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# H09- Gas Laws-solutions.pdf

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University of Texas at Austin

Chemistry

CH 301

Sutcliffe

Spring

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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

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