Comprehensive Study Notes on Gas Laws: Boyle, Charles, Gay-Lussac, and Combined Law

Core Concepts and Vocabulary of Gas Laws

• Focus Question: How are a gas's temperature, pressure, and volume related?
• Scientific Law: A relationship in nature that is supported by many experiments.
• Absolute Zero: Zero on the Kelvin scale ($0\,K$). It is the lowest possible temperature where nothing could be colder, and no heat energy remains in a substance. Theoretically, the motion of atoms and molecules stops completely at this point, and they exist in their lowest possible energy state.     * Equivalent temperatures: $-273.15^{\circ}C$ or $-459.67^{\circ}F$.
• New Vocabulary List:     * Absolute zero     * Boyle's law     * Charles's law     * Gay-Lussac's law     * Combined gas law

Boyle’s Law: Pressure and Volume Relationship

• Definition: Boyle’s law states that the volume of a fixed amount of gas held at a constant temperature varies inversely with the pressure.
• Mathematical Formula:     * $P_1V_1 = P_2V_2$     * Where $P$ represents pressure and $V$ represents volume.
• Graphical Representation: In a pressure-volume relationship, plotting volume versus pressure results in a downward curve, indicating an inversely proportional relationship.
• Experimental Evidence:     * Condition 1: $(1.0\,atm, 10\,L)$ at $25.0^{\circ}C$. Product: $P_1V_1 = (1\,atm) \times (10\,L) = 10\,atm\cdot L$ (constant).     * Condition 2: $(2.0\,atm, 5\,L)$ at $25.0^{\circ}C$. Product: $P_2V_2 = (2\,atm) \times (5\,L) = 10\,atm\cdot L$ (constant).     * Condition 3: $(4.0\,atm, 2.5\,L)$ at $25.0^{\circ}C$. Product: $P_3V_3 = (4\,atm) \times (2.5\,L) = 10\,atm\cdot L$ (constant).

Boyle’s Law: Problem Solving and Practice

• Example Problem 1: Bubble Volume Underwater     * Problem: A diver blows a $0.75\,L$ air bubble $10\,m$ underwater. As it rises to the surface, the pressure changes from $2.25\,atm$ to $1.03\,atm$. What will be the volume of air in the bubble at the surface?     * Analyze: A decrease in pressure must result in an increase in volume.     * Known variables: $V_1 = 0.75\,L$, $P_1 = 2.25\,atm$, $P_2 = 1.03\,atm$.     * Solution: $V_2 = V_1 \times \frac{P_1}{P_2} = 0.75\,L \times \frac{2.25\,atm}{1.03\,atm} = 1.6\,L$.     * Evaluation: The pressure decreased by roughly half ($2.25$ to $1.03$), so the volume roughly doubled ($0.75$ to $1.6$).

• Practice Problem 1: The volume of a gas at $99.0\,kPa$ is $300.0\,mL$. If pressure increases to $188\,kPa$, what is the new volume?     * $V_2 = \frac{V_1P_1}{P_2} = \frac{300.0\,mL \times 99.0\,kPa}{188\,kPa} = 158\,mL$

• Practice Problem 2: Pressure of helium in a $1.00\,L$ container is $0.988\,atm$. What is the new pressure in a $2.00\,L$ container?     * $P_2 = \frac{V_1P_1}{V_2} = \frac{1.00\,L \times 0.988\,atm}{2.00\,L} = 0.494\,atm$

• Practice Problem 3 (Challenge): Air in a piston occupies $145.7\,mL$ at $1.08\,atm$. What is the volume when pressure increases by $25\%$?     * $P_2 = 1.08\,atm \times 1.25 = 1.35\,atm$     * $V_2 = \frac{V_1P_1}{P_2} = \frac{145.7\,mL \times 1.08\,atm}{1.35\,atm} = 117\,mL$

Charles’s Law: Temperature and Volume Relationship

• Definition: Charles’s law states that the volume of a given amount of gas is directly proportional to its Kelvin temperature at constant pressure.
• Direct Proportionality: As temperature increases, so does the volume of a gas sample if amount and pressure are constant.
• Mathematical Formula:     * $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
• Absolute Zero and Graphing:     * A temperature of $0\,K$ corresponds to $0\,mL$.     * Doubling the temperature doubles the volume.     * Absolute zero is the theoretical point where molecular motion stops.

Charles’s Law: Problem Solving and Practice

• Example Problem 2: Helium Balloon in a Car     * Problem: A balloon occupies $2.32\,L$ at $40.0^{\circ}C$. If the car temperature rises to $75.0^{\circ}C$, what is the new volume?     * Step 1: Temperature Conversion: $T_1 = 40.0^{\circ}C + 273 = 313.0\,K$; $T_2 = 75.0^{\circ}C + 273 = 348.0\,K$.     * Step 2: Solve: $V_2 = V_1 \times \frac{T_2}{T_1} = 2.32\,L \times \frac{348.0\,K}{313.0\,K} = 2.58\,L$.

• Practice Problem 4: What volume will gas in a balloon ($4.3\,L$ at $350\,K$) occupy at $250\,K$?     * $V_2 = \frac{V_1T_2}{T_1} = \frac{4.3\,L \times 250\,K}{350\,K} = 3.1\,L$

• Practice Problem 5: Gas at $89^{\circ}C$ occupies $0.67\,L$. At what Celsius temperature will volume increase to $1.12\,L$?     * $T_1 = 89^{\circ}C + 273 = 362\,K$     * $T_2 = \frac{T_1V_2}{V_1} = \frac{362\,K \times 1.12\,L}{0.67\,L} = 605\,K$     * $T_{Celsius} = 605 - 273 = 332^{\circ}C$ (Rounded to $330^{\circ}C$ for sig figs).

• Practice Problem 6: A $3.00\,L$ sample lowered from $80.0^{\circ}C$ to $30.0^{\circ}C$. Resulting volume?     * $T_1 = 353\,K$, $T_2 = 303\,K$     * $V_2 = \frac{3.00\,L \times 303\,K}{353\,K} = 2.58\,L$

• Practice Problem 7 (Challenge): Gas occupies $0.67\,L$ at $350\,K$. What temperature is required to reduce volume by $45\%$?     * $V_2 = 0.67\,L - (0.67\,L \times 0.45) = 0.37\,L$     * $T_2 = \frac{350\,K \times 0.37\,L}{0.67\,L} = 190\,K$

Gay-Lussac’s Law: Temperature and Pressure Relationship

• Definition: Gay-Lussac’s law states that the pressure of a fixed amount of gas varies directly with the Kelvin temperature when volume remains constant.
• Mathematical Formula:     * $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
• Experimental Example:     * Condition A: $1.5\,atm$ at $150\,K$. Ratio: $\frac{1.5}{150} = 0.01\,atm/K$.     * Condition B: $3.0\,atm$ at $300\,K$. Ratio: $\frac{3.0}{300} = 0.01\,atm/K$.

Gay-Lussac’s Law: Problem Solving and Practice

• Example Problem 3: Oxygen Canister on Mt. Everest     * Problem: Oxygen pressure is $5.00\,atm$ at $25.0^{\circ}C$. Temperature falls to $-10.0^{\circ}C$. What is the new pressure?     * Temperature Conversion: $T_1 = 298.0\,K$; $T_2 = 263.0\,K$.     * Solution: $P_2 = P_1 \times \frac{T_2}{T_1} = 5.00\,atm \times \frac{263.0\,K}{298.0\,K} = 4.41\,atm$.

• Practice Problem 8: Tire pressure is $1.88\,atm$ at $25.0^{\circ}C$. Pressure at $37.0^{\circ}C$?     * $T_1 = 298\,K$, $T_2 = 310\,K$     * $P_2 = \frac{1.88\,atm \times 310\,K}{298\,K} = 1.96\,atm$

• Practice Problem 9: Helium in $2.00\,L$ cylinder at $1.12\,atm$. At $36.5^{\circ}C$, pressure is $2.56\,atm$. Initial temperature in Celsius?     * $T_2 = 36.5^{\circ}C + 273 = 309.5\,K$     * $T_1 = \frac{T_2P_1}{P_2} = \frac{309.5\,K \times 1.12\,atm}{2.56\,atm} = 135\,K$     * $T_{Celsius} = 135 - 273 = -138^{\circ}C$

• Practice Problem 10 (Challenge): Pressure is $30.7\,kPa$ at $0.00^{\circ}C$. By how many degrees Celsius must it increase to double the pressure?     * $T_1 = 273\,K$, $P_2 = 61.4\,kPa$     * $T_2 = \frac{273\,K \times 61.4\,kPa}{30.7\,kPa} = 546\,K$     * $T_{Celsius} = 546 - 273 = 273^{\circ}C$. The temperature must increase by $273^{\circ}C$.

The Combined Gas Law

• Definition: The combined gas law expresses the relationship between pressure, temperature, and volume for a fixed amount of gas.
• Principle: For a fixed amount of gas, the product of pressure and volume divided by the Kelvin temperature is constant.
• Mathematical Formula:     * $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

The Combined Gas Law: Problem Solving and Practice

• Example Problem 4: Changing Multi-Variables     * Problem: Gas at $110\,kPa$ and $30.0^{\circ}C$ fills a flexible $2.00\,L$ container. If temp rises to $80.0^{\circ}C$ and pressure reaches $440\,kPa$, what is the new volume?     * Analyze: Pressure quadruples, while temperature increases slightly. Volume should decrease.     * Kelvin conversion: $T_1 = 303.0\,K$, $T_2 = 353.0\,K$.     * Solution: $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.00\,L \times \frac{110\,kPa}{440\,kPa} \times \frac{353.0\,K}{303.0\,K} = 0.58\,L$.

• Practice Problem 11: Air in syringe at $1.02\,atm$ and $22.0^{\circ}C$. Placed in boiling water ($100.0^{\circ}C$), pressure increased to $1.23\,atm$, volume reduced to $0.224\,mL$. Initial volume?     * $T_1 = 295\,K$, $T_2 = 373\,K$     * $V_1 = \frac{V_2T_1P_2}{T_2P_1} = \frac{0.224\,mL \times 295\,K \times 1.23\,atm}{373\,K \times 1.02\,atm} = 0.214\,mL$

• Practice Problem 12: Balloon with $146.0\,mL$ at $1.30\,atm$ and $5.0^{\circ}C$. If pressure doubles and temp decreases to $2.0^{\circ}C$, what is the volume?     * $T_1 = 278\,K$, $T_2 = 275\,K$, $P_2 = 2.60\,atm$     * $V_2 = \frac{P_1T_2V_1}{P_2T_1} = \frac{1.30\,atm \times 275\,K \times 146.0\,mL}{2.60\,atm \times 278\,K} = 72\,mL$

• Practice Problem 13 (Challenge): Gas cylinder at $0.00^{\circ}C$, $1.00\,atm$, and $30.0\,mL$. If temp increases to $30.0^{\circ}C$ and pressure to $1.20\,atm$, does the piston move up or down?     * $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 30.0\,mL \times \frac{1.00\,atm}{1.20\,atm} \times \frac{303\,K}{273\,K} = 27.7\,mL$     * $\frac{V_2}{V_1} \approx 0.92$. Since $V_2 < V_1$, the piston moves down.

Evaluative Quiz

1. Question: According to Boyle’s law, what happens to the volume of a fixed amount of gas at a constant temperature when the pressure doubles?     * Answer: It decreases by half.
2. Question: Which is NOT true about absolute zero?     * Answer Options: (A) Potential lowest theoretical temp (B) Atoms in lowest energy state (C) Zero on Celsius scale (D) Zero on Kelvin scale.     * Correct Response: It is zero on the Celsius scale (False statement).
3. Question: Which law states that the volume of a given amount of gas is directly proportional to its Kelvin temperature at constant pressure?     * Answer: Charles’s law.
4. Question: Which variables remain constant when applying the formula for Gay-Lussac’s law?     * Answer: Amount of gas and volume.

Lesson Review and Progress Check

• Relationships: The combined gas law summarizes the relationships: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. If temperature increases, either volume or pressure (or both) must increase.
• Proportionality:     * Directly Proportional: Pressure and Volume are both directly proportional to Temperature.     * Inversely Proportional: Pressure and Volume are inversely proportional to each other.
• Predicting Volume: To predict volume at a final altitude (like a weather balloon), one must know the final temperature and final pressure to apply the combined gas law.
• Safety and Compressed Gases:     * Compressed gases must be shielded from heat because increasing temperature increases pressure, which could cause cylinders to explode.     * Oxygen must be decompressed before inhalation.
• Specific Calculation (Methane): Methane at $660\,torr$ and $22.0^{\circ}C$ ($295\,K$) in a $1.00\,L$ rigid container. Pressure at $44.6^{\circ}C$ ($318\,K$)?     * $P_2 = \frac{660\,torr \times 318\,K}{295\,K} = 711\,torr$.
• Concept Mapping Summary:     * Constant amount of gas is the baseline for all these laws.     * Boyle's Law: Temperature held constant ($P_1V_1 = P_2V_2$).     * Charles's Law: Pressure held constant ($\frac{V_1}{T_1} = \frac{V_2}{T_2}$).     * Gay-Lussac's Law: Volume held constant ($\frac{P_1}{T_1} = \frac{P_2}{T_2}$).