soln5A_08W_term1

# soln5A_08W_term1 - \A’Dhework 5A(11 “ Model Assume the...

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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 expel-tent '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 ideal-gas-law 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 Pa-m6)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 ideal-gas law, v A nRi': _ (0.10 moi)(8.3| J/mol K)(423 K) , _ =1.Is7x|0’-‘ 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 ideal-gas law once again as follows: V a 7“ 3 '1} =1;£f__r=(423 K) 91p; m =346.9 K‘s-14°C pi Vi pi 1.157340-3 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 .r-a |.<: 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 10-3 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 =(9.81x10-3 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 = m-2 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, Miriam-mg : 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") : 4-9 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, {8-l2x10“ 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 i i i i 'l i ...
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