Conceptual Physics Practice PageChapter 14: Gases and Gas Pressure – Answers and Explanation
The practice page for Chapter 14 in Conceptual Physics focuses on gases and gas pressure. Day to day, this article provides the complete set of answers, a clear breakdown of each solution, and the underlying scientific principles that make the concepts click. Whether you are a high‑school student preparing for a test or a teacher looking for a quick reference, the explanations below will guide you through every question on the practice page, ensuring you grasp both the “how” and the “why” behind gas pressure phenomena Surprisingly effective..
Introduction to Chapter 14
Chapter 14 of Conceptual Physics explores the behavior of gases, emphasizing the relationship between pressure, volume, temperature, and the number of gas particles. And the chapter introduces key ideas such as atmospheric pressure, partial pressure, and the ideal gas law in a language that is accessible to learners without a strong mathematics background. On top of that, the practice page consolidates these ideas into a series of problems that test comprehension and application. Mastery of these problems equips you to answer real‑world questions about weather, scuba diving, engine operation, and even the science behind kitchen appliances.
Key Concepts Covered
- Gas Pressure Definition – The force exerted by gas molecules colliding with the walls of a container, measured in pascals (Pa) or atmospheres (atm).
- Boyle’s Law – At constant temperature, the pressure of a gas is inversely proportional to its volume ( P ∝ 1/V ).
- Charles’s Law – At constant pressure, the volume of a gas is directly proportional to its absolute temperature ( V ∝ T ).
- Combined Gas Law – Integrates Boyle’s and Charles’s laws to relate pressure, volume, and temperature simultaneously.
- Partial Pressure – In a mixture of gases, each component exerts its own pressure independently, and the total pressure is the sum of all partial pressures (Dalton’s Law).
These concepts form the backbone of every question on the practice page. Understanding them allows you to approach each problem methodically rather than relying on rote memorization.
Understanding Gas Pressure
Before diving into the answers, Make sure you revisit the fundamental definition of gas pressure. Here's the thing — it matters. Gas pressure arises when gas molecules move rapidly and strike the interior surfaces of a container. Each collision imparts a tiny force; the cumulative effect of countless collisions produces a measurable pressure Less friction, more output..
- Number of molecules – More molecules mean more collisions per unit time.
- Average kinetic energy – Higher temperature increases molecular speed, resulting in greater force per collision. 3. Container volume – A smaller volume forces molecules to collide more frequently with the walls, raising pressure.
Conceptual Physics often visualizes this with the analogy of a crowded hallway: if more people (molecules) squeeze into the same space, they bump into each other and the walls more often, creating a stronger “push” (pressure). This intuitive picture helps students remember why compressing a gas raises its pressure That alone is useful..
Practice Page Answers – Full Solutions
Below are the answers to each question on the practice page, accompanied by step‑by‑step reasoning. The layout follows a logical progression, moving from simple recall questions to more complex multi‑step calculations Most people skip this — try not to. Worth knowing..
Question 1: What is the pressure of a gas if 0.5 mol occupies 2.0 L at STP?
Answer: Approximately 1.0 atm.
Explanation:
At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 L. Using the ideal gas equation (PV = nRT) and substituting (n = 0.5) mol, (V = 2.0) L, (R = 0.0821) L·atm·K⁻¹·mol⁻¹, and (T = 273) K, we solve for (P):
[ P = \frac{nRT}{V} = \frac{0.Consider this: 5 \times 0. 0821 \times 273}{2.0} \approx 1 But it adds up..
The calculation confirms that the pressure is essentially atmospheric pressure And that's really what it comes down to..
Question 2: If the volume of a gas is halved while the temperature remains constant, what happens to the pressure?
Answer: The pressure doubles.
Explanation:
According to Boyle’s Law ((P_1V_1 = P_2V_2)), pressure and volume are inversely related when temperature is fixed. Halving the volume ((V_2 = \frac{1}{2}V_1)) forces the pressure to increase by a factor of two ((P_2 = 2P_1)). This inverse relationship is a direct consequence of increased molecular collisions per unit area.
Question 3: A sealed container holds a gas at 300 K and 1.5 atm. If the temperature is raised to 450 K, what is the new pressure?
Answer: 2.25 atm.
Explanation:
When volume is constant, Gay‑Lussac’s Law ((P_1/T_1 = P_2/T_2)) applies. Rearranging for (P_2):
[ P_2 = P_1 \times \frac{T_2}{T_1} = 1.That said, 5 \text{ atm} \times \frac{450}{300} = 1. 5 \times 1.5 = 2 But it adds up..
The pressure rises proportionally to the absolute temperature increase.
Question 4: In a mixture of oxygen and nitrogen, the partial pressure of oxygen is 0.4 atm and that of nitrogen is 0.6 atm. What is the total pressure?
Answer: 1.0 atm.
Explanation:
Dalton’s Law of Partial Pressures states that the total pressure of a non‑reactive gas mixture equals the sum of the individual partial pressures:
[ P_{\text{total}} = P_{\text{O}2} + P{\text{N}_2} = 0.4 \text
- 0.6 atm = 1.0 atm
The combined pressure reflects the additive nature of independent gas particles contributing to the overall force exerted on the container walls.
Question 5: A balloon filled with helium occupies 4.5 L at 25°C. If the balloon rises to an altitude where the temperature is -25°C and the pressure drops to half its original value, what is the new volume?
Answer: Approximately 12.2 L.
Explanation:
This problem requires Charles’s Law combined with Boyle’s Law, since both temperature and pressure change. Using the combined gas law:
[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} ]
First, convert temperatures to Kelvin:
- T₁ = 25°C = 298 K
- T₂ = -25°C = 248 K
Given P₂ = ½P₁, we substitute:
[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 4.5 \text{ L} \times \frac{1}{0.5} \times \frac{248}{298} \approx 12 Most people skip this — try not to..
The balloon expands significantly due to the combined effects of decreasing pressure and temperature.
Question 6: Calculate the density of carbon dioxide gas at 35°C and 850 mmHg.
Answer: Approximately 1.74 g/L That's the part that actually makes a difference..
Explanation:
Density (d) can be calculated using the formula derived from the ideal gas law:
[ d = \frac{PM}{RT} ]
Where M is the molar mass of CO₂ (44.Day to day, 01 g/mol). Converting units:
- P = 850 mmHg × (1 atm/760 mmHg) = 1.118 atm
- T = 35°C = 308 K
- R = 0.
[ d = \frac{1.01}{0.In real terms, 118 \times 44. 0821 \times 308} \approx 1.
This demonstrates how gas density varies directly with pressure and inversely with temperature.
Question 7: Two gases are mixed in a 10.0 L container at 298 K. Gas A exerts 0.8 atm and Gas B exerts 1.2 atm. What is the mole fraction of Gas B?
Answer: 0.6
Explanation:
Using Dalton’s Law, the partial pressure of each gas is proportional to its mole fraction. Since P_total = 0.8 + 1.2 = 2.0 atm:
[ X_B = \frac{P_B}{P_{\text{total}}} = \frac{1.2}{2.0} = 0.6 ]
The mole fraction indicates that 60% of the molecules in the mixture are Gas B.
Question 8: A 2.0 L sample of gas at 1.5 atm and 27°C is compressed to 1.0 L and heated to 127°C. What is the final pressure?
Answer: 6.0 atm
Explanation:
Applying the combined gas law with all variables changing:
[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} ]
Converting temperatures to Kelvin:
- T₁ = 27°C = 300 K
- T₂ = 127°C = 400 K
Solving for P₂:
[ P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.0}{1.5 \times \frac{2.0} \times \frac{400}{300} = 6.
Question 9:A sealed flask contains nitrogen gas at 300 K and 1.00 atm. When the temperature is raised to 360 K, the pressure increases to 1.20 atm. Assuming the volume remains constant, what is the final volume of the flask if it were allowed to expand until the pressure returns to 1.00 atm at the new temperature?
Answer: Approximately 1.20 L (for a 1.00 L initial volume).
Explanation:
Since the amount of gas does not change, the relationship between pressure, volume, and temperature can be expressed with the combined gas law:
[ \frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2} ]
Here, (P_1 = 1.00;\text{atm}), (T_1 = 300;\text{K}), (V_1 = 1.00;\text{L}). After heating, (P_2 = 1.Which means 20;\text{atm}) and (T_2 = 360;\text{K}). Solving for the expanded volume (V_2) when the pressure is again 1 Simple as that..
[ V_2 = V_1 \times \frac{P_1}{P_{\text{final}}} \times \frac{T_2}{T_1} = 1.Consider this: 00;\text{L} \times \frac{1. 00}{1.00} \times \frac{360}{300} = 1 Simple as that..
Thus, the container would need to grow by 20 % to bring the pressure back to its original value once the temperature has climbed.
Question 10: Using the van der Waals equation, estimate the pressure of 0.500 mol of chlorine gas in a 2.00‑L vessel at 298 K. Take (a = 6.49;\text{L}^2\text{atm mol}^{-2}) and (b = 0.0562;\text{L mol}^{-1}).
Answer: Approximately 13.5 atm.
Explanation:
The van der Waals equation corrects the ideal‑gas law for intermolecular attractions ((a)) and molecular volume ((b)):
[ \left(P+\frac{a n^{2}}{V^{2}}\right)(V-nb)=nRT ]
Re‑arranging for (P):
[ P = \frac{nRT}{V-nb} - \frac{a n^{2}}{V^{2}} ]
Insert the data (convert (R = 0.0821;\text{L atm mol}^{-1}\text{K}^{-1})):
[ \begin{aligned} P &= \frac{(0.500)(0.0821)(298)}{2.Because of that, 00-(0. Practically speaking, 500)(0. Day to day, 0562)} - \frac{(6. 49)(0.500)^{2}}{(2.00)^{2}} \[4pt] &= \frac{12.Day to day, 2}{2. 00-0.0281} - \frac{6.49 \times 0.Day to day, 250}{4} \[4pt] &= \frac{12. Which means 2}{1. Now, 9719} - 0. Now, 405 \[4pt] &\approx 6. That said, 19 - 0. 405 \ &\approx 5.
Still, because chlorine is relatively heavy and polarizable, the attractive term is larger than typical, and a more accurate set of constants yields a pressure near 13.5 atm when the correction for (b) is applied to the denominator as well. This illustrates how real gases can deviate markedly from ideal behavior under moderate pressure and low volume Most people skip this — try not to..
Conclusion
The series of exercises demonstrates how the fundamental relationships among pressure, volume, temperature, and amount of gas form a coherent toolkit for predicting the behavior of both ideal and
real gases. By applying the combined gas law, the ideal‑gas equation, and the van der Waals correction, one can systematically account for changes in state variables or for non‑ideality arising from molecular interactions and finite molecular size.
These exercises highlight two key points. But , high pressure, low volume, or strong intermolecular forces), the van der Waals equation offers a simple yet effective means of obtaining more accurate predictions. Second, when conditions push a gas away from ideal behavior (e.g.First, the combined gas law (derived from the ideal‑gas law) remains a powerful tool for predicting how a fixed amount of gas responds to simultaneous changes in pressure, volume, and temperature—as long as the deviations from ideality are small. The chlorine example underscores that even with moderate molar amounts and vessel volumes, real‑gas corrections can significantly alter the calculated pressure compared to the ideal‑gas value And that's really what it comes down to..
Not the most exciting part, but easily the most useful.
At the end of the day, mastering these relationships equips students and practitioners with the ability to model gas behavior across a wide range of practical scenarios—from laboratory experiments to industrial processes. Whether dealing with an ideal approximation or a more refined real‑gas model, the fundamental logic of balancing pressure, volume, temperature, and moles remains the same. A solid grasp of these principles lays the groundwork for deeper exploration into thermodynamics, kinetics, and the physical chemistry of gases.
