# 110a1h1k - Problem gel*1” Soluiw‘oms P1.2 Consider a...

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Unformatted text preview: Problem gel *1” Soluiw‘oms P1.2) Consider a gas mixture in a 2.00-dm3 ﬂask at 27.0°C. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole perCent. a) 1.00 g H2 and 1.00 g 02 b) 1.00 gN2 and 1.00 g 02 c)1.00 g CH4 and 1.00 g NH3 a) RT . . 1 . '1 ‘1 PH =nH2 =100/2016mox8314l3m31 K X3OOK=624X105Pa I V 2.00x10 m RT . I —l —1 P0 =n02 =1.00/3200molx8314{3m(:l K ><300K=390><104Pa 2 V 2.00x10 In em, = 6.57x105Pa 1H . . mol%H2=100x—-EB—Z~=100x - 100/2016 =94,1% mol H2 + mol o2 1.00/2.016 +1.00/32.oo o . . . mol%02=100x ml 2 =100x 100/32 00 =5.9% mole-i-molOz 1.00/2.016+1.00/32.00 b) RT —1 —1 N =nNZ =1.00/28.02mol><8.31413rm:1 K ><3001<1=445x104Pal = V 2.00x10 m - RT _ . -l —1 P =noz =1.00/3200mol><8314{3m<:l K x300K=3.90x104Pa 02 V 2.00x10 m 13m, = 8.35x104Pa mol N 1.00/28.02 I‘VN =100x——-—2—=100x =53.3°/ m0 ° 2 molN2+m0102 1.00/28.02+1.00/32.00 ° mol 0 1.00/32.00 w o =100x——2—=100x =46.7°/ m0 ° 2 molN2+m0102 1.00/28.02+1.00/32.00 ° c) 71 RT _ . ‘1 '1 W = NH: =1.OO/1703mol><8314J-3m<:1 K ><300K=732x104Pa = V 2.00><10 rn nCH RT 1.00/16.04 mol><8.314Jmol‘1K"x300K 4 = 4 =————————~____=7.77x10 P CH4 V 2.00X10'3m3 a 12m, =1.51x105Pa mol NH 1.00/17.03 NV NH =100x—3—=100x—-—ﬁ=4s.5ty m0 ° 3 molNH3+m01CH4 1.00/17.03+1.00/16.04 ° mol % o2 =100x “‘01 CH4 =100>< 1'00/16'04 = 51.5% mol NH3 + mol CH4 1.00/17.03 +1.00/16.04 P1.5) A balloon ﬁlled with 10.50 L of Ar at 180°C and 1 atm rises to a height in the atmosphere Where the pressure is 248 Torr and the temperature is —30.5°C. What is the ﬁnal volume of the balloon? g T, V _ 760 Toer(273.15—30.5)K — i———————-X10.50L=26.8L P T. 248 Torr><(273.15+18.0)K P1.10) A sample of propane (C3H3) placed in a closed vessel together with an amount of 02 that is 3.00 times the amount needed to completely oxidize the propane to C02 and H20 at constant temperature. Calculate the mole ﬁaction of each component in the resulting mixture after oxidation assuming that the H20 is present as a gas. The reaction is C3Hg(g) + 502(g) —-> 3C02(g) + 4H2O(g). If m mol of propene are present initially, there must be 15m mol of 02. After the reaction is complete, there are 3m mol of CO2, 4m mol of H20, and 10m mol of 02. xco: ﬁg: 0.176; xﬂzo =14; =0.235; x02 =i—(7)%= 0.588 Pl.12) The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.400 atm when measured at the same values of T and Vas the original mixture. What was the composition of the original mixture in mole percent? 2H2(g) + 020;) -+ 2H20(1) initial moles n; no 0 1 at equilibrium 11;,z — 20! n5 —a 2a If the O2 is completely consumed, n51 -— a = 0 or a: n52. The number of moles of H2 remaining is 11;,z — 2a: n}: — 2n52. Let P1 be the initial total pressure and P2 be the total pressure aﬁer all the 02 is consumed. . .. RT . . RT P1 = ("H2 +"0.)? an“’2 = ("Hz “2’10. )7 Dividing the second equation by the ﬁrst P n" n" —2-=°—HZ—q_2—02—.=XH 2x0 =1"'xo “2x0 = ‘3x0 P1 an +rzo2 n”: +nc,2 1 1 1 1 P 1.15) The barometric pressure falls off with height above sea level in the Earth's Mi? atmosphere as R = B°e RT where P; is the partial pressure at the height 2, R." is the partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, and T is the absolute temperature. Consider an atmosphere that has the ‘ composition xN2 = 0.600 and xcol = 0.400 and that T = 300 K. Near sea level, the total pressure is 1.00 bar. Calculate the mole fractions of the two components at a height of 50.0 km. Why is the composition different from its value at sea level? M 28.04x10‘3kgx9.81 m s'2x50x103m] _M lg: - PN =P£e RT =0.600xl.0125><lOsPaexp 1 1 ‘ 2 2 8.314 Jmol' K“ X300 K = 242 Pa M __4§. —3 —2 3 PW: =Pgoze RT =0.400x1.0125x105Pa exp —h_‘_44'04xw ng9-5f1mls X50X10 “1 8.314 Jmol' K‘ x300 K, = 6.93 Pa x _ Pcoz _ 6.93 C01 PCOI+PN2 6.93+242 The mole fraction of C02 at the high altitude is much reduced relative to its value at sea level because it has a larger molecular mass than N2. for a molar volume 1.42 L at 300 K using 1 ulate the pressure exerted by Ar t1; 62:1) dc: SNaals equation of state. The van der Waals parameters a and b for AI are 1.355 bar dm6 mol‘2 and 0.0320 dm3 mol“1, respectivelyL) Is the attractive or repulswe portion of the potential dominant under these conditions. RT _ _a_ = Vm —b V; 6 4 8.314x10‘2bar dm3mol'1K" x300K _ngz_ :1” bar = 1.42 dm3mor‘ - 0.0321dm3mol'1 (1_42dm3m01-1) RT _W=mbax I?de =7’ 1.42L Because P < Pideal, the attractive part of the potential dominates. P2.2) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using a constant external pressure of 5.00 atm. Calculate q, w, AU, and AH. w=2.03><10‘J;AU=0 andAH=0 q=—2.03x104J P23) A system consisting of 57.5 g of liquid water at 298 K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 1.50 A passes through the 10.0—ohm resistor for 150 5, what is the ﬁnal temperature of the water? The heat capacity for water can be found in Appendix A. 1th + ncpyng nC Pm (1.50A)2 X10.0 ohmxlSO s +—L'5g——_T>< 75.291 Jmol‘lK'5< 298 K 18.02 gmol _. = T, __.______——-————————-———— 312K ——§7—'5—‘—°’—_—1x75.291 moi-1K-1 18.02 g mol q=12Rt=nCP1m(Tf—Z);Tf = P2.7) Calculate w for the adiabatic expansion of 1 mol of an ideal gas at an initial pressure of 2.00 bar from an initial temperature of 450 K to a ﬁnal temperature of 300 K. Write an expression for the work done in the isothermal reversible expansion of the gas at 300 K from an initial pressure of 2.00 bar. What value of the ﬁnal pressure would give the same value of w as the ﬁrst part of this problem? Assume that Cg," = 5/2R. wad = —AU = n(CP‘m —R)AT=—%X8.3l4 Jmol-‘K‘1 x150 K = -l.87><1031 . p _ ' Wmmible = "nRTlnfLﬂn-L = 1M Pf Pf nRT P‘ "RT 1.87x103J III—I- : = —1 —1 = 0.7497 Pf wrzvmible 1 Jmol K K P, = 0.47213. = 0.944 bar P2.12) An ideal gas described by T,- = 300 K, P; = 1.00 bar, and V,- = 10.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed-cycle process in a P-V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction? Chapter 2/ Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics 20 40 60 80 100 120 V(L) B. ,. 1.00 bar><10.0 L "2—: I; 8.3 145X10'2L bar mol"K‘l ><300 K = 0.401 mol The process can be described by step 1: Pi,V,-,T,- —> P1 = 10.0 bar, Vi, T1 step 2: P1,Vi, T1-—>P,-,V2 T1 step 3: Pi, V2, T1—> Pi,Vi,Ti. In step 1, Pi,V.-,T,~ -—> P1,Vi, T1, w = 0 because Vis constant. In step 2, P1,Vi, T1—>Pi, V2, T1 Before calculating the work in step 2, we ﬁrst calculate T1. T1 = 354300 leo'o bar =3000 K I; 1.00 bar V P w=—nRT1 1n—’-=-nRT1 1n—' V,- Pf = —0.401 mol>< 8.314 mol-1K-I x 3000 len 10'0 bar = —23.0x103J 1.00 bar In step 3, 13V.- =P.-V2; V2 =¥=IOK =100 L 5 —3 3 w=—PWW,AV=-1.00 bar><10 Pa 10 m ><(10 L—100L)>< =9.00X103J bar we?” = O—23.0x103J +9.00X103J = —14.0><103J If the cycle were traversed in the opposite direction, the magnitude of each work term would be unchanged, but all signs would change. 2-8 ...
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