Class Notes (808,489)
United States (313,114)
Chemistry (599)
CH 301 (158)
Sutcliffe (5)
Lecture

# H09- Gas Laws-solutions.pdf

7 Pages
65 Views

School
University of Texas at Austin
Department
Chemistry
Course
CH 301
Professor
Sutcliffe
Semester
Spring

Description
ngac (ttn793) – H09: Gas Laws – mccord – (51600) 1 This print-out should have 22 questions. Multiple-choice questions may continue on 2. 24 mm Hg the next column or page – ﬁnd all choices before answering. 3. 2400 mm Hg correct 001 10.0 points 4. 600 mm Hg Divers know that the pressure exerted by Explanation: the water increases about 100kPa with every V =10.0L V =5.0L 10.2m of depth. This means that at 10.2m 1 2 below the surface, the pressure is 201kPa; P 11200mmHg at 20.4m below the surface, the pressure is Boyle’s law relates the volume and pressure 301kPa; and so forth. If the volume of a of a sample of gas: balloon is 3.4tSTPandheempaue of the water remains the same, what is the P1V 1 P V2 2 volume 42.99 m below the water’s surface? P1V 1 (1200 mm Hg)(10.0L) P 2 = V2 5L Correct answer: 0.659377 L. =2400mmHg Explanation: P1=1atm Depth=42 .99 m V1=3 .4L V2=? 003 10.0 points 101.325 kPa = 1 atm At standard temperature, a gas has a volume ForP : of 298 mL. The temperature is then increased 2 to 133 C, and the pressure is held constant. 10.2m = 42.99 m What is the new volume? 100kPa x (10.2m)( x)=(42 .99 m)(100kPa) Correct answer: 443.179 mL. (42.99 m)(100kPa) Explanation: x = 10.2m =421 .471 kPa ▯ T1=0 C+2▯3=273K V 1298mL T2=133 C+273=406K V2=? P2=101kPa+421 .471 kPa 1atm =522 .471 kPa ▯ 101.325kPa V1 = V2 =5 .15638 atm T1 T2 V T (298 mL)(406 K) Applying Boyle’s law, V2= 1 2 = T 1 273 K P V = P V =443 .179 mL 1 1 2 2 P1V 1 (1 atm)(3.4L) V2= P = 5.15638 atm 2 004 10.0 points =0 .659377 L Aamlfgsi ldtrta temperature of 87 Candapsref7am ▯ is heated to 341 C. What pressure does the 002 10.0 points gas exert at the higher temperature? Agsildi a. Lkat00 mm Hg pressure. Which of the following is Correct answer: 11.9389 atm. aeoal lertepsrewenhe gas is pumped into a 5.00 L vessel? Explanation: T1=87 C+273=360K P1=7atm T =341 ▯C+273=614K P =? 1. 0.042 mm Hg 2 2 ngac (ttn793) – H09: Gas Laws – mccord – (51600) 2 Applying the Gay-Lussac law, Using the Combined Gas Law, P1 P2 P1V 1 P2V 2 = = T1 T2 T1 T 2 P T (7 atm)(614 K) P 2 1 2 = and recalling that STP implies standard tem- T1 360 K perature (273.15 K) and pressure (1 atm or =11 .9389 atm 760 torr), we have V = P 1 1 2 2 P2T 1 005 10.0 points Asga1 .71 ▯ 106 Pa and 19 Cu▯ -o (740 torr)(250 mL)(273.15 K) pies a volume of 387-m cta.tA = (760 torr)(298.15 K) 3 perature would the gas occupy 542 cm at =223 .01 mL 2.84 ▯ 10 Pa? 007 10.0 points Correct answer: 406.193 C. What is the percent yield if 4.51 moles of4CH Explanation: produces 16 L of CO2at STP? 6 6 P1=1 .71 ▯ 10 Pa P2=2 .84 ▯ 10 Pa CH +2O ▯ CO +2H O V1=387cm 3 T1=19 C+273=292K 4 2 2 2 3 V2=542cm T2=? 1. 6.31% P V P V 1 1 = 2 2 T1 T2 2. 15.8% correct T = P2V 2 1 2 P1V 1 3. 1.26% 6 3 (2.84 ▯ 10 Pa) (542 cm )(292K) = (1.71 ▯ 10 Pa) (387 cm ) 4. 36.4% =679 .193 K = 406.193 C 5. 78.1% Explanation: 006 10.0 points ▯ nCH4 =4.51mol VCO 2=16L Asampleofidealgasoccupies250mLat25 C At STP we can use the standard molar and 740 torr. 22.4L volume, to convert L of CO2to moles: What is its volume at STP? mol 1mol 1. 223 mL correct ?mol=16L ▯ =0 .714 mol 22.4L 2. 250 mL 0.714 moles is the actual yield. Now calculate the theoretical yield: 3. 235 mL 1molCO 2 ?molCO 2=4 .51 mol CH ▯4 4. 280 mL 1molCH 4 =4 .51 mol CO 2 5. 266 mL 0.714 %yield= ▯ 100% = 15.8% Explanation: 4.51 P =740torr P =760torr 1 2 V1=250mL T2=273.15K 008 10.0 points T1=25 C+273.15=298.15K If I have 44.8 liters of nitrogen gas at standard ngac (ttn793) – H09: Gas Laws – mccord – (51600) 3 temperature and pressure, how much will it Applying the ideal gas law equation, weigh? PV = nRT 1. 28 g nRT P = V ▯ ▯ 2. 28 kg (2.82 mol) 0.08206 mol·K (588 K) P = 3. 44.8 g 0.075 L · 760 torr 1atm 4. 56 g correct 6 =1 .37883 ▯ 10 torr 5. 56 u 010 10.0 points Explanation: How many moles of hydrogen are present in P =1atm V =44 .8L a2Lcotirtapeueof0trat 145 C? T =273K PV = nRT Correct answer: 0.0920088 mol. PV Explanation: n = V =12L T =145 ▯C+273=418K RT 1atm n = ▯ (1 atm)(44▯8L) · 28 g P =200torr · =0 .263 atm 0.08206 L·atm (273 K) 1mol 760 torr K·mol Applying the ideal gas law equation, =55 .9941 g PV = nRT n = PV 009 10.0 points RT A75mLm all il i . gf (0.263 atm)(12 L) CH and 3.23 g of NH la.tha n = ▯ ▯ 4 3 ▯ 0.0821 mol·K (418 K) pressure in the bulb at 315 C? =0 .0920088 mol 1. 7.38 ▯ 10 torr 011 10.0 points 2. There is no observable pressure in the We mix 146 grams of oxygen gas with bulb because the metal bulb does not allow 133 grams of argon gas in a volume of 1737 mL ▯ the exterior atmosphere to exert a force. at 23 C. What will be the ﬁnal pressure of the gas mixture? 3. 1.81 ▯ 10 torr Correct answer: 110.467 atm. 6 4. 1.38 ▯ 10 torr correct Explanation: mol 5. 2.39 torr nO2 =146grams · =4 .5625 mol 32 g mol Explanation: nAr =133grams · =3 .32933 mol mol 39.948 g nCH 4=42 .1g · 16 g =2 .63 mol ntotal7 .89183 mol ▯ mol V =1737mL=1 .737 L T =23 C=296K nNH =3 .23 g · =0 .19 mol
More Less

Related notes for CH 301

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.