# Boyle's Law

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## Boyle's Law Answer Key

**Instructions:**
Read each question carefully. Choose the answer that best fits the question. Short answer response questions must be responded to in complete sentences. If the question involves calculations, you must show all your math work.

1.

A gas sample has a volume of 450 ml when it is under a pressure of 780 torr. What will be the volume of the same gas sample if the pressure is increased to 850 torr?

- 413 ml

2.

A gas has a volume of 350 cc at 740 mmHg. How many cubic centimeters will the gas occupy at a pressure of 900 mmHG if the temperature remains constant?

- 288 mmHg

3.

A gas is compressed at constant temperature from a volume of 540 ml to 320 ml. If the initial pressure was 475 mmHg, what is the final pressure in mmHg?

- 802 mmHg

4.

Under constant temperature, the pressure of a gas is inversely proportional to the volume, according to which gas law?

- Charles' Law
- Boyle's Law
- Gay-Lusaac's Law
- Combined Gas Law

5.

According to Boyle's Law, in a gas at a constant temperature, the change in pressure is inversely proportional to to the change in volume. This relationship can be expressed as an equation, [math]P_1//P_2=V_2//V_1[/math]. If 250mL of a gas, at a pressure of 214mmHg and a constant temperature, is forced into a 105mL rigid container, what will the final pressure be in mmHg?

- [math]P_1V_1=P_2V_2[/math]

[math]214mmHg*0.250L=P_2*0.105L[/math]

[math]P_2=(53.5L*mmHg)/(0.105L)[/math]

[math]P_2=509.52mmHg[/math]

6.

Use Boyle's Law, [math]P_1V_1 = P_2V_2[/math], to solve the following problems.

A.

A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters?

- [math]P_1V_1 = P_2V_2[/math]

[math](125kPa)(3.50L)=(P_2)(2.00L)[/math]

[math]P_2=((125kPa)(3.50L))/(2.00L)=218.75 kPa[/math]

B.

A 2.0 liter container of nitrogen had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm?

- [math]P_1V_1 = P_2V_2[/math]

[math](3.2atm)(2.0L)=(1.0atm)V_2[/math]

[math]P_2=((3.2atm)(2.0L))/(1.0atm)=6.4L[/math]

C.

A 175 mL sample of neon had its pressure changed from 75 kPa to 150 kPa. What is its new volume?

- [math]P_1V_1 = P_2V_2[/math]

[math](75kPa)(0.175L)=(150kPa)V_2[/math]

[math]V_2=((75kPa)(0.175L))/(150kPa)=0.0875L[/math]

D.

A sample of oxygen gas occupies a volume of 250.0mL at 740.0 torr pressure. What volume will it occupy at 800.0 torr pressure?

- [math]P_1V_1 = P_2V_2[/math]

[math]V_2=(P_1V_1)/(P_2)=((740.0"torr")(0.250L))/(800.0"torr")=0.23125L[/math]

E.

A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing a new volume of 750 mL. What was its original volume?

- [math]P_1V_1 = P_2V_2[/math]

[math]V_1=(P_2V_2)/P_1=((0.50atm)(0.750L))/(1.5atm)=0.25L[/math]

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