Ch6a: Gas Laws

Introduction to Pressure

• Gas molecules are in constant motion in all directions, possessing force and velocity.

• When gas molecules collide with a surface, they apply a force upon it.

• Definition of Pressure ($P$): A measure obtained by dividing the average of all molecular forces by the area of the container's surface with which they collide.

  • Alternatively, pressure is defined as force exerted per unit area by gas molecules as they strike the surrounding surface.

• Formula: $P = F/A$, where $F$ is force and $A$ is area.

• Concept: Gas pressure is a direct result of the continuous movement of gas molecules and their constant collisions with the surfaces around them.

Factors Affecting Gas Pressure

• The pressure of a gas is dependent on three main factors:

  1. Number of gas particles in a given volume (Concentration):

     • Higher concentration of particles leads to more molecules exerting force, resulting in higher pressure.

     • Conversely, a low density of gas particles results in low pressure, while a high density results in high pressure.

  2. Volume of the container:

     • As the volume of the container increases, the concentration of gas molecules decreases (assuming the number of molecules remains constant).

     • This leads to fewer molecular collisions with the container walls, thus lowering the pressure.

     • Therefore, if the volume of the container expands, the pressure will decrease.

  3. Average speed of the gas particles (Temperature):

     • (This factor is explored in detail in Charles's Law).

• Summary:

  • Adding more gas molecules to a fixed volume increases pressure.

  • Expanding the volume of the container decreases pressure.

• Concentration: Can be thought of as the number of moles of gas per unit volume of the gas.

Pressure Detection in the Human Body: The Eardrums

• Our eardrums serve as natural pressure detectors.

• The eardrum is a membrane within the ear canal that is constantly subjected to collisions from air molecules.

• Sound Perception: The ability to hear sound is due to vibrations of gas molecules. Sound itself is essentially vibrations (changes in pressure) in the gas, caused by modulations (e.g., from a voice).

• Airplane Discomfort: When flying, changes in ambient pressure (e.g., during ascent or descent) can cause discomfort. People often "pop" their ears (e.g., by pinching nose and exhaling) to adjust the pressure between the inside of the ear and the outside environment.

• Auditory and Orientation Role: The vibrations of the eardrum are transmitted to two small bones, which then send these vibrations to the semicircular channels. This process enables hearing and contributes to our sense of orientation (up and down).

• Space Travel Connection: In space, the absence of regulated pressure (in addition to lack of gravity) makes it difficult for astronauts to perceive "up" and "down."

Measuring Pressure

There are two primary instruments for measuring pressure:

• 1. Barometer:

  • Purpose: Measures atmospheric pressure.

  • Inventor: Italian scientist Torricelli (after whom the unit "Torr" is named).

  • Mechanism: A tube filled with mercury is inverted into a container of mercury. As some mercury flows out, a vacuum is created at the top of the inverted tube.

  • Atmospheric pressure pushing down on the mercury in the open container forces mercury upwards into the tube, while gravity simultaneously pulls the mercury column down.

  • The height of the mercury column in the tube indicates the atmospheric pressure.

  • Why Mercury? Mercury is approximately $13$ times denser than water. Using water would require a significantly taller (gigantic) and impractical device to measure atmospheric pressure.

  • Standard Atmospheric Pressure: The atmospheric pressure can support a column of mercury that is about $0.76$ meters ($760$ millimeters) tall, or equivalently, $29.92$ inches tall.

  • Unit of Measurement: Atmospheric pressure is often defined as $760 \text{ mm Hg}$ (millimeters of mercury), based on the height of the mercury column in a barometer.

• 2. Manometer:

  • Purpose: Measures the pressure of a gas trapped within a contained system (e.g., a tank).

  • Mechanism: Consists of a U-shaped tube, often containing mercury. One end of the tube is connected to the gas container, and the other end is open to the atmosphere.

  • Principle: A competition exists between the pressure of the gas pushing from one end and the atmospheric pressure pushing from the other.

  • Measurement: The difference in the liquid levels (mercury height) within the U-tube indicates the pressure difference between the contained gas and the atmosphere.

  • If the gas pressure is higher than atmospheric pressure, it pushes the mercury column down on its side, creating a height difference ($\Delta h$). The gas pressure would be $P_{\text{atm}} + \Delta h$.

  • If the gas pressure is lower, the atmospheric pressure pushes the mercury column down on its side, and the gas pressure would be $P_{\text{atm}} - \Delta h$.

  • Units are typically expressed in millimeters of mercury ($mm \text{ Hg}$).

Units of Pressure

Pressure units are derived from the ratio of force over area ($F/A$), which fundamentally relates to mass, length, and time.

• Pascal ($Pa$):

  • The International System of Units (SI) unit for pressure.

  • Defined as one Newton per square meter ($1 \text{ N/m}^2$).

  • Atmospheric pressure at sea level: Approximately $101,325 \text{ Pa}$ or $101.3 \text{ kilopascals } (kPa)$.

• Pounds per Square Inch ($PSI$):

  • A common unit in some engineering contexts.

  • Represents force in pounds over area in square inches.

  • Atmospheric pressure at sea level: Approximately $14.7 \text{ PSI}$.

  • Equivalence: $101.3 \text{ kPa} \approx 14.7 \text{ PSI}$.

• Torr ($Torr$):

  • Defined as exactly one millimeter of mercury ($1 \text{ mm \text{ Hg}}$).

  • Named after Torricelli.

  • Atmospheric pressure: Exactly $760 \text{ Torr}$ (this exact value is not subject to significant figures rules).

• Inches of Mercury ($in \text{ Hg}$):

  • Another unit based on the height of a mercury column.

  • Atmospheric pressure: Approximately $29.92 \text{ in \text{ Hg}}$.

• Atmospheres ($atm$):

  • A widely used standard unit for pressure.

  • $1 \text{ atmosphere} = 1 \text{ ATM}$.

• Bar ($bar$):

  • Used in more advanced physical chemistry.

  • Approximately equal to one atmosphere ($1 \text{ bar} \approx 1 \text{ atm}$; more precisely, $1 \text{ bar} = 100 \text{ kPa}$).

• Reason for Multiple Units: There is no universal agreement on the "best" unit, and different units have arisen from historical context and diverse practical applications across various fields.

Introduction to Gas Laws

To fully characterize a gas and understand its behavior, four fundamental properties must be considered. These properties are interrelated, meaning a change in one will affect the others.

• Four Basic Properties of Gases:

  1. **Pressure ($P$)

  2. **Volume ($V$)

  3. **Temperature ($T$)

  4. **Amount of matter ($n$), typically measured in moles.

• Gas Laws: Simple gas laws describe the relationships between pairs of these properties, and eventually, one comprehensive equation combines them all.

Boyle's Law (Volume-Pressure Relationship)

• Discoverer: Robert Boyle (though Robert Hooke invented the apparatus used).

• Concept: States that for a fixed amount of gas at a constant temperature, the volume of the gas is inversely proportional to its pressure.

• Inverse Relationship: If one property (e.g., pressure) increases, the other (e.g., volume) decreases proportionally.

• Mathematical Representation:

  • $V \propto 1/P$ (Volume is inversely proportional to Pressure).

  • This implies that $V = k/P$, where $k$ is a proportionality constant.

  • Rearranging gives the core equation: $PV = k$ (Pressure times Volume equals a constant).

• Practical Application: For a given gas sample under constant temperature and amount, the initial conditions multiply to the same constant as the final conditions:

  • P1V1=P2V2

• Graphical Representation:

  • A plot of pressure ($P$) versus volume ($V$) yields an inverse curve (hyperbola).

  • A plot of pressure ($P$) versus the inverse of volume ($1/V$) yields a straight line.

• Explanation: As the volume of a gas sample decreases, the gas molecules are confined to a smaller space. This increases the frequency of collisions with the container walls, proportionally raising the pressure.

• Scuba Diving Example (Real-World Implication):

  • Pressure in water increases significantly with depth. For every $3$ meters of depth, an additional atmosphere of pressure is experienced. For example, a $20$-meter dive subjects a diver to approximately $3$ atmospheres of pressure.

  • Danger of Rapid Ascent: If a diver rises to the surface too quickly (going from high to low pressure), the gas (primarily nitrogen, oxygen, carbon dioxide) dissolved in their body tissues will expand rapidly.

  • Consequences: This rapid expansion can cause severe damage to internal organs and tissues, leading to a condition known as "the bends," which can be fatal. For instance, a three-fold decrease in external pressure would cause a threefold expansion in lung volume if a diver holds their breath, causing internal damage.

  • Solution: Divers must ascend slowly or use a decompression chamber to allow the dissolved gases to be released gradually.

• Problem Example 1: A person's lungs contain $2.75 \text{ L}$ of air at $1.02 \text{ atm}$. If they increase their lung volume to $3.25 \text{ L}$ without inhaling more air, what is the new pressure in their lungs?

  • Given: P1 = 1.02 atm$,$ V1 = 2.75 L$,$V2 = 3.25 L

  • Using Boyle's Law: P1V1=P2V2

  • Solve for P2$:$ P2 = (P1V1) / V2

  • $P_2 = (1.02 \text{ atm} \times 2.75 \text{ L}) / 3.25 \text{ L}$

  • $P_2 \approx 0.863 \text{ atm}$. The pressure in the lungs decreases as the volume expands.

• Problem Example 2: A gas is contained in a cylinder with an initial height of $6 \text{ cm}$ at $1 \text{ atm}$. If the height is reduced to $3 \text{ cm}$ (halving the volume), what is the final pressure?

  • Given: P1 = 1 atm, h1 = 6 cm (volume is proportional to height, $V=A \times h$), $h_2 = 3 \text{ cm}$.

  • Using Boyle's Law: P1V1 = P2V2 implies P1(A x h1) = P2(A x h2)

  • Since area ($A$) is constant, P1h1 = P2h2

  • Solve for P2$:$P2 = (P1h1) / h2

  • $P_2 = (1 \text{ atm} \times 6 \text{ cm}) / 3 \text{ cm}$

  • $P_2 = 2 \text{ atm}$. Halving the volume doubles the pressure.

Charles's Law (Volume-Temperature Relationship)

• Discoverer: Jacques Charles (French scientist).

• Concept: States that for a fixed amount of gas at a constant pressure, the volume of the gas increases linearly with increasing temperature.

• Direct Proportionality: If one property (e.g., temperature) increases, the other (e.g., volume) increases proportionally.

• Crucial Condition for Temperature: Temperature must always be expressed in Kelvin ($K$).

  • Kelvin Conversion: $T(\text{K}) = T(^\circ\text{C}) + 273.15$.

  • Absolute Scale: Kelvin is an absolute temperature scale and is not expressed with degrees (e.g., $273 \text{ K}$, not $273^\circ\text{K}$).

• Mathematical Representation:

  • $V \propto T$ (Volume is directly proportional to Temperature).

  • This implies that $V=k^{}T$ , where $k'$ is a proportionality constant.

  • Rearranging gives the core equation: $V/T=k^{}$ (Volume divided by Temperature equals a constant).

• Practical Application: For a given gas sample under constant pressure and amount, the ratio of initial volume and temperature equals the ratio of final volume and temperature:

  • V1/T1 = V2/T2

• Discovery of Absolute Zero:

  • When plots of volume versus temperature (at constant pressure and for different gas amounts) are extrapolated to zero volume, all lines converge at a single point: $-273.15^\circ\text{C}$.

  • This temperature is defined as absolute zero ($0 \text{ K}$ ), the theoretical lowest possible temperature where molecular motion ceases.

  • Gases cannot be experimentally measured at absolute zero because they condense into liquid or solid states before reaching this point.

• Practical Example:

  • A balloon placed in an ice water bath will shrink. When moved to a boiling water bath, its volume expands.

  • A balloon exposed to liquid nitrogen (extremely cold) will significantly shrink as the gas inside cools, causing the particles to move slower and occupy less collective space.

Sample Problem - Charles' Law

• Initial Setup:

  • A gas sample has a volume of 2.8 liters at an unknown temperature.

  • When placed in ice water (0 °C or 273.15 K), the volume decreases to 2.57 liters.

  • Objective: Determine the initial temperature (T₁).

• Volume 1 (V₁) = 2.8 L

• Volume 2 (V₂) = 2.57 L

• Final Temperature (T₂) = 0 °C = 273.15 K

• Gas temperature must always be in Kelvin for calculations.

• Charles' Law formula: {V1}/{T1} = {V2}/{T2}

• Rearranged to find T₁: T1 = {V1 T2}/{V2}

• Substitute the values:

  • $T_1 = \frac{2.8 \times 273.15}{2.57}$

  • Performing the calculation:

  • $T_1 = 299.97 K$

  • Converting back to Celsius:

  • T1 (°C) = T1 (K) - 273.15 = 26.82 °C

  • Rounding to three significant figures: $T_1 = 26.8 °C$

Avogadro's Law (Volume-Moles/Molecules Relationship

Definition: Volume of a gas is directly proportional to the number of gas molecules (n). Volume is directly proportional to the number of gas molecules in moles

• More gas molecules = larger volume

• When the amount of gas in a sample increases at constant temperature and pressure, its volume increases in direct proportion because the greater number of gas particles fill more space.

Formula: v1/n1 = v2/n2

Implication: Increasing the number of gas molecules increases the volume—like inflating a balloon.

Sample Problem - Gas Exhalation

• Setup: An athlete has a lung volume of 6.15 L and contains 2.54 moles of air.

• On exhalation, the lung volume reduces to 2.55 L.

• Objective: Determine how many moles of gas were exhaled.

• Variables:

  • Initial moles (n₁) = 2.54 mol

  • Initial volume (V₁) = 6.15 L

  • Final volume (V₂) = 2.55 L

• Formula: n2 = {V2 n1}/{V1}

• Substitute the values:

  • $n_2 = \frac{2.55 \times 2.54}{6.15}$

  • Calculate to find excess moles exhaled.

Density of Gases

• Formula: $d = \frac{m}{V}$ where d is density, m is mass, and V is volume.

• For a gas, $d = \frac{PM}{RT}$, where P is pressure, M is molar mass, R is gas constant, and T is temperature.

Example Problem - Density of Nitrogen Gas

• Given: Temperature = 131°C, Pressure = 755 mmHg.

• Convert temperature:

  • $T = 131 + 273.15 = 404.15 K$

• Convert pressure:

  • $P = \frac{755}{760} \approx 0.993 atm$

• Calculate density using $d = \frac{PM}{RT}$ where M for nitrogen is 28.02 g/mol

• Substitute values and solve.

Combined Gas Laws

Boyle's Law: P1 V1 = P2 V2 (relationship between pressure and volume)

Charles's Law: {V1}/{T1} = {V2}/{T2}

Avogadro's Law: {V1}/{n1} = {V2}/{n2}

Gas Constant (R)

• R can take on different values depending on the units of measurement used:

  • $R = 0.08206 \frac{L atm}{mol K}$ or 0.08206 L∙atm∙mol^-1∙K^-1 when using atmospheres for pressure and liters for volume.

Ideal Gas Law

• Central formula: $PV = nRT$

• Used to calculate volume, pressure, or temperature if the other variables are known.

• Standard conditions defined: 1 atm pressure and 0 °C (273.15 K).

• At STP, one mole of any gas occupies 22.4 L regardless of the type of gas.

• Molar Volume at STP: 22.4 L for one mole of any gas at STP.

Ideal Gas Law: finding molar mass example

• Given the parameters of pressure, volume, temperature, and number of moles, apply the Ideal Gas Law (PV = nRT) to find unknown values.

• We can determine the molar mass of an unknown substance by:

• Measuring its mass and volume under conditions of known pressure and temperature.

• Then, we determine the amount of the gas in moles from the ideal gas law.

• Finally, we calculate the molar mass by dividing the mass (in grams) by the amount (in moles).

• molar mass = (mass)/(moles)

Example: A sample of gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 ºC and a pressure of 886 mmHg. Find its molar mass.

PV = nRT

R = 0.08206 L∙atm∙mol–1∙K–1

1 atm = 760 mmHg

Step 1: convert P, T

• P = 886 mmHg (1 atm/760 mmHg)

• P = 1.1658 atm

• T = 55 + 273 K

• T = 328 K

Step 2: solve for n

• n = (PV)/(RT)

• n = (1.1658 atm) (0.225 L) /(0.08206 L∙atm∙mol–1∙K–1)(328 K)

• n = 0.009745 mol = 9.745×10^–3 mol

Step 3: solve for ℳ

• ℳ = mass/moles

• ℳ = "0.311 g" /"0.009745 mol" = 31.9 g mol–1

• ℳ = 31.9 g mol^-1 (Ar)