Gas Law Practice
2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure?
(740.0 mmHg) (2.00 L) =(760.0 mmHg) (x) X =(740 mmHg) (2.00L) / (760.0 mmHg)
2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume was determined to be 8.00 L. What was the unknown pressure?
(x) (2.50 L) = (101.325 kPa) (8.00 L) (101.325 kPa) (8.00 L) / (2.50 L) Answer: notice the units of the pressure were not specified, so any can be used. If this were a test question, you might want to inquire of the teacher as to a possible omission of desired units. Let's use kPa since the other two units were used above. Once again, insert into P1V1 = P2V2 for the solution.
5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L?
1.08 atm) (5.00 L) =(x) (10.0 L) X = (1.08 atm) (5.00 L) / (10.0 L)
A soda bottle is flexible enough that the volume of the bottle can change when without opening it. If you have an empty soda bottle with a volume of 2.0 L at room temperature (25˚C), what volume will the bottle occupy when in the freezer at a temperature of -4.0˚C?
1.8 L
A sample of carbon dioxide gas occupies a volume of 3.50 L at 125 kPa. Assuming a constant temperature, what pressure is exerted on the gas if it has a volume of 2.00 L? 6)
219kPa
A sample of oxygen gas occupies a volume of 250. mL at a pressure of 740. torr. What volume will the gas occupy at a pressure of 800. torr if temperature is held constant?
231 mL
2) A sample of nitrogen occupies a volume of 250 mL at 25°C. What volume will it occupy at 95°C?
310ml
A sample of oxygen gas has a volume of 36.7 L at 145 kPa and 65.˚C. What volume will the sample have at STP?
42 l
5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What is the new temperature in order to maintain the same pressure? :
5.00/100K=20.0L/xK
A sample of helium occupies a volume of 3.8 L at -45°C. What volume will it occupy at 45°C?
5.3 L
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 L 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, what will be the volume of the bubble?
72 l
A sample of gas has a volume of 256 mL at 720 torr and 25°C. What pressure will the gas exert at 50.°C and 245 mL?
820 torr
10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?
Answer: change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325. Insert values into the equation 97.0kPa/298.0 K=101.325kPa/x The answer is 311.3 K, but the question asks for Celsius, so you subtract 273 to get the final answer of 38.3°C,
4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?
Answer: convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug into the equation and solve for x, like this: 4.40L/323K=x/298K Note2.20 L is the wrong answer. Sometimes one is tempted to look at the temperature being cut in half and reason that the volume must also be cut in half. That would be true if the temperature was in Kelvin. However, in this problem the Celsius is cut in half, not the Kelvin.
5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the temperature is changed to standard conditions, what is the new pressure?
Answer: convert to Kelvin and insert 745.0mmHg/295K=x/273.0 K:
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP ( standard temperature and pressure) STP is P 745.ommHg V 2.00 L T 298 K
P1 745.ommHg V1 2.00 L T1 298 K p2 760mmHg V2 X T2 278 k (745 mmHg)(2.00l)/298K= (760mmHg) X/278 K
A gas at STP is cooled to -185°C. What pressure in atmospheres will it have at this temperature (volume remains constant)?
P1 = 1.00 atm T1 = 273 K P2 = ? T2 = -185°C = 88 K Gay-lussac's Law P2=0.32atm
Chlorine gas has a pressure of 1.05 atm at 25°C. What pressure will it exert at 75°C?
P1 = 1.05 atm T1 = 25°C = 298 K P2 = ? T2 = 75°C = 348 K Gay-lussac's law P2 = 1.23 atm
A gas at 2.5 atm and 25 °C expands to 750 mL after being cooled to 0 °C and depressurized to 122 kPa. What was the original volume of the gas?
P1 = 2.5 atm T1 = 25°C = 298 K V2 = 750 mL T2 = 0 °C = 273 K P2=122 kPa =1.20 atm V1 = ? Combined Gas Law V1 = 390ml
At 27°C, fluorine occupies a volume of 0.500 L. To what temperature in degrees Celsius should it be lowered to bring the volume to 200.0 mL?
T1 = 27ºC = 300 K V1 = 0.500 L T2 = ?°C V2=200.0 mL =0.2000 L Charle's Law t2=-153*C (120 K) ( the last Zero is underlined
Helium occupies 3.8 L at -45°C. What volume will it occupy at 45°C?
Use Charle's Law V1 = 3.8 L T1 = -45°C = 228 K V2 = ? T2 = 45°C = 318 K V2=5.3L
A gas occupies 1.5 L at 850.0 mm Hg and 15°C. At what pressure will this gas occupy 2.5 L at 30.0°C?
V1 = 1.5 L P1 = 850.0 mm Hg T1 = 15°C = 288 K P2 = ? V2 = 2.5 L T2=30.0°C =303 K Combined Gas Law P2+540 mm Hg
A gas occupies 125 mL at 125 kPa. After being cooled to 75°C and depressurized to 100.0 kPa, it occupies 0.100 L. What was the original temperature of the gas?
V1 = 125 mL P1 = 125 kPa T2 = 75°C = 348 K P2 = 100.0 kPa V2 = 0.100 L = 100 mL T1 = ? Combined Gas Law T1=271*C (544K)
A gas occupies 256 mL at 720 torr and 25°C. What will its volume be at STP?
V1 = 256 mL P1 = 720 torr T1 = 25°C = 298 K V2 = ? P2 = 760.0 torr T2 = 273 K Combined Gas Law V2 = 220 mL
A 3.20-L sample of gas has a pressure of 102 kPa. If the volume is reduced to 0.650 L, what pressure will the gas exert?
V1 = 3.20 L P1 = 102 kPa V2 = 0.650 L P2 = ? Boyle's Law P2+502 kPa
Ammonia gas occupies a volume of 450.0 mL at 720.0 mm Hg. What volume will it occupy at standard pressure?
V1 = 450.0 mL P1 = 720.0 mm Hg V2 = ? P2 = 760.0 mm Hg Boyle's Law v2=425.3ml
A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature?
convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this: 2.85l/298k=x/273K
