Gas Laws and Properties

Gas Laws and Kinetic Molecular Theory

Multiple Choice Questions

Question 1: Pressure Change with Added Gas

• Scenario: 4 moles of gas are added to a container already holding 1 mole of gas. Volume remains constant.
• Question: How does the pressure change?
• Options:
  • a. The final pressure will be four times higher than the initial pressure.
  • b. The final pressure will be double the initial pressure.
  • c. The final pressure will be five times greater than the initial pressure.
  • d. The final pressure will be the same as the initial pressure.
• Correct Answer: c. The final pressure will be five times greater than the initial pressure.
• Explanation: According to the ideal gas law, PV = nRT. If the volume V, the gas constant R, and the temperature T are constant, then the pressure P is directly proportional to the number of moles n. If the number of moles increases from 1 to 5 (1 + 4), the pressure will increase by a factor of 5.

Question 2: Partial Pressure of Oxygen and Temperature

• Question: What happens to the partial pressure of oxygen in a sample of air when the temperature is increased?
• Options:
  • a. It increases.
  • b. It stays the same.
  • c. It decreases.
  • d. The change cannot be determined.
• Correct Answer: a. It increases.
• Explanation: According to the ideal gas law, PV = nRT. If the number of moles n and the volume V are constant, then the pressure P is directly proportional to the temperature T. Therefore, as temperature increases, the partial pressure of oxygen increases.

Question 3: Boyle's Law Application

• Question: Boyle's Law can be used for which of the following?
• Options:
  • a. Predicting the expected volumes of two party balloons.
  • b. Predicting the relative pressures inside a hot air balloon.
  • c. Predicting the change in volume of an inflatable toy from summer to winter.
  • d. Predicting the height of a mercury barometer column in a low pressure system.
• Correct Answer: c. Predicting the change in volume of an inflatable toy from summer to winter.
• Explanation: Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional (P1V1 = P2V2). This law is best applied to scenarios where the amount of gas and temperature are constant, such as predicting the volume change of an inflatable toy due to temperature-induced pressure changes from summer to winter.

Question 4: Volume Change with Temperature

• Scenario: A sample of argon gas occupies 50 L at standard temperature.
• Condition: Constant pressure.
• Question: What volume will the gas occupy if the temperature is doubled?
• Options:
  • a. 25 L
  • b. 50 L
  • c. 75 L
  • d. 100 L
• Correct Answer: d. 100 L
• Explanation: According to Charles's Law, at constant pressure, the volume of a gas is directly proportional to its temperature (V1/T1 = V2/T2). If the temperature is doubled, the volume will also double. Hence, if the initial volume is 50 L, the new volume will be 100 L.

Question 5: Kinetic Molecular Theory Exceptions

• Question: All of the following statements underline the kinetic molecular theory of gases EXCEPT:
• Options:
  • a. Gas molecules have no intermolecular forces.
  • b. The average kinetic energy is proportional to the temperature of the gas in ºC.
  • c. The collisions between gas particles are elastic.
  • d. Gas particles are in constant random motion.
• Correct Answer: b. The average kinetic energy is proportional to the temperature of the gas in ºC.
• Explanation: The kinetic molecular theory states that the average kinetic energy of gas molecules is proportional to the absolute temperature (in Kelvin), not Celsius. The other statements are fundamental postulates of the kinetic molecular theory.

Question 6: Cause of Observable Pressure

• Question: The random collisions of molecules upon a surface are the cause of the observable:
• Options:
  • a. Temperature of
  • b. Pressure on
  • c. Volume of
  • d. Condensation on
• Correct Answer: b. Pressure on
• Explanation: Pressure is defined as the force exerted per unit area. In gases, this force is due to the countless collisions of gas molecules against the walls of the container.

Question 7: Definition of Diffusion

• Question: What is diffusion?
• Options:
  • a. the tendency of particles to move from areas of higher concentration to areas of lower concentration
  • b. the rate at which a gas escapes through a pinhole into a vacuum
  • c. the amount of time it takes for a gas to fill a room
  • d. the rate at which a vapor disperses through a vacuum
• Correct Answer: a. the tendency of particles to move from areas of higher concentration to areas of lower concentration
• Explanation: Diffusion is the process by which particles spread out from areas of high concentration to areas of low concentration due to their random motion. Option b describes effusion.

Question 8: Perfume Smell Explanation

• Scenario: A student opens a bottle of perfume in the back of the classroom, and the teacher in the front can smell it a few seconds later.
• Question: This can be explained in terms of:
• Options:
  • a. diffusion only
  • b. evaporation only
  • c. both evaporation and diffusion
  • d. neither evaporation nor diffusion
• Correct Answer: c. both evaporation and diffusion
• Explanation: First, the perfume evaporates, turning from a liquid to a gas. Then, the gas molecules diffuse throughout the room, moving from an area of higher concentration (near the bottle) to an area of lower concentration (the front of the room).

Question 9: Reason for Gas Diffusion

• Question: Gases diffuse throughout available space because of the _ motion of molecules.
• Options:
  • a. Random
  • b. Slow
  • c. Quick
  • d. Calculated
• Correct Answer: a. Random
• Explanation: Gas molecules are in constant, random motion. This random motion causes them to spread out and fill any available space, leading to diffusion.

Question 10: Volume Increase Effect on Temperature and Pressure

• Scenario: An ideal gas is in a container, and the volume is increased.
• Question: What would be the effect on the temperature and pressure of the gas?
• Options:
  • a. Both temperature and pressure increase.
  • b. temperature increases and pressure decreases
  • c. temperature decreases and pressure increases
  • d. Both temperature and pressure decrease.
• Correct Answer: d. Both temperature and pressure decrease.
• Explanation: If the volume of a container holding an ideal gas is increased without adding energy (i.e., adiabatically), the gas will expand and do work against the external pressure. This expansion causes the internal energy of the gas (and thus its temperature) to decrease. Furthermore, increasing the volume at a constant number of moles and decreasing temperature will result in decreased pressure.

SHOW YOUR WORK Problems

Problem 11: Volume Change with Pressure Change

• Initial conditions:
  • V_1 = 425 \, \text{mL}
  • P_1 = 515 \, \text{mm Hg}
• Final condition:
  • P_2 = 130 \, \text{kPa}
• Equation: Boyle's Law: P1V1 = P2V2
• Conversion: Convert P_2 to mm Hg: 130 \, \text{kPa}

\times \frac{760 \, \text{mm Hg}}{101.325 \, \text{kPa}} = 975.9 \, \text{mm Hg}

• Calculation:
  • V2 = \frac{P1V1}{P2} = \frac{515 \, \text{mm Hg} \times 425 \, \text{mL}}{975.9 \, \text{mm Hg}} = 224.4 \, \text{mL}
• Answer: V_2 = 224 \, \text{mL} (with correct significant figures)

Problem 12: Volume Change with Temperature and Pressure Change

• Initial conditions:
  • V_1 = 3.17 \, \text{L}
  • P_1 = 1 \, \text{atm} (standard pressure)
  • T_1 = 45 \, ^\circ\text{C} = 318.15 \, \text{K}
• Final conditions:
  • P_2 = 1.4 \, \text{atm}
  • T_2 = 0 \, ^\circ\text{C} = 273.15 \, \text{K} (standard temperature)
• Equation: Combined Gas Law: \frac{P1V1}{T1} = \frac{P2V2}{T2}
• Calculation:
  • V2 = \frac{P1V1T2}{P2T1} = \frac{1 \, \text{atm} \times 3.17 \, \text{L} \times 273.15 \, \text{K}}{1.4 \, \text{atm} \times 318.15 \, \text{K}} = 1.93 \, \text{L}
• Answer: V_2 = 1.93 \, \text{L} (with correct significant figures)

Problem 13: Pressure Change with Temperature Change

• Initial conditions:
  • P_1 = 1.31 \, \text{atm}
  • T_1 = 515 \, \text{K}
• Final condition:
  • T_2 = 850 \, \text{K}
• Condition: Volume is constant.
• Equation: Gay-Lussac's Law: \frac{P1}{T1} = \frac{P2}{T2}
• Calculation:
  • P2 = \frac{P1T2}{T1} = \frac{1.31 \, \text{atm} \times 850 \, \text{K}}{515 \, \text{K}} = 2.16 \, \text{atm}
• Answer: P_2 = 2.16 \, \text{atm} (with correct significant figures)

Problem 14: Volume Change with Temperature Change (Balloon)

• Initial conditions:
  • V_1 = 12.0 \, \text{L}
  • T_1 = 25.0 \, ^\circ\text{C} = 298.15 \, \text{K}
• Final condition:
  • T_2 = 50.0 \, ^\circ\text{C} = 323.15 \, \text{K}
• Condition: Pressure does not change.
• Equation: Charles's Law: \frac{V1}{T1} = \frac{V2}{T2}
• Calculation:
  • V2 = \frac{V1T2}{T1} = \frac{12.0 \, \text{L} \times 323.15 \, \text{K}}{298.15 \, \text{K}} = 13.0 \, \text{L}
• Answer: V_2 = 13.0 \, \text{L} (with correct significant figures)