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Unformatted text preview: \A’Dhework 5A (11 “ Model: Assume the gas is ideal. The process is not adiabatic, despite Equation I138, because the expeltent
'3' given in the problem is not equal to 'y. The gas is identified as diatomic which means that 'y = 1.40, but the exponent is explicitly given as 2, so while we are given sz = constant, it is not true that pVY =constant. We will use the fact that the gas is diatomic to deduce Cv = §R = 20.8Jlmol . k , although we won‘t speciﬁcally need '3! = 1.40  Visualize: We are given n =0.020mo , 7; = 293K , K : l500cm" , and Vf #500ij . Solve: (a) If we use the idealgaslaw expression p = nRT/V in {TV2 = constant we get TV = constant . 3
Ti =£l£= (293 [06500 cm )=879 K'=606°C VI 500 cm“
(b) The strategy to find Q will be to use the first law Q = 1350. —W . We will use £0. = nCvAT and W = — IpdV. 05m = nCVAT = (0.020 mol){20.3 Jlmol  k )(379 K — 293 K) = 243.8 1 To do the W=—J'pdV integral we need to know what the constant is in pV2 =constant . Use the ideal gas law to compute pi.
p; g n37; = (0.020 mol)(8.3l Jgnol; K)(293 K) #1460 Pa
; 1.5x“) m"
constant = pit/i2 =(32,460Pa)(1.5x10‘‘ m‘f = 0.0730 Pa ~ m6
w = — [pdv = — :mgfmdv = —(0.0730 Pam6)j:;::'"3% = (0.0730 Pa 111%;1 msom’ = 97.4 1
. MKHSMJ 9:05", ~W=243.8J—97.4} :146412 1501 (c) We also need pf to complete the pV diagram. We could use the ideal—gas law again to compute Pr , but let’s do it another way. ps‘viz=ptvr2
2 2
V. 500ch
:9 =pj _' : i. —ﬁ =9 .2292,
‘ [11,] p[500cm3] 1" 200%
p(kPa>
290  f
32 i
0 ——. .w— .— V(cm3) 1—
500 1000 [500 2000
Assess: Both Q and Ware positive because heat was added to the system and work was done on the system. :Qq. * Model: We assume the gas is an ideal gas and y = 1.40 fora diatomic gas.
Solve: Using the idealgas law, v A nRi': _ (0.10 moi)(8.3 J/mol K)(423 K) , _ =1.Is7x0’‘ m"
p, (3x1.013x10’ Pa) For an adiabatic process,
at" = m Ur aim)
5V; =Va[£] =(l.l57x10’3 m‘)[i—] =‘1.Eu"><10‘3 m"
P; 0. To ﬁnd the final temperature, we use the idealgas law once again as follows: V a 7“ 3
'1} =1;£f__r=(423 K) 91p; m =346.9 K‘s14°C
pi Vi pi 1.1573403 m3 (QB '. Model: Assume that the gas isanideal gas.
Visualize: The volume of container A is a constant. On the other. hand hea ‘ '
pressure remains the samc‘ , ting container B causes the volume to change, but the Solve: (a) For the healing of the gas in container A, M"‘ = QA inCY . Similarly, for the gas in container B,
A7], = Q” fnCl, . Because QA = QB and C? > CV, we see that AT_ > AT”. The gases started at the same temperature, so TA) TB.
03) V (c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is, , [OR 9.8 1
pn=pm+wm =1.013x1051>a+—___( 3” "1(3) _' = [.08 x £06133
AM“ 1.0x") m” Container A has the same volume, temperature, and number of moles of gas as container B, so I: = P“ = 1.08xl0“ Pa. (d) The heating of container 8 is isobaric, so :1 [s
II
.ra .<:
U
E
II
_<
H in We have Ti = 293 K, and T. can be obtained from
Q = Pm = nc,,(1} —T) The number of moles of gas is n : PiVi {RI : 0.355 mol. Thus
(25 mm s) : (0.355 mo:)(20.8 J/moi K)(T,. ,293 K) :11. =344' K 2V, =(8_0x]0_‘ m3)(344 K/293 K) :939 x l0'4 m3=939 crn3 ”940 an? G4?  Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoricr processes. Solve: (a) The initial conditions are ,01 = 30 atm = 304,000 Pa, 1/. = 100 cm3 = 1.0 x 10“ m3, and T1 2 100°C = 373 K.
The number of moles of gas is _ ply. _ (304,ooopa)(r.0x104 m3) n 0.
R7; (8.3: .l/rnol K)(373 K) = 9.81x10’3 mol At point 2 we have p2 = pl 2 3.0 atm and V: = 300 em3 = 3V]. This is an isobaric process, so V. V. v
i:—:>T =—lT=3 373K =1119
T2 Ti 2 Vi I ( .) K The gas is heated to raise the temperature from T1 to T3. The amount of heat required is
Q = no.0?" = (9.8 ix 103 mo)(20.8 J/mol K)(I 1 19 K H373 K) = 152 J This amount of heat is added during process I “e 2.
(b) Point 3 returns to T3 = [00°C = 373 K. This is an isochoric process, so Q = ncxgar =(9.81x103 mol)(12.5 J/rnol K)(373 K —l I19 K) = —91.5 J ' This amount of heat is removed during process 2 a 3. Q95 I Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules.
Visualize: _ Please refer to Figure P1730.
Solve: (a) The piston is ﬂoating in static equilibrium, so the downward force of gravity on the piston’s mass must exactly balance the upward force of the gas, Fm = pA where A = m2 is the area of the face of the piston. Since the upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus, Miriammg : pCquisumg
A A Mmg = pA :> p = = pmgh = (8920 kg/m3)(9.80 tn/sz)(0.040 m) = 3500 Pa (h) The gas volume is V' = 3'sz = JI(0.030)2(0.20 m) = 5.65 x 10‘4 m“ The number of moles is 3500 P 5.65 104 ‘
n=—p'vi =————————( a)( x m )=8.i2x10_‘ moi
RT. (8.31 J/Inol K)(293 K) The number of molecules is N = RNA = (8.12 x 104‘ mol)(6.02 x 103 moi") : 49 x to?" (c) The pressure in the gas is determined simpiy by the'weight of the pistOn. That will not change as heat is added, so the
heating takes place at constant pressure with Q = nCpAT. The temperature increase is Q , 2.0}
nCl, {8l2x10“ mol)(29.l J/mol K) AT= =85K This raises the gas temperature to T3 = T, + AT= 378 K = 105°C.
(d) Noting that the volume of a cylinder is V = drill, and that r doesn't change, the idealigas relationship for an isobaric process is 378 K
293 K LH is :3 _<:
‘3‘1 Li“ :4 lb _>L_,=:3[1—[ ]200m=25.8cm
1 (e) The work done by the gas is Wgas = FwAy . The force exerted on the piston by the gas is F '=pA=errﬁ=9.90N This force is applied through Ay = 5.8 em = 0.053 In, so the work done is WE... = (9.90 N)(0.058 m) = 0.574 3 50.57 r
Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is 0.57 J. t
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 Fall '09
 BILLMCUTCHEON
 Physics

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