Question

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14a A mixture of nitrogen and krypton gases, in a 6.64 L flask at 13 °C, contains 8.14 grams of nitrogen and 7.40 grams of krypton. The partial pressure of krypton in the flask is  atm and the total pressure in the flask is atm.  

14b A mixture of methane and helium gases is maintained in a 5.88 L flask at a pressure of 1.90 atm and a temperature of 86 °C. If the gas mixture contains 4.12 grams of methane, the number of grams of helium in the mixture is g.

14c The stopcock connecting a 2.38 L bulb containing carbon dioxide gas at a pressure of 8.95 atm, and a 7.89 L bulb containing argon gas at a pressure of 1.67 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.

15a A mixture of nitrogen and krypton gases, at a total pressure of 775 mm Hg, contains 6.20 grams of nitrogen and 5.63 grams of krypton. What is the partial pressure of each gas in the mixture?

P_N2 = mm Hg
P_Kr = mm Hg

15b

A mixture of methane and helium gases contains methane at a partial pressure of 489 mm Hg and helium at a partial pressure of 232 mm Hg. What is the mole fraction of each gas in the mixture?

X_CH4 =
X_He =

15c A mixture of carbon dioxide and argon gases at a total pressure of 935 mm Hg contains carbon dioxide at a partial pressure of 280 mm Hg. If the gas mixture contains 3.89 grams of carbon dioxide, how many grams of argon are present?

g Ar

Answer 1

isla) Na 15(c) COL O ner=5.63 and ker and Ar 6.209m 5.639m 3.89gn War= ? Protal = 775 mmng Pr = 935 mm NN₂= 6.20 0.221 mole Pco, = 280 mm nco₂=3.89 088 mole ५५ 0.067 mole 83.8 Xco₂= POOL 280 0.299 XN₂ = NN2 0.221 Pr 935 hN, there 0.221+0.067 noz 0.299 XN₂ -0.767 nco, than Acc. to Daltons law 0.088 0.29 9 PN, = XN₂ Pr 0.088thar 70.767 X 775 594.70 0.088 than 0.294 Pkur = Pr - PN. Naro.206 mole 775-594.70= 180.3 0.206 PN, =) 594-70 mm of Hg Pkr 180.3 mm of Hg War= 0.2068 39.948 War= 8.242 gm 15(b) Pony = 489 mm Pne = 232 mm Pr=489+232 = 721 Per Xcnu = Penu - 489 Pr 721 Xchu =) 0.68 Xchyt Xne = 1 Xne=1-0.68 = 0.32 De دم کیا T Mar 14a) N and Ker Let the moles of He = ne 8 149 7. Yogu ..PV= (hehut ne RT N = 6.64L Nenut ne PV T= -13°C = 286 15k RT MN₂7 8.14 0.291 mole 0.258+ Mae = 1.985-88 28 0.0821x359.15 n kr = 7.40 -0.088 mole 0.258thre = 0.379 83.8 Nue = 0.121 mole use ideal gol PV=nRT Whe o 121 PV = (MNathkr) RT Mre P = (nN₂ + Ner) RT Whe= 0.121x4 Whe) 0.484 gen P = (0.29170-88) 0-0821X 286.15 6.64 14(c) Co Ar Protal= 0.388 0.0821 x 2 86.15 6.64 Protal= 1.342 atm 2.382 7.89 L 8.95 atm 1.67atu X kor = ner for CO2 in 8.95X2-38=co. BT Nether 0.291+0.088 for Ar=) 7.89x167 = NAIRT X kor = 0.768 Final mixture Acc. to Daltons law P(2.38+7.89)= (n co tham) RT Pkr = X kr Pi from ea and ② put -0.768 X 1.342 value of moles Pkr 1.03 atm P(10.72) = /21.301, 13.176) PT RT 1462 Chu and ne P-34.477 3.22 4. 12m Whe gen 10.72 v=5-88L p=1.90 atm Print 3.22 atm T=86°C = 359.15k nenu = 4.12 ) 0.258 mole 16 7 0.291 RT V X chu y 0.68 Xne = 0.32