Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Can you help me with the number 4 question? Can you explain step by step including the formula (Boyle’s law)? I need to plug in the fraction to give the correct answer.
Transcribed Image Text:V₁=550.0mL
T₁ = 77°c=(77+273)k
= 350k
Gas Laws Worksheet
-86°C = 86+273) K = 359K
Charles's Law (temperature, volume)
V₁ = √₂
1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure
= Constant
remains constant.
T Ta
Va= Vita :(550.0 mL) x (359k)
350k
V₁=1.00L
2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C?
+₁=0,00°C = 10.00+273) K=273k
Ta=333.0°C (333.0 +273) K = 606K
Va v
V₂= V₁Ta - (1.00 L) x (6061) _
273K
=
564 ml
2.22 L
Ta
V₂ = 564m²
Boyle's Law (pressure, volume) Pv=nr + >constant
3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm.
P.V₁ = Pay₂
V
n
P₁ = 63.0 atm √2=2013X10³/2
V₁ = 3382
21300 L or 2.13 x 10 L
1/₂ = ₁V₁ = (63.0 atm) x (338L) = (2.13×10¹ L Pa = 1.00A+m
Pa
7,00 atm
Tatm-760mm Hg
5730 mmHg
4) A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mm Hg. Find the volume of nitrogen
when its pressure is changed to 400.0 torr while the temperature is held constant.
V₁ = √₂
Initial volume nitregen (V₁) = 14.0L
Initial pressure nitregen (Pi) = 760.0mmits
Final volume a) =
na
26.6 L
^₁
760.0mm Hg x 14.0 L
400.0mm Hg
-_Pressure (pa) = 400.0 torr
= 26.6
V₂= P₁V₁
Pa
5) What pressure (mm Hg) is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder
whose volume is 26.0 liters? P₂ = P₁V₁
Va
P₁V₁=lava 7.540+m V₁ = 196.0L
1.00atm x 196.0L
26.0L
P₁ = 19+m
-564 ML
= 2.22L
T₁=37% 25+273,15 K = 298,15K
318 K 12 = XT = 16.0 atm X 298,15 K
P.
1510 atm
7.54 atms 7.54 atm (160mmitty)
Gay-Lussac's Law (temperature, pressure
(5730mm Hg)
6) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C?
P₂ = P₁T2=0,03700mx 273 k temy (+₁) = 50.00 (
Pressure (P=0.0370 atm
.0313 atm
318.0266667
Ta-318K
r=1mm 17g
4000 torr = 400,0mmity
P₁ V₁ = Pav₂
323 K
Pa = 0.031272 446
0.0313atm
7) If a gas in a closed container, with an original temperature of 25.0 °C, is pressurized from 15.0
atmospheres to 16.0 atmospheres, what would the final temperature
of the gas be?
V2=2610L
P2=5 130 mm Hg
(S0+273) k
T2=0.00°C = (+373) K = 273 k
= 323K
Initial temp = 25°C
Initial pressure (P)=15.0 atm
Final pressure (Pa): 16.0 atm
Final temperature (Ta):
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