UNIT 13 – Gas Laws Keys
Avogadro’s Law Worksheet
1. Avogadro’s law states that equal volumes of different gases, at the same
temperature and _pressure_, contain the same _number__ of moles of gas
particles.
2. According to Avogadro’s law, how will the number of molecules in 2 liters of
hydrogen gas compare with the number of molecules in 2 liters of oxygen gas at
the same temperature and pressure? _equal number of molecules___
3. The molar volume of a gas at standard temperature and pressure is __22.4L_____
liters.
V = Volume
n = number moles
V1 V2

n1 n 2
or
V 1  n 2 = V2  n 1
Solve the following problems assuming a constant pressure and temperature. Express all answers to 3
significant figures. Show all work for credit.
4. n1= 3.40 mol
V1= 3.50 L
n2 = 1.50 mol V2= ?
(3.50)(1.50) = (X)(3.40)
x= 1.54L
5. n1= 1.50 mol
(30)(X) = (10)(1.5)
V1= 30.0 L
n2 = ? mol
x= 0.500mol
V2= 10.0 L
6. n1= 1.10 mol
V1= 7.70 L
n2 = 3.30 mol V2= ?
(7.70)(3.30) = (X)(1.10)
X= 23.1L
7. n1= 0.500 mol
V1= 11.2 L
n2 = ? mol
(11.2)(X) = (44.8)(.500) X= 2mol
V2= 44.8 L
8. If 0.50 mol of oxygen occupies a volume of 7.20 L at STP, what volume does
1.5 moles of O2 occupy at STP?
(7.2)(1.5) = (X)(.5) X= 21.6L
9. If 3.25 mole of argon gas occupies a volume of 1.00 L at STP, what volume does
14.15 mol of argon occupy?
(1)(14.15) = (X)(3.25)
X= 4.35L
10. If 12.40 mol of CO2 occupies a volume of 96.8 L, how many moles occupy 72.6 L?
(96.8)(X) = (72.6)(12.40) X=9.30moles
11. If 0.40 g of Helium has a volume of 6.2 L, how many grams of helium would
occupy 15.5 L?
.40g/4g = .1mole
(6.2)(X) = (15.5)(.1) = .25mole x 4 = 1gram
* you need to convert to moles first and then at the end convert back to grams!
12. If 14.0 g of nitrogen occupies a volume of 16.8 L, what volume will 49.0 g of
nitrogen have?
n1= 14g N2 / 28g = .5mole
n2= 49g N2 / 28g = 1.75mole
(16.8)(1.75) = (X)(.5) = 58.8L
Chemistry: The Combined Gas Law
Solve the following problems. As always, include enough work and show the units to ensure full credit.
1. The pressure of a gas changes from 120 kPa to 50 kPa. The volume changes from 45 L to 40 L. If the initial
temperature is 81oC, what is the final temperature in oC?
P1= 120
P2= 50
V1= 45
V2=40
T1 = 81+273= 354 T2= X
(120)(45) = (50)(40)
(354)
(X)
(120)(45)(X) = (50)(40)(354) X = 131K – 273 = -142oC
cross multiply and divide
2. A sample of nitrogen goes from 21 m3 to 14 m3 and its pressure increases from 100 kPa to 150 kPa. The final
temperature is 300 K. What was the initial temperature in Kelvins?
P1= 100
P2= 150
V1= 21
V2=14
T1 = X
T2= 300K
(100)(21) = (150)(14)
(X)
(300)
cross multiply and divide
(100)(21)(300) = (150)(14)(X) X = 300K
3. A sample of argon goes from 500 K to 350 K and its pressure changes from 280 kPa to 380 kPa. If the initial volume
is 18 dm3 (this is a volume unit), what is the final volume?
P1= 280
P2= 380
V1= 18
V2=X
T1 = 500K
T2= 350K
(280)(18) = (380)(X)
(500)
(350)
cross multiply and divide
(280)(18)(350) = (380)(X)(500) X = 9.3dm3
4. A sample of neon experiences a pressure drop from 75 kPa to 53 kPa. The temperature increases from 27 oC to
93oC. If the initial volume is 12 L, what is the final volume?
P1= 75
P2= 53
V1= 12
V2=X
T1 = 27+273=300K T2= 93+273=263K
(75)(12) = (53)(X)
(300)
(263)
cross multiply and divide
(75)(12)(263) = (53)(X)(300) X = 20.7L
5. The volume of a sample of helium increases from 5 L to 25 L and its temperature drops from 2000 K to 1750 K. If the
initial pressure is 1500 mm Hg, what is the final pressure?
P1= 1500
V1= 5
T1 = 2000K
P2= X
V2=25
T2= 1750K
(1500)(5) = (X)(25)
(2000)
(1750)
cross multiply and divide
(1500)(5)(1750) = (X)(25)(2000) X = 262.5mmHg
6. The temperature of a gas increases from 212 oC to 380oC. The volume goes from 30 dm3 to 18 dm3. If the final
pressure is 1.85 atm, what was the initial pressure?
P1= X
V1= 30
T1 = 212+273= 485K
P2= 1.85
V2=18
T2 = 380+273=653K
(X)(30) = (1.85)(18)
(485)
(653)
cross multiply and divide
(X)(30)(653) = (1.85)(18)(485) X = .82atm
Charles’ Law Problems
Show all work.
1)
The temperature inside my refrigerator is about 4 0 Celsius. If I place a balloon in my fridge that initially has a
temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by
my refrigerator?
T1= 22+273 = 295K V1=.5L T2= 4+273=277K V2 =X
.5L
= X
cross multiply and divide X = .47L
295K
277K
2)
A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C,
what will the volume of the balloon be after he heats it to a temperature of 250 0C?
T1 = 20+ 273 = 293K V1= .4L
T2= 250+ 273 = 523K V2= X
.4L
= X
cross multiply and divide X = .71L
293K
523K
3)
On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been
opened. If I have a 250 mL bag at a temperature of 19 0C, and I leave it in my car which has a temperature of 60 0
C, what will the new volume of the bag be?
T1 = 19+ 273 = 292K V1= 250mL
T2= 60+ 273 = 333K V2= X
250mL
= X
cross multiply and divide X = 283mL
292K
333K
4)
A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an
empty soda bottle (volume of 2 L) at room temperature (25 0C), what will the new volume be if you put it in your
freezer (-4 0C)?
T1 = 25+ 273 = 298K
2L
= X
298K
269K
V1= 2L
T2= -4+ 273 = 269K V2= X
cross multiply and divide X = 1.8L
Boyle’s Law
1)
1.00 L of a gas at standard
new pressure of the gas?
temperature and pressure is compressed to 473 mL.
What is the
V1= 1.00L P1= 1atm V2= 473mL/.473L P2= X (need to have same units)
(1atm)(1L) = (X)(.473L) solve for X = 2.11atm
2)
In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 106 atm.
When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a
pressure of 1.00 atm. What is the volume of the gas after the explosion?
V1= 0.05L P1= 4x10 6atm V2= X P2= 1atm (need to have same units)
(4x10 6atm)(0.05L) = (X)(1atm) solve for X = 200 000L
3)
Synthetic diamonds can be manufactured at pressures of 6.00 x 104 atm. If we took 2.00 liters of gas at 1.00
atm and compressed it to a pressure of 6.00 x 104 atm, what would the volume of that gas be?
V1= 2.00L P1= 1atm V2= X P2= 6x10 4atm (need to have same units)
(1atm)(2.00L) = (X)(6x10 4atm) solve for X = 3.3x10 -5L
4)
Submarines need to be extremely strong to withstand the extremely high pressure of water pushing down on
them. An experimental research submarine with a volume of 15,000 liters has an internal pressure of 1.2 atm. If
the pressure of the ocean breaks the submarine forming a bubble with a pressure of 250 atm pushing on it, how
big will that bubble be?
V1= 15 000L P1= 1.2atm V2= X P2= 250atm (need to have same units)
(1.2atm)(15 000L) = (X)(250atm) solve for X = 72L
Chemistry: Unit Conversions for the Gas Laws
Directions: Complete the following tables, showing your work for each lettered box beside the corresponding letter below.
Include units on your work, and write your final answers in the tables.
TEMPERATURE
K
373 K
o
C
(D) 100
PRESSURE
mm Hg
890 mm Hg
o
(A)329K
56 C
(G)922
(B)427k
154oC
(H)486
128 K
(E)-145
800 K
(F)527
(C)263
3140 mm Hg
(I)1786
o
–10 C
(J)188
(A)
(J)
(B)
(K)
(C)
(L)
(D)
(M)
(E)
(N)
(F)
(O)
(G)
(P)
(H)
(Q)
kPa
(K)119kpa
123 kPa
(L)65
(M)419
(N)238
25 kPa
atm
(O)1.17atm
(P)1.2
0.64 atm
(Q)4.13
2.35 atm
(R).25
(I)
(R)