Gas Laws Compiled Act - Answer Keys
Gas Laws Compiled Act - Answer Keys
Gas Laws Compiled Act - Answer Keys
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1. Two hundred mL (200 cm3) of gas is contained in a vessel under a pressure of 850 mmHg. What would
be the new volume of the gas if the pressure is changed to 1000 mmHg? Assume that the temperature
remains constant.
2. A 2.5 L sample of a gas is collected at a pressure of 1.25 atm. Calculate the pressure needed to reduce
the volume of the gas to 2.0 L. The temperature remains unchanged.
(10 pts.) 2. A balloon is filled to a volume of 500 mL at a temperature of 30.0 °C. The balloon is then
heated to a temperature of 75.0 °C. Find the new volume of the balloon in Liters.
(10 pts.) 3. A 113L sample of Helium at 27°C is cooled to -78°C. Calculate the new volume of the
Helium.
(10 pts.) 2. If a gas in a closed container is pressurized from 15.0 atmospheres (atm) to 16.0 atmospheres
(atm) and its original temperature was 25.0 °C, what would the final temperature of the gas be in degrees
Celsius?
(15 pts.) 2. A cylinder contains 32.4 L of oxygen gas at a pressure of 1,748 mmHg and a temperature of
298 K. How much gas (in moles) is in the cylinder? (1 atm= 760 mmHg)
Answer Key
Boyle’s Law Quiz (20 pts.)
1. Two hundred mL (200 cm3) of gas is contained in a vessel under a pressure of 850 mmHg. What would
be the new volume of the gas if the pressure is changed to 1000 mmHg? Assume that the temperature
remains constant.
Given: V1 = 200 mL / 200 cm3 P1 V 1
Equation: P1 V 1=P2 V 2 ≫ = V2
P1 = 850 mmHg P2
*complete process of derivation must be shown such as cancellation of
P2 = 1000 mmHg necessary variables
Required: V2 or final volume Solution: (850 mmHg x 200 mL) / 1000 mmHg
Answer: 170 mL or 170 cm3
2. A 2.5 L sample of a gas is collected at a pressure of 1.25 atm. Calculate the pressure needed to reduce
the volume of the gas to 2.0 L. The temperature remains unchanged.
Given: V1 = 2.5 L P1 V 1
Equation: P1 V 1=P2 V 2 ≫ = P2
P1 = 1.25 atm V2
*complete process of derivation must be shown such as cancellation of
V2 = 2.0 L necessary variables
(10 pts.) 3. A 113L sample of Helium at 27°C is cooled to -78°C. Calculate the new volume of the
Helium.
Given: T1 = 27 0C + 273 = 300 K V1 V2 V1T2
Equation: = ≫ =V 2
V1 = 113 L T 1 T2 T1
*complete process of derivation must be shown such as cancellation of necessary
T2 = -78 0C + 273 = 195 K variables
(10 pts.) 2. If a gas in a closed container is pressurized from 15.0 atmospheres (atm) to 16.0 atmospheres
(atm) and its original temperature was 25.0 °C, what would the final temperature of the gas be in degrees
Celsius?
Combined Gas Law (Assignment) Submitted on 4/25 (15 pts.)
1. A sample of gas occupies a volume of 2.40 L at 685 mmHg and 130 0C. What would the pressure be (in
mmHg) if the volume were increased to 5.00 L and temperature increased to 210 0C? Use GRESA
Method.
Given: V1 = 2.40 L, V2 = 5.00 L
P1 = 685 mmHg
T1 = 130 0C + 273 = 403 K
T2 = 210 0C + 273 = 483 K
Required: P2?
Equation: (P1V1) / T1 = (P2V2) / T2 >> (P1V1T2) / V2T1 = P2
Solution: (685 mmHg x 2.40 L x 483 K) / (5.00 L x 403 K) = P2
Answer: 394 mmHg
(15 pts.) 2. A cylinder contains 32.4 L of oxygen gas at a pressure of 1,748 mmHg and a temperature of
298 K. How much gas (in moles) is in the cylinder? (1 atm= 760 mmHg)
Given: V = 32.4 L PV
Equation: PV = nRT >> =n
RT
P = 1748 mmHg x 1 atm/760 mmHg =
2.3 atm Solution: (2.3 atm x 32.4 L) / (0.0821
L.atm/mol.K) x (298 K)
T = 298 K
Answer: 3.05 mol
Required: number of moles of gas?