equation for )5” as a ﬁmction of V for the real .. .93 + '"
b) (15pts) 0!“ *' GVJTCW W V2.
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dagcin 4V MASK r.th dis! aaus oil/‘(zcm‘ V (VI “3 €29 6'9 012 on? FM DO EITHER #4 or #5 Cross off the one you do not want graded. If no decision is made we
will grade only #5 . 4)(20 pts) a) (18)Water has an unusually high molar heat capacity at constant pressure. This is
very important for cellular and ecological reasons in preventing rapid changes in temperature.
Given the Cp of liquid water is 4.18 J/gK and that of ice is 1.94 .1ng at 0C. Suppose you were
studying ﬁeezing cells cryogenically at 80C in dry ice— acetone mix. Assuming the Cp’s don’t
change much over this temperature range compute AH for the following reaction at 80C.
9H 02,; we: “r907?! HgOﬂ) —> H20 (3) AH (0C) a 333 Jig 5 ' T ’ ’gf/ 4.
F Hie/wee) ash/(goo): :7 {TIES/+5?! 3:?
_ ELF/5 /
with x _F_. 5 ’1 ’1 C 5
$27 i [Sh/(6:15 zeHéﬂ "i SWT+ FEBHQB #435! magi/3:72P @4639) 3 ‘333; 1— 0.94  ‘11, i8>(goco> 5 ._ [54f f/ b) (2 pts) Comment on the sizeoof the new AH compared to the original 6‘
"333 "C459 : evil? game We? eta/v are” I g 22 M5 a m, 6! as H ekesaw t“ M @755 5:98am“ Ea? wagea “'0 $573325”de #5 (20pts) In a very short range of T and P the volume of a liquid may be expressed as V= K eKPeuT Assuming a and K are constant in this range and V=V{P,T), Show that is a function of state Hr / K
ea):_+<l<e“’<°“ MK; 51” ‘ P T (=2 «K? "9(7—
/ / .ﬁ‘fﬂq‘ v “RWKQ e
u 5 i7 T 4L 5
i\ 93% _/ 1% VII/S. FW’VC77aeJ 6F 6) (30pts) In your own words glve a very bnef explanation of the followmg dramngs may help: a) Cp> Cv foranidealgas byR. ,U N A775¢irds #39773”; 2’72 773“) 5“ U/éb/M T22“) & F {ff/66.5: “an”? CV {MUCDLUE'ZS ‘C/‘gif a: (U i ﬂM/N‘ST ,t.iT‘Px_;g;3 72?: <3? M, ea new"? r“ UM 4” ' '
’P cave?" ~
in) CV for N2 at room temperature is much lower than the value of 7/2 R you would estimate
74K ﬁfﬁﬁgtwg‘j A (at foo 5/ 73497" as 0/8 (.545, {9625731 j, a? @7724; W: Twas? agar" .237 T2— 293 mu? ame‘NoTsa'cV x‘gmw
Oman: 0) Why is (the vander Waals “b” much larger than the actual volume of one mole of gas b : ggcmmu (/Oéowmi m5 £5,424 madam ca? 5537)»)
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d) U(1000K)> U(100K) . .. a . 
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e) W(rev) > w(irrev) C i 50 7336.41“ M, {away NJ)
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 Fall '07
 Wesolowski

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