Boyle's Sample Problems
Boyle’s Law
1.Example #1: 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)
Answer :
2. 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)
Answer:
3. 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:
4. A sample gas occupies a volum of 225 ml at a pressure of 720.0 torr and a temperature of 20.0 celcius. Calculate the new pressure if the volume is increased to 350.0 mL, at constant temperature.
* P2 = P1V1/V2 = (720.0 torr )(225 mL)/350.0mL
Answer: 463 torr
5. A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a change in temperature, what would be the volume of the balloon?
* Vf = (2.0 L)(3 atm)/(0.5 atm)
Vf = 6 L/0.5
Answer: 12 L * Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? * Answer: convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this: * * Example #2: 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: * * Example #3: 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 (as required by Charles' Law)? * Answer
:
1.Example #1: 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)
Answer :
2. 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)
Answer:
3. 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:
4. A sample gas occupies a volum of 225 ml at a pressure of 720.0 torr and a temperature of 20.0 celcius. Calculate the new pressure if the volume is increased to 350.0 mL, at constant temperature.
* P2 = P1V1/V2 = (720.0 torr )(225 mL)/350.0mL
Answer: 463 torr
5. A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5 atmospheres without a change in temperature, what would be the volume of the balloon?
* Vf = (2.0 L)(3 atm)/(0.5 atm)
Vf = 6 L/0.5
Answer: 12 L * Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its volume at standard temperature? * Answer: convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We plug into our equation like this: * * Example #2: 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: * * Example #3: 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 (as required by Charles' Law)? * Answer
: