Gases Test
Review Guide
Gases Review
𝟏
1. Explain the relationship between pressure and volume in Boyle’s Law. 𝑷 𝜶 , as one increases the
𝑽
other must decrease.
2. Explain the relationship between volume and temperature in Charles’ Law. V α T, as one increases the
other must increase or vice versa.
3. Fill in the blank regarding the Kinetic Molecular Theory:
a) All gases have __mass__________________.
b) Gas particles are widely __spread apart_______________.
c) Gas particles must be in __constant_____, ___random__________ motion.
d) Gases exert __pressure/force___ because they frequently collide with the walls of their container.
e) The average kinetic energy of the gas depends only on the __temperature________ of the gas.
f) Gases are neither __attracted_________ or ___repelled__________ by one another.
4. What are the four variables for gas laws and what unit do we use with them when R is present in the
equation?
P = Pressure (atm - atmospheres)
V= Volume (L - Liters)
n=Amount (mol - moles)
T=Temperature (K – Kelvin)
5. What is the value and unit on R? 𝟎. 𝟎𝟖𝟐𝟏
𝒂𝒕𝒎(𝑳)
𝒎𝒐𝒍(𝑲)
6. If a gas is at 45.6 oC, what is that temperature in Kelvin? 45.6 + 273 = 318.6 K
7. 1 atm = _760____ torr = _101,325__ Pa = _760____ mm Hg = __101.325___ kPa = _14.70___ lb/in2
8. What does the term STP stand for and list them? Standard temperature (273K) and standard
pressure (1 atm).
9. What is the difference between the use of a manometer and a barometer? A barometer measure the
atmospheric pressure. A manometer compares the pressure in an inclosed container to
the atmospheric pressure.
10. A manometer contains a sample of nitrogen gas at a pressure of 78.3 kPa. The level of mercury in the Utube is 11.4 mm lower on the end open to the atmosphere. What is the atmospheric pressure in kPa? What is
the pressure in torr?
𝟏𝟏. 𝟒 𝒎𝒎 𝑯𝒈 𝑿
𝟏𝟎𝟏. 𝟑𝟐𝟓 𝒌𝑷𝒂
= 𝟏. 𝟓𝟏𝟗𝟖𝟕𝟓 𝒌𝑷𝒂
𝟕𝟔𝟎 𝒎𝒎 𝑯𝒈
78.3 kPa + 1.519875 kPa = 79.8 kPa
𝟕𝟗. 𝟖𝟐𝟎 𝒌𝑷𝒂 ×
𝟕𝟔𝟎 𝒕𝒐𝒓𝒓
𝟏𝟎𝟏.𝟑𝟐𝟓 𝒌𝑷𝒂
= 𝟓𝟗𝟗 𝒕𝒐𝒓𝒓
11. A gas container is fitted with a manometer. The level of mercury is 10. mm higher on the open side. Using
a laboratory barometer you find the atmospheric pressure is 750. mm Hg. What is the pressure of the gas in the
container in mm Hg? What is the pressure in atmospheres?
𝟕𝟔𝟎 𝒎𝒎 𝑯𝒈 𝑿
750 mm Hg + 10 mm Hg = 760 mm Hg
𝟏 𝒂𝒕𝒎
𝟕𝟔𝟎 𝒎𝒎 𝑯𝒈
= 𝟏. 𝟎 𝒂𝒕𝒎
12. A gas at a pressure of 456 mm Hg is held in a container with a volume of .344 L. The volume of the
container is then increased to 670 mL without a change in temperature. What is the new pressure of the gas?
P1 = 456 mm Hg
V1 = .344 L
V2 = 670 mL = .670 L
P2 = ?
𝑷𝟏 𝑽𝟏 = 𝑷𝟐 𝑽𝟐
𝟒𝟓𝟔(. 𝟑𝟒𝟒) =. 𝟔𝟕𝟎𝑷𝟐
𝑷𝟐 = 𝟐𝟑𝟎 𝒎𝒎 𝑯𝒈
13. A gas occupies a volume of 2.45 L at a pressure of 1.58 atm. If the temperature remains constant and the
pressure changes to 722 mm Hg, what is the new volume of the gas?
P1 = 1.58 atm
𝟕𝟐𝟐 𝒎𝒎 𝑯𝒈 ×
V1 = 2.45 L
V2 = ?
P2 = 722 mm Hg
𝑷𝟏 𝑽𝟏 = 𝑷𝟐 𝑽𝟐
𝟏 𝒂𝒕𝒎
𝟕𝟔𝟎 𝒎𝒎 𝑯𝒈
= 𝟎. 𝟗𝟓 𝒂𝒕𝒎
𝟏. 𝟓𝟖(𝟐. 𝟒𝟓) = 𝟎. 𝟗𝟓𝑽𝟐
𝑽𝟐 = 𝟒. 𝟎𝟕 𝑳
14. A gas sample at 83°C occupies a volume of 140 L. At what temperature will it occupy .120 mL?
T1 = 83°C + 273 = 356 K
𝑽𝟏
𝑽
V1 = 140 L
= 𝟐
𝑻𝟏
𝟏𝟒𝟎
𝑻𝟐
𝟑𝟓𝟔
.𝟎𝟎𝟎𝟏𝟐𝟎
=
𝑻𝟐
𝟏𝟒𝟎𝑻𝟐 = . 𝟎𝟒𝟐𝟕𝟐
𝑻𝟐 = 𝟑. 𝟏 𝑿 𝟏𝟎−𝟒 𝑲
T2 = ?
V2 = .120 mL = .000120 L
15. What will be the volume of a gas sample at 335 K if its volume at 256 K is 7.66 L?
T1 = 335 K
V1 = ?
𝑽𝟏
𝑻𝟏
=
𝑽𝟐
𝑽𝟏
𝑻𝟐
𝟑𝟑𝟓
=
𝟕.𝟔𝟔
𝟐𝟓𝟔
𝟐𝟓𝟔𝑽𝟏 = 𝟐𝟓𝟔𝟔. 𝟏
𝑽𝟏 = 𝟏𝟎. 𝟎 𝑳
T2 = 256 K
V2 = 7.66 L
16. If the atmospheric pressure is 104 kPa, what is the partial pressure of oxygen gas if it makes up 22% of the
air?
𝑷𝑻 × (%) = 𝑷𝑶𝟐
𝟏𝟎𝟒 × . 𝟐𝟐 = 𝟐𝟑 𝒌𝑷𝒂
17. What is the pressure of a mixture of nitrogen, oxygen, and helium if their partial pressures are 650 mm Hg,
123 mm Hg, and 98 mm Hg?
N2 = 650 mm Hg
O2 = 123 mm Hg
He = 98 mm Hg
PT = P1 + P2 + P3
PT = 650+123+98
PT = 870 mm Hg
18. An air-filled balloon has a volume of 225 L at 0.94 atm and 25oC. Soon after, the pressure changes to 0.99
atm and the temperature changes to 0oC. What is the new volume of the balloon?
V1 = 225 L
P1 = .94 atm
T1 = 25°C + 273 = 298 K
P2 = .99 atm
T2 = 0°C + 273 = 273 K
V2 = ?
𝑷𝟏 𝑽𝟏
𝑷𝟐 𝑽𝟐
=
𝒏𝑹𝑻𝟏
.𝟗𝟒(𝟐𝟐𝟓)
𝒏𝑹𝑻𝟐
.𝟗𝟗𝑽𝟐
=
𝟐𝟗𝟖
𝟐𝟕𝟑
. 𝟗𝟒(𝟐𝟐𝟓)(𝟐𝟕𝟑) = 𝟐𝟗𝟖(. 𝟗𝟗)𝑽𝟐
̅𝟎 𝑳
𝑽𝟐 = 𝟐𝟎
19. What pressure is exerted by 1.24 moles of H2 gas in a 0.0454 cm3 container at -24.0oC?
P=?
PV=nRT
3
-5
V = 0.0454 cm = 0.0454 mL = 4.54 X 10 L
T = -24.0°C + 273 = 249 K
4.54 X 10-5 P=1.24(0.0821)(249)
N = 1.24 mol
R = 0.0821
P = 558,000atm
20. A balloon is inflated with 0.358 g of hydrogen gas to a pressure of .98 atm. If the desired volume of the
balloon is 1.250 L, what must the temperature be in °C?
𝟎. 𝟑𝟓𝟖 𝒈 𝑯𝟐 ×
P = .98 atm
V = 1.250 L
T=?
m = .358 g H2
R = 0.0821
𝟏 𝒎𝒐𝒍 𝑯𝟐
𝟐.𝟎 𝒈 𝑯𝟐
= 𝟎. 𝟏𝟕𝟗 𝒎𝒐𝒍 𝑯𝟐
PV=nRT
.98(1.250)=0.179(0.0821)T
T = 83.357 K – 273 = -190°C
21. A gas confined in a 515 ml container exerts a pressure of 95.8 kPa at 25.8°C. At what Celsius temperature
will it exert a pressure of 205.8 kPa if it is placed into a 644 mL container?
V1 = 515 mL
P1 = 95.8 kPa
T1 = 25.8°C + 273 = 298.8 K
P2 = 205.8 kPa
T2 = ?
V2 = 644 mL
𝑷𝟏 𝑽𝟏
𝑷𝟐 𝑽𝟐
=
𝒏𝑹𝑻𝟏
𝟗𝟓.𝟖(𝟓𝟏𝟓)
𝒏𝑹𝑻𝟐
𝟐𝟎𝟓.𝟖(𝟔𝟒𝟒)
=
𝟐𝟗𝟖.𝟖
𝑻𝟐
𝟐𝟗𝟖. 𝟖(𝟐𝟎𝟓. 𝟖)(𝟔𝟒𝟒) = 𝟗𝟓. 𝟖(𝟓𝟏𝟓)𝑻𝟐
𝑻𝟐 = 𝟖𝟎𝟐. 𝟔𝟕𝟑 𝑲 − 𝟐𝟕𝟑 = 𝟓𝟑𝟎. °𝑪
18. An air-filled balloon has a volume of 225 L at 0.94 atm and 25oC. Soon after, the pressure changes to 0.99
atm and the temperature changes to 0oC. What is the new volume of the balloon?
V1 = 225 L
P1 = .94 atm
T1 = 25°C + 273 = 298 K
P2 = .99 atm
T2 = 0°C + 273 = 273 K
V2 = ?
𝑷𝟏 𝑽𝟏
𝑷𝟐 𝑽𝟐
=
𝒏𝑹𝑻𝟏
.𝟗𝟒(𝟐𝟐𝟓)
𝒏𝑹𝑻𝟐
.𝟗𝟗𝑽𝟐
=
𝟐𝟗𝟖
𝟐𝟕𝟑
. 𝟗𝟒(𝟐𝟐𝟓)(𝟐𝟕𝟑) = 𝟐𝟗𝟖(. 𝟗𝟗)𝑽𝟐
̅𝟎 𝑳
𝑽𝟐 = 𝟐𝟎
19. What pressure is exerted by 1.24 moles of H2 gas in a 0.0454 cm3 container at -24.0oC?
P=?
PV=nRT
3
-5
V = 0.0454 cm = 0.0454 mL = 4.54 X 10 L
T = -24.0°C + 273 = 249 K
4.54 X 10-5 P=1.24(0.0821)(249)
N = 1.24 mol
R = 0.0821
P = 558,000atm
20. A balloon is inflated with 0.358 g of hydrogen gas to a pressure of .98 atm. If the desired volume of the
balloon is 1.250 L, what must the temperature be in °C?
𝟎. 𝟑𝟓𝟖 𝒈 𝑯𝟐 ×
P = .98 atm
V = 1.250 L
T=?
m = .358 g H2
R = 0.0821
𝟏 𝒎𝒐𝒍 𝑯𝟐
𝟐.𝟎 𝒈 𝑯𝟐
= 𝟎. 𝟏𝟕𝟗 𝒎𝒐𝒍 𝑯𝟐
PV=nRT
.98(1.250)=0.179(0.0821)T
T = 83.357 K – 273 = -190°C
21. A gas confined in a 515 ml container exerts a pressure of 95.8 kPa at 25.8°C. At what Celsius temperature
will it exert a pressure of 205.8 kPa if it is placed into a 644 mL container?
V1 = 515 mL
P1 = 95.8 kPa
T1 = 25.8°C + 273 = 298.8 K
P2 = 205.8 kPa
T2 = ?
V2 = 644 mL
𝑷𝟏 𝑽𝟏
𝑷𝟐 𝑽𝟐
=
𝒏𝑹𝑻𝟏
𝟗𝟓.𝟖(𝟓𝟏𝟓)
𝒏𝑹𝑻𝟐
𝟐𝟎𝟓.𝟖(𝟔𝟒𝟒)
=
𝟐𝟗𝟖.𝟖
𝑻𝟐
𝟐𝟗𝟖. 𝟖(𝟐𝟎𝟓. 𝟖)(𝟔𝟒𝟒) = 𝟗𝟓. 𝟖(𝟓𝟏𝟓)𝑻𝟐
𝑻𝟐 = 𝟖𝟎𝟐. 𝟔𝟕𝟑 𝑲 − 𝟐𝟕𝟑 = 𝟓𝟑𝟎. °𝑪