5.1 Gas Laws & the Ideal Gas Equation (PV = nRT)

5.1 Gas Laws & the Ideal Gas Equation (PV = nRT)

Gases lack fixed shape or volume, expanding uniformly to fill containers, compressing under pressure, and mixing spontaneously in all proportions. Four macroscopic state variables govern gas behavior: pressure ($P$), volume ($V$), temperature ($T$), and molar amount ($n$).

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Gas Pressure & Measurement Units

Pressure is defined as perpendicular force per unit area: $P = \frac{F}{A}$

Evangelista Torricelli developed the mercury barometer in 1643, balancing atmospheric pressure against a liquid mercury column ($d = 13.595\text{ g/cm}^3$) in an evacuated tube. At sea level, standard atmospheric pressure supports a $760.0\text{ mm}$ mercury column. Manometers measure differential gas pressures relative to atmospheric or evacuated references.

Common pressure units and conversions:

• Atmosphere ($\text{atm}$): Standard sea-level benchmark.
• Millimeter of mercury ($\text{mmHg}$): Direct height displacement.
• Torr ($\text{torr}$): Equivalent to $\text{mmHg}$ ($1\text{ torr} \equiv 1\text{ mmHg}$).
• Pascal ($\text{Pa}$): SI derived unit ($1\text{ N/m}^2 = 1\text{ kg}/(\text{m}\cdot\text{s}^2)$).
• Kilopascal ($\text{kPa}$): Metric unit ($1\text{ kPa} = 10^3\text{ Pa}$).
• Bar ($\text{bar}$): Thermodynamic unit ($1\text{ bar} = 10^5\text{ Pa} = 100\text{ kPa}$).

$1.000\text{ atm} = 760.0\text{ mmHg} = 760.0\text{ torr} = 101,325\text{ Pa} = 101.325\text{ kPa} = 1.01325\text{ bar}$

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Empirical Gas Laws

Historical experiments isolated pairwise relationships between two variables while holding two constant.

Boyle's Law: Pressure and Volume

Robert Boyle (1662) showed that at constant $n$ and $T$, gas volume is inversely proportional to pressure: $P_1 V_1 = P_2 V_2 \quad (n, T\text{ constant})$ A plot of $V$ versus $P$ yields a hyperbola; $V$ versus $1/P$ yields a straight line through the origin.

Charles's Law: Volume and Temperature

Jacques Charles (1787) found that at constant $n$ and $P$, gas volume is directly proportional to absolute temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad (n, P\text{ constant})$ Extrapolating volume isobars across gases intersects zero volume at $-273.15^\circ\text{C}$, defining absolute zero ($0\text{ K}$) and the Kelvin scale ($T(\text{K}) = T(^\circ\text{C}) + 273.15$). Gas calculations strictly require Kelvin.

Avogadro's Law: Volume and Amount

Amedeo Avogadro (1811) stated that equal volumes of gases at identical temperature and pressure contain equal numbers of particles: $\frac{V_1}{n_1} = \frac{V_2}{n_2} \quad (P, T\text{ constant})$ At Standard Temperature and Pressure (STP: $0^\circ\text{C} = 273.15\text{ K}$, $1.000\text{ atm}$), one mole of ideal gas occupies standard molar volume ($V_m$): $V_m = \frac{V}{n} = 22.414\text{ L/mol}$

Gay-Lussac's Law: Pressure and Temperature

At constant $n$ and $V$, gas pressure is directly proportional to absolute temperature: $\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad (n, V\text{ constant})$

Combined Gas Law

For a fixed quantity of gas ($n = \text{constant}$), these empirical relations combine into: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

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Gas Laws Summary Table

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The Ideal Gas Law ($PV = nRT$)

Combining the empirical relationships yields the ideal gas equation of state: $PV = nRT$

Universal gas constant $R$ values depend on operational units:

• $R = 0.082057\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$: Standard for volumetric stoichiometry ($P$ in $\text{atm}$, $V$ in $\text{L}$).
• $R = 8.3145\text{ J}/(\text{mol}\cdot\text{K}) = 8.3145\text{ Pa}\cdot\text{m}^3/(\text{mol}\cdot\text{K})$: Used for thermodynamic work, energy, and molecular velocities.
• $R = 62.364\text{ L}\cdot\text{torr}/(\text{mol}\cdot\text{K})$: Used directly with pressures measured in $\text{torr}$ or $\text{mmHg}$.

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Gas Density & Molar Mass Calculations

Because moles $n = m / \mathcal{M}$ (where $m$ is mass and $\mathcal{M}$ is molar mass): $PV = \left(\frac{m}{\mathcal{M}}\right)RT \quad \implies \quad P\mathcal{M} = \left(\frac{m}{V}\right)RT = dRT$

Solving for gas density ($d = m/V$): $d = \frac{P\mathcal{M}}{RT}$

Solving for molar mass ($\mathcal{M}$): $\mathcal{M} = \frac{dRT}{P} = \frac{mRT}{PV}$

The Dumas Method

In the Dumas vapor density method, an unknown volatile liquid is vaporized in a flask of volume $V$ immersed in a boiling water bath at temperature $T$ under barometric pressure $P$. Condensing and weighing the vapor gives mass $m$, yielding molar mass via $\mathcal{M} = mRT / (PV)$.

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Worked Numerical Examples

Example 1: Gas Density at Non-Standard Conditions

Calculate the density of sulfur dioxide gas ($\text{SO}_2, \mathcal{M} = 64.07\text{ g/mol}$) at $37.0^\circ\text{C}$ and $742.0\text{ torr}$.

1. Convert units:
   • $T = 37.0 + 273.15 = 310.15\text{ K}$
   • $P = 742.0\text{ torr} / 760.0\text{ torr/atm} = 0.9763\text{ atm}$
2. Compute density: $d = \frac{(0.9763\text{ atm})(64.07\text{ g/mol})}{(0.08206\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(310.15\text{ K})} = 2.46\text{ g/L}$

Example 2: Dumas Method Molar Mass

A student vaporizes an unknown volatile liquid in a $254.0\text{ mL}$ flask at $99.2^\circ\text{C}$ under $751.2\text{ mmHg}$. Condensed vapor mass is $0.718\text{ g}$. Find the molar mass.

1. Convert units: $V = 0.2540\text{ L}$, $T = 372.35\text{ K}$, $P = 751.2 / 760.0 = 0.9884\text{ atm}$, $m = 0.718\text{ g}$.
2. Compute molar mass: $\mathcal{M} = \frac{(0.718\text{ g})(0.08206)(372.35\text{ K})}{(0.9884\text{ atm})(0.2540\text{ L})} = 87.4\text{ g/mol}$