Properties of Ideal Gases Worksheet with Explanations and Answers

5. List five properties of gases.

Compressibility: Gases can be compressed easily due to the large spaces between particles.
Expansion: Gases expand to fill the entire volume of their container.
Low Density: Gases have lower density compared to solids and liquids because of the large distances between particles.
Diffusion: Gas particles move freely and mix evenly without external agitation.
Exertion of Pressure: Gas particles exert pressure on the walls of their container due to collisions.

6. Define the Kinetic Molecular Theory.

The Kinetic Molecular Theory states that gases are composed of small particles in constant, random motion. The theory explains the behavior of gases in terms of:

Particle movement.
Energy of particles.
Collisions between particles and container walls.

7. What are the four assumptions that, if met, describe the behavior of Ideal Gases?

Gas particles are in constant, random motion.
Collisions between gas particles and with container walls are perfectly elastic (no energy is lost during collisions).
The volume of gas particles is negligible compared to the volume of the container.
There are no intermolecular forces between gas particles.

8. Define Absolute Zero.

Absolute zero is the lowest theoretical temperature, $0 , \text{K}$ (or $-273.15^\circ , \text{C}$ ), at which the motion of particles theoretically ceases. At this point, a gas would have no kinetic energy.

9. Temperature is defined as the…

…measure of the average kinetic energy of the particles in a substance. It determines how fast the particles are moving on average.

10. A gas at 200 K has four times as much $E_k$ (kinetic energy) as a gas at…

If kinetic energy is proportional to temperature, then:

$T_2 = \frac{T_1}{4}$

where $T_1 = 200 , \text{K}$ .

Thus:

$T_2 = \frac{200}{4} = 50 , \text{K}$ .

Answer: The gas at 200 K has four times as much kinetic energy as a gas at 50 K.

11. Volume is defined as…

…the amount of three-dimensional space occupied by a substance or object. For gases, volume is typically measured in liters ( $L$ ) or cubic meters ( $m^3$ ).

12. The pressure of a gas is caused by the…

…collisions of gas particles with the walls of the container. The frequency and force of these collisions determine the pressure.

13. Explain why, as the temperature of a gas increases, the pressure of the gas will also increase, so long as the volume is held constant.

When the temperature of a gas increases:

The average kinetic energy of gas particles increases.
Particles move faster and collide with the walls of the container more frequently and with greater force.
Since the volume is constant, the increased force and frequency of collisions result in higher pressure.

Example: Imagine 1 mole of $\text{O}_2(g)$ molecules in a sealed, rigid metal canister. As the temperature rises, the particles gain more energy, leading to higher pressure.

14. Does $\text{CO}_2(g)$ behave more like an Ideal Gas at $20^\circ \text{C}$ or $150^\circ \text{C}$ ? Explain your answer.

$\text{CO}_2(g)$ behaves more like an Ideal Gas at $150^\circ \text{C}$ .

At higher temperatures, gas particles have greater kinetic energy, which reduces the effect of intermolecular attractions.
This makes the gas behavior align more closely with the Ideal Gas assumptions.

15. Does $\text{CH}_4(g)$ behave more like an Ideal Gas at $1.5 , \text{atm}$ or $40 , \text{atm}$ ? Explain your answer.

$\text{CH}_4(g)$ behaves more like an Ideal Gas at $1.5 , \text{atm}$ .

At lower pressures, gas particles are farther apart, minimizing the effect of intermolecular forces.
High pressures force particles closer together, causing deviations from Ideal Gas behavior.

16. Why does $\text{H}_2(g)$ behave more like an Ideal Gas than $\text{I}_2(g)$ ?

$\text{H}_2(g)$ behaves more like an Ideal Gas than $\text{I}_2(g)$ because:

Smaller Molecular Size: $\text{H}_2(g)$ molecules are smaller, making their volume negligible compared to the container’s volume.
Weaker Intermolecular Forces: $\text{H}_2(g)$ has weaker van der Waals forces due to its small size and non-polar nature.
In contrast, $\text{I}_2(g)$ has significant intermolecular forces and a larger molecular size, leading to greater deviations from Ideal Gas behavior.

Here are the detailed solutions to your Boyle’s Law questions, using

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17. The relationship that exists between pressure and volume is This means that as one variable increases, the other one will

The relationship is inverse.

According to Boyle’s Law: $P_1 V_1 = P_2 V_2$ , where $P$ is pressure and $V$ is volume.
This means that as one variable increases, the other one decreases, provided the temperature and the amount of gas remain constant.

18. A sample of gas is initially at a pressure of 53 kPa and has a volume of 1.8 L. Determine the new pressure if the volume of the gas decreases to 1.2 L.

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$53 , \text{kPa} \times 1.8 , \text{L} = P_2 \times 1.2 , \text{L}$

Solve for $P_2$ :

$P_2 = \frac{53 \times 1.8}{1.2} = 79.5 , \text{kPa}$

Answer: $P_2 = 80 , \text{kPa}$

19. A sample of gas is initially at a pressure of 1.3 atm and has a volume of 2.3 mL. Determine the new volume if the pressure of the gas decreases to 1.1 atm.

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$1.3 , \text{atm} \times 2.3 , \text{mL} = 1.1 , \text{atm} \times V_2$

Solve for $V_2$ :

$V_2 = \frac{1.3 \times 2.3}{1.1} = 2.7 , \text{mL}$

Answer: $V_2 = 2.7 , \text{mL}$

20. A sample of gas has a volume of 5.0 L at 1.0 atm. Determine the new volume of the gas if the pressure is tripled.

If the pressure is tripled, $P_2 = 3.0 , \text{atm}$ . Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$1.0 , \text{atm} \times 5.0 , \text{L} = 3.0 , \text{atm} \times V_2$

Solve for $V_2$ :

$V_2 = \frac{1.0 \times 5.0}{3.0} = 1.7 , \text{L}$

Answer: $V_2 = 1.7 , \text{L}$

21. What is the new volume if a 1.5 L balloon is increased in pressure from 1.0 atm to 850 mmHg?

Convert $850 , \text{mmHg}$ to atm:

$850 , \text{mmHg} \times \frac{1 , \text{atm}}{760 , \text{mmHg}} = 1.118 , \text{atm}$

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$1.0 , \text{atm} \times 1.5 , \text{L} = 1.118 , \text{atm} \times V_2$

Solve for $V_2$ :

$V_2 = \frac{1.0 \times 1.5}{1.118} = 1.3 , \text{L}$

Answer: $V_2 = 1.3 , \text{L}$

22. A small oxygen canister contains 110 mL of oxygen gas at a pressure of 3.0 atm. All of the oxygen is released into a balloon with a final pressure of 2.0 atm.

a. Explain, using the kinetic molecular theory, if the final volume of the balloon will be larger or smaller than its initial volume.

According to the kinetic molecular theory:

A decrease in pressure allows gas particles to spread out more, leading to an increase in volume.
Therefore, the final volume of the balloon will be larger than its initial volume.

b. Calculate the final volume of the balloon.

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$3.0 , \text{atm} \times 110 , \text{mL} = 2.0 , \text{atm} \times V_2$

Solve for $V_2$ :

$V_2 = \frac{3.0 \times 110}{2.0} = 165 , \text{mL} = 0.165 , \text{L}$

Answer: $V_2 = 0.17 , \text{L}$

23. A diving bell contains 32.0 kL of air at a pressure of 98 kPa at the surface of the lake. About 5 m below the surface, the volume of air trapped inside the bell is 21 kL. What is the pressure of the air in the bell, assuming temperature remains constant?

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$98 , \text{kPa} \times 32.0 , \text{kL} = P_2 \times 21.0 , \text{kL}$

Solve for $P_2$ :

$P_2 = \frac{98 \times 32.0}{21.0} = 149.33 , \text{kPa}$

Answer: $P_2 = 1.5 \times 10^2 , \text{kPa}$

24. A 5.0 L air sample is at a pressure of 760 mmHg. What is the new volume if the pressure is increased to 800 kPa?

Convert $760 , \text{mmHg}$ to kPa:

$760 , \text{mmHg} \times \frac{101.325 , \text{kPa}}{760 , \text{mmHg}} = 101.325 , \text{kPa}$

Using Boyle’s Law:

$P_1 V_1 = P_2 V_2$

Substitute the known values:

$101.325 , \text{kPa} \times 5.0 , \text{L} = 800 , \text{kPa} \times V_2$

Solve for $V_2$ :

$V_2 = \frac{101.325 \times 5.0}{800} = 0.633 , \text{L}$

Answer: $V_2 = 0.63 , \text{L}$

2. Pressure Unit Conversions

a. Convert 3.21 atm into units of kPa.

1 atm = 101.325 kPa.

$3.21 , \text{atm} \times 101.325 , \text{kPa} = 325.25 , \text{kPa}$

Answer: $3.21 , \text{atm} = 325.25 , \text{kPa}$

b. Convert 0.78 bar into units of kPa.

1 bar = 100 kPa.

$0.78 , \text{bar} \times 100 , \text{kPa} = 78 , \text{kPa}$

Answer: $0.78 , \text{bar} = 78 , \text{kPa}$

c. Convert 670 mm Hg into units of torr.

1 mm Hg = 1 torr (by definition).

$670 , \text{mm Hg} = 670 , \text{torr}$

Answer: $670 , \text{mm Hg} = 670 , \text{torr}$

d. Convert 167 kPa into units of atm.

1 atm = 101.325 kPa.

$167 , \text{kPa} \times \frac{1 , \text{atm}}{101.325 , \text{kPa}} = 1.65 , \text{atm}$

Answer: $167 , \text{kPa} = 1.65 , \text{atm}$

e. Convert 3.27 psi into mm Hg.

1 psi = 51.715 mm Hg.

$3.27 , \text{psi} \times 51.715 , \text{mm Hg} = 169.1 , \text{mm Hg}$

Answer: $3.27 , \text{psi} = 169.1 , \text{mm Hg}$

f. Convert 250 mm Hg into units of kPa.

1 mm Hg = 0.133322 kPa.

$250 , \text{mm Hg} \times 0.133322 , \text{kPa} = 33.33 , \text{kPa}$

Answer: $250 , \text{mm Hg} = 33.33 , \text{kPa}$

3. Volume Unit Conversions

a. Convert 347 mL into litres.

1 mL = 0.001 L.

$347 , \text{mL} \times 0.001 , \text{L/mL} = 0.347 , \text{L}$

Answer: $347 , \text{mL} = 0.347 , \text{L}$

b. Convert 0.0926 L into mL.

1 L = 1000 mL.

$0.0926 , \text{L} \times 1000 , \text{mL/L} = 92.6 , \text{mL}$

Answer: $0.0926 , \text{L} = 92.6 , \text{mL}$

c. Convert $3 \times 10^{-3}$ L into mL.

$1 , \text{L} = 1000 , \text{mL}$ .

$3 \times 10^{-3} , \text{L} \times 1000 , \text{mL/L} = 3.0 , \text{mL}$

Answer: $3 \times 10^{-3} , \text{L} = 3.0 , \text{mL}$

d. Convert 12 mL into litres.

$1 , \text{mL} = 0.001 , \text{L}$ .

$12 , \text{mL} \times 0.001 , \text{L/mL} = 0.012 , \text{L}$

Answer: $12 , \text{mL} = 0.012 , \text{L}$

4. Pressure Unit Conversion Table

4. Pressure Unit Conversion Table (Using Fractions)

Pressure Unit	Equivalent in kPa	Equivalent in atm	Equivalent in mm Hg	Equivalent in torr	Equivalent in bar
1 kPa	1	$\frac{1}{101.325}$	$\frac{760}{101.325}$	$\frac{760}{101.325}$	$\frac{1}{100}$
1 atm	101.325	1	760	760	$\frac{101.325}{100}$
1 mm Hg	$\frac{1}{7.5}$	$\frac{1}{760}$	1	1	$\frac{1}{750}$
1 torr	$\frac{1}{7.5}$	$\frac{1}{760}$	1	1	$\frac{1}{750}$
1 bar	100	$\frac{1}{1.01325}$	$\frac{750}{1.01325}$	$\frac{750}{1.01325}$	1

Explanation of Key Fractions:

1 kPa:
- Converts to atm as $\frac{1}{101.325}$ since 1 atm = 101.325 kPa.
- Converts to mm Hg or torr as $\frac{760}{101.325}$ .
- Converts to bar as $\frac{1}{100}$ .
1 atm:
- Converts to kPa as 101.325.
- Converts to mm Hg or torr as 760.
1 mm Hg (or 1 torr):
- Converts to kPa as $\frac{1}{7.5}$ since 7.5 mm Hg = 1 kPa.