Gas Laws Review Practice Flashcards

Units and Fundamental Gas Concepts

Pressure Definition: Pressure is fundamentally caused by gas particles bouncing off the interior walls of a container. Two specific factors determine the magnitude of the pressure: - The frequency with which the particles hit the container sides (how often collisions occur). - The amount of force with which the particles strike the sides.
Pressure Unit Equivalencies: At sea level and at a temperature of $273\,K$ , the following units represent the standard atmospheric pressure: - $1\,atm$ - $760\,mm\,Hg$ - $760\,Torr$ - $101.3\,kPa$ - $14.7\,psi$
Temperature Requirements: In gas law calculations, temperature must always be expressed in Kelvin ( $K$ ). - Temperature is a measure of the average kinetic energy of the particles in a substance. - Absolute Zero: This is identified as $0\,K$ , the point at which matter completely stops moving. - Conversion Formula: to convert Celsius ( $^{\circ}C$ ) to Kelvin ( $K$ ), use the formula $\text{K} = \text{C} + 273$ . - Conversion Examples: - At $0^{\circ}C$ , the temperature is $273\,K$ . - To find Celsius from $298\,K$ : $298 - 273 = 25^{\circ}C$ . - To find Kelvin from $65^{\circ}C$ : $65 + 273 = 338\,K$ .
Volume: Volume can be measured in milliliters ( $mL$ ) or liters ( $L$ ). The primary rule is that the units used must be consistent when performing calculations. Note that $1\,mL = 1\,cm^3$ .
Number of Particles: In conceptual terms, this is often referred to as the number of "puffs." These are relative amounts used to focus on the concepts rather than complex stoichiometry. - For example: "10 puffs" of particles contains exactly twice as many particles as "5 puffs." - While moles ( $n$ ) can be used, "puffs" is used here for simplicity.

Principal Gas Laws

Boyle's Law: - Definition: This law states that as the volume of a gas decreases, the pressure increases, provided that the temperature and the number of particles are held constant. - Relationship: This is an inverse relationship, meaning pressure is inversely proportional to volume. In the lab, it is observed that if the pressure doubles, the volume decreases by half. - Mathematical Representation: If $k = P_1 \times V_1$ and $k = P_2 \times V_2$ , then: $P_1 \times V_1 = P_2 \times V_2$
Avogadro's Hypothesis (Law): - Definition: Equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. - Key Concept: The number of particles increases the pressure regardless of the size of the individual molecules, assuming temperature and volume remain constant. - Mathematical Representation: Since pressure and number are in direct proportion: $\frac{P_1}{n_1} = \frac{P_2}{n_2}$
Gay-Lussac's Law: - Definition: For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature. - Mathematical Representation: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Charles's Law: - Definition: For a given mass and constant pressure of an ideal gas, the absolute temperature is directly proportional to the volume of the gas. - Mathematical Representation: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Combined Gas Law: This combines the relationships of pressure, volume, and temperature into a single formula: $\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}$

Kinetic Molecular Theory and Gas Measure

Kinetic Molecular Theory (KMT): This theory explains the behavior of gases at the molecular level. It is based on specific assumptions regarding a theoretical construct known as an "ideal gas."
Measurement Instruments: - Manometer: An instrument used to measure gas pressure within a container. - Barometer: An instrument used to measure atmospheric pressure.
Atmospheric Pressure Characteristics: - The Earth's atmosphere has weight, which creates atmospheric pressure. - At $1\,atm$ , the height of the mercury ( $Hg$ ) in a barometer is exactly $760\,mm$ .
Pressure Calculation Scenario: - A block of wood weighing $18\,Newtons$ ( $N$ ) rests on a table. The block is $3\,cm$ long and $2\,cm$ wide. - Area Calculation: $3\,cm \times 2\,cm = 6\,cm^2$ . - Pressure Calculation: $P = \frac{18\,N}{6\,cm^2} = 3\,N/cm^2$ . - (Alternative unit conversion noted in transcript: $6\,cm^2 = 0.0006\,m^2$ ).

Conceptual Applications of Kinetic Theory

Shrinking Helium Balloons: When a balloon is moved from a warm room to the cold outdoors, the temperature decreases. According to KMT, the particles slow down and hit the walls with less force. To maintain constant pressure, the volume decreases (illustrating Charles's Law).
Popping Bubble Wrap: Squeezing a bubble decreases its volume. Following Boyle's Law, this decrease in volume causes an increase in pressure until the plastic material can no longer contain the force, resulting in a pop.
Increasing Tire Pressure: Adding more "puffs" of air increases the number of gas particles. This leads to a higher frequency of collisions against the tire walls, which results in higher pressure.

Gas Law Problem Sets

Problem 1 (Charles's Law): A sample occupies $240.\,mL$ at $20.0^{\circ}C$ . What is the volume at $40.0^{\circ}C$ ? - $T_1 = 20 + 273 = 293\,K$ - $T_2 = 40 + 273 = 313\,K$ - Calculation: $\frac{240\,mL}{293\,K} = \frac{V_2}{313\,K}$ - Result: $V_2 = 256.4\,mL$
Problem 2 (Boyle's Law): A sample has a volume of $180\,L$ at $740\,mmHg$ . What is the volume at $800.\,mmHg$ ? - Calculation: $V_2 = \frac{180\,L \times 740\,mmHg}{800.\,mmHg}$ - Result: $V_2 = 166.5\,L$ (Note: Transcript calculation shows $164.6\,L$ ).
Problem 3 (Boyle's Law): A sample occupies $800.\,mL$ at $680\,mmHg$ . What pressure results in $700.\,mL$ ? - Calculation: $P_2 = \frac{800\,mL \times 680\,mmHg}{700.\,mL}$ - Result: $P_2 = 777.1\,mmHg$ (Note: Transcript shows a different value entry of $595\,mmHg$ , possibly from a different digit set).
Problem 4 (Charles's Law): A sample occupies $400.\,mL$ at $10.0^{\circ}C$ . Find the temperature for $250\,mL$ . - $T_1 = 10 + 273 = 283\,K$ - Calculation: $T_2 = \frac{250\,mL \times 283\,K}{400.\,mL}$ - Result: $T_2 = 176.8\,K$
Problem 5 (Combined Gas Law): Volume of $200.\,mL$ at $27.0^{\circ}C$ and $700.\,mmHg$ . Find volume at $127^{\circ}C$ and $800.\,mmHg$ . - $T_1 = 300\,K; T_2 = 400\,K$ - Calculation: $V_2 = 200\,mL \times \left(\frac{700}{800}\right) \times \left(\frac{400}{300}\right)$ - Result: $V_2 = 233.3\,mL$ (Note: Transcript notes $266.667$ ).
Problem 6 (Gay-Lussac's Law): Pressure is $2.00\,atm$ at $25.0^{\circ}C$ . Find temp for $1.50\,atm$ . - $T_1 = 298\,K$ - Calculation: $T_2 = \frac{1.50\,atm \times 298\,K}{2.00\,atm}$ - Result: $T_2 = 223.5\,K$
Problem 7 (Combined Gas Law): Volume $260\,L$ at $85.0^{\circ}C$ ( $358\,K$ ) and $750\,torr$ . Volume becomes $240\,L$ at $25.0^{\circ}C$ ( $298\,K$ ). Find new pressure. - Calculation: $P_2 = \frac{750\,torr \times 260\,L \times 298\,K}{358\,K \times 240\,L}$ - Result: $P_2 = 676.2\,torr$ (Note: Transcript calculation result listed as $632.6\,torr$ ).
Problem 8 (Gay-Lussac's Law): Temp of $60.0^{\circ}C$ ( $333\,K$ ) and pressure of $15.1\,PSI$ . Find pressure at $80.0^{\circ}C$ ( $353\,K$ ). - Calculation: $P_2 = \frac{15.1\,PSI \times 353\,K}{333\,K}$ - Result: $P_2 = 16\,PSI$