Comprehensive Gas Laws Study Guide and Worksheet Solutions

Gas Laws Fundamental Principles and Temperature Conversion

• Governing Definitions:     * Charles' Law: This law states that the volume of a gas varies directly with the Kelvin temperature, provided that the pressure remains constant.     * Absolute Temperature Conversion: To solve gas law problems, temperatures must be converted from Celsius ($^{\circ}C$) to Kelvin ($K$). The standard formula for this conversion is:         * $K = {^{\circ}C} + 273$
• Charles' Law Mathematical Representations:     * Fraction form: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$     * Product form: $V_1 \times T_2 = V_2 \times T_1$, where $V_1$ and $T_1$ represent the initial volume and temperature, and $V_2$ and $T_2$ represent the final volume and temperature.

Gas Laws Worksheet 2: Problems 1 through 3

• Problem 1: Temperature-Volume Relationship (Charles' Law)     * Task: Calculate the decrease in temperature when $6.00\,L$ at $20.0\,^{\circ}C$ is compressed to $4.00\,L$.     * Mandatory Step: Convert Celsius to Kelvin: $20.0\,^{\circ}C + 273 = 293\,K$.     * Work Shown:         * Equation Setup: $\frac{6.00\,L}{293\,K} = \frac{4.00\,L}{T_2}$         * Cross-multiplication: $6.00(T_2) = 1172$         * Division: $T_2 = \frac{1172}{6.00}$         * Final Answer: $200\,K$ (Note: Work shows values 1122 and 200k in handwritten text).     * Identity of the Law: Charles' Law.
• Problem 2: Pressure-Volume Relationship (Boyle's Law)     * Task: Determine the new volume if $22.5\,L$ of nitrogen at $748\,mmHg$ are compressed to $725\,mmHg$ at constant temperature.     * Condition: Use the pressure unit $mmHg$; do not convert to atmospheres.     * Work Shown:         * Equation: $(748\,mmHg) \times (22.5\,L) = (725\,mmHg) \times V_2$         * Result: $V_2 = 23.2\,L$     * Identity of the Law: Boyle's Law.
• Problem 3: Pressure-Volume Relationship (Boyle's Law)     * Task: A gas with a volume of $4.0\,L$ at a pressure of $205\,kPa$ expands to a volume of $12.0\,L$. Calculate the new pressure in $kPa$ if temperature remains constant.     * Work Shown:         * Equation: $(205\,kPa) \times (4.0\,L) = P_2 \times (12.0\,L)$         * Result: $P_2 = 68\,kPa$     * Identity of the Law: Boyle's Law (Note: Transcript mark box indicates Charles' Law, which is a conceptual error).

Gas Laws Worksheet 2: Problems 4 through 6

• Problem 4: Pressure-Volume Relationship (Boyle's Law)     * Task: Calculate the pressure required to compress $196.0\,L$ of air at $1.00\,atm$ into a cylinder with a volume of $2,600\,mL$.     * Mandatory Step: Convert milliliters to liters: $2,600\,mL = 2.6\,L$.     * Work Shown:         * Equation: $(1.00\,atm) \times (196.0\,L) = P_2 \times (2.6\,L)$         * Result: $P_2 = 75\,atm$     * Identity of the Law: Boyle's Law.
• Problem 5: Pressure-Temperature Relationship (Gay-Lussac's Law)     * Task: A container of gas is initially at $0.500\,atm$ and $25\,^{\circ}C$. Calculate the pressure at $125\,^{\circ}C$.     * Mandatory Step: Convert $^{\circ}C$ to $K$:         * $T_1 = 25\,^{\circ}C + 273 = 298\,K$         * $T_2 = 125\,^{\circ}C + 273 = 398\,K$     * Work Shown:         * Equation: $\frac{0.500\,atm}{298\,K} = \frac{P_2}{398\,K}$         * Cross-multiplication: $298(P_2) = 199$         * Division: $P_2 = \frac{199}{298}$         * Final Answer: $0.668\,atm$     * Identity of the Law: Gay-Lussac's Law.
• Problem 6: Volume-Temperature Relationship (Charles' Law)     * Task: A container with $5.00\,L$ of gas is collected at $100\,K$ and allowed to expand to $20.0\,L$. Calculate the new temperature required to maintain constant pressure.     * Work Shown:         * Equation: $\frac{5.00\,L}{100\,K} = \frac{20.0\,L}{T_2}$         * Cross-multiplication: $5.00(T_2) = 2000$         * Division: $T_2 = \frac{2000}{5.00}$         * Final Answer: $400\,K$     * Identity of the Law: Charles' Law.

Gas Laws Worksheet 2: Problems 7 through 9

• Problem 7 (Numbered 6 in transcript): Volume-Temperature Relationship (Charles' Law)     * Task: If $15.0\,L$ of neon at $25.0\,^{\circ}C$ is allowed to expand to $45.0\,L$, find the new temperature at constant pressure.     * Mandatory Step: Convert Celsius to Kelvin: $25.0\,^{\circ}C + 273 = 298\,K$.     * Work Shown:         * Equation: $\frac{15.0\,L}{298\,K} = \frac{45.0\,L}{T_2}$         * Cross-multiplication: $15.0(T_2) = 13410$         * Division: $T_2 = \frac{13410}{15.0}$         * Final Answer: $894\,K$     * Identity of the Law: Charles' Law.
• Problem 8: Pressure-Temperature Relationship (Gay-Lussac's Law)     * Task: A container is at $47\,mmHg$ and $77\,K$ (liquid nitrogen temperature). Determine the pressure when it warms to room temperature ($25\,^{\circ}C$).     * Mandatory Step: Convert room temperature to Kelvin: $25\,^{\circ}C + 273 = 298\,K$.     * Work Shown:         * Equation: $\frac{47\,mmHg}{77\,K} = \frac{P_2}{298\,K}$         * Cross-multiplication: $77(P_2) = 14006$         * Division: $P_2 = \frac{14006}{77}$         * Final Answer: $180\,mmHg$     * Identity of the Law: Gay-Lussac's Law.
• Problem 9: Pressure-Temperature Relationship (Gay-Lussac's Law)     * Task: A gas thermometer reads $248\,kPa$ at $273\,^{\circ}C$. Find the temperature when the pressure reads $3.41\,atm$.     * Mandatory Step 1: Convert pressure units: $\text{atm} = \frac{248\,kPa}{101.325\,kPa/atm} = 2.447569701\,atm$.     * Mandatory Step 2: Convert temperature to Kelvin: $273\,^{\circ}C + 273 = 546\,K$.     * Work Shown:         * Equation: $\frac{2.447\dots\,atm}{546\,K} = \frac{3.41\,atm}{T_2}$         * Final Result (Written): $654.4\,K$ (Work in transcript contains a decimal typo as $65.4\,k$).     * Identity of the Law: Gay-Lussac's Law.

Charles' Law Practice Problems (Instructional Fair, Inc.)

• Problem 1: A nitrogen sample occupies $250\,mL$ at $25\,^{\circ}C$. Calculate volume at $95\,^{\circ}C$.     * $T_1 = 298\,K$, $T_2 = 368\,K$.     * Work: $\frac{250\,mL}{298\,K} = \frac{V_2}{368\,K}$.     * Results: $0.838\,mL/K \times 368\,K = 310\,mL$ (rounded to 310).
• Problem 2: Oxygen occupies $2.3\,L$ at $40\,^{\circ}C$. Find temperature required to occupy $6.5\,L$.     * $T_1 = 40 + 273 = 313\,K$.     * Work: $\frac{2.3\,L}{313\,K} = \frac{6.5\,L}{T_2}$.     * Results: $T_2 = 884.6\,K$.
• Problem 3: Hydrogen cooled from $150\,^{\circ}C$ to $50\,^{\circ}C$. New volume is $75\,mL$. Find original volume.
• Problem 4: Chlorine occupies $25\,mL$ at $300\,K$. Find volume at $600\,K$.
• Problem 5: Neon gas at $50\,^{\circ}C$ and $2.5\,L$ cooled to $25\,^{\circ}C$. Calculate new volume.
• Problem 6: Fluorine at $300\,K$ occupies $500\,mL$. Find temperature for $300\,mL$.
• Problem 7: Helium at $3.8\,L$ and $-45\,^{\circ}C$. Find volume at $45\,^{\circ}C$.
• Problem 8: Argon volume moves from $380\,mL$ to $250\,mL$. Final temperature was $-55\,^{\circ}C$. Find original temperature.