Mastering the Gas Laws: Ideal, Combined, Partial Pressures, and Real Gases

Gases are unique among the states of matter. Unlike solids and liquids, gases expand to fill their containers, are highly compressible, and form homogenous mixtures. Whether you are calculating the lift of a weather balloon or engineering a high-pressure chemical reactor, understanding the behavior of gases is fundamental to thermodynamics.

In this guide, we break down the progression from the simple Ideal Gas Law to complex Real Gas corrections. We will move beyond memorizing formulas and focus on applying them to solving real-world chemistry problems.

The Ideal Gas Law ($PV=nRT$): Fundamentals and Calculations

The cornerstone of gas stoichiometry is the Ideal Gas Law. It relates four macroscopic variables: pressure ($P$), volume ($V$), temperature ($T$), and the amount of substance in moles ($n$).

$$ PV = nRT $$

Key Variables and Units

Pressure ($P$): Often measured in atmospheres (atm), pascals (Pa), or mmHg (torr).
Volume ($V$): Usually in Liters (L) or cubic meters ($m^3$).
Temperature ($T$): Must always be in Kelvin (K). Convert Celsius to Kelvin using $T_K = T_C + 273.15$.
Gas Constant ($R$): The value of $R$ depends on the units of $P$ and $V$.
- If using L and atm: $R \approx 0.08206 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$
- If using $m^3$ and Pa (SI units): $R \approx 8.314 \, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$

Calculating Gas Density

One of the most useful applications of the Ideal Gas Law is determining the density ($\rho$) of a gas. By substituting moles $n = \frac{\text{mass}}{M}$ (where $M$ is molar mass) into the ideal gas equation, we derive:

$$ \rho = \frac{P \cdot M}{R \cdot T} $$

Illustrative Example: Density of Ammonia

Problem: Determine the density of ammonia gas ($NH_3$) at $800$ torr and $25^\circ\text{C}$.

Step-by-Step Solution:

Convert units to match $R$ ($0.08206 \, \text{L atm/mol K}$):
- Pressure: $800 \, \text{torr} \div 760 \, \text{torr/atm} = 1.053 \, \text{atm}$
- Temperature: $25^\circ\text{C} + 273.15 = 298.15 \, \text{K}$
Calculate Molar Mass ($M$) of $NH_3$:
$N (14.01) + 3 \times H (1.008) = 17.03 \, \text{g/mol}$
Apply the Density Formula:
$$ \rho = \frac{(1.053 \, \text{atm})(17.03 \, \text{g/mol})}{(0.08206 \, \text{L atm/K mol})(298.15 \, \text{K})} $$
Compute:
$$ \rho \approx \frac{17.93}{24.47} \approx 0.73 \, \text{g/L} $$

Answer: The density is approximately 0.73 g/L.

Beyond Ideal: Boyle's, Charles's, and the Combined Gas Law

Often, you need to calculate how a gas changes when conditions shift (e.g., heating a balloon or compressing a piston). If the number of moles ($n$) remains constant, we can combine Boyle’s Law ($P \propto 1/V$), Charles’s Law ($V \propto T$), and Gay-Lussac’s Law ($P \propto T$) into a single equation.

The Combined Gas Law

$$ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $$

This equation applies when a fixed amount of gas undergoes a change in conditions. Note that units for $P$ and $V$ can be anything (as long as they are consistent on both sides), but $T$ must be in Kelvin.

Practice Problem: The Weather Balloon

Scenario: A helium weather balloon has a volume of $534 \, \text{L}$ at $14.9^\circ\text{C}$ and $755 \, \text{mmHg}$. It rises to an altitude where the pressure drops to $278 \, \text{mmHg}$ and the temperature is $-36.1^\circ\text{C}$. What is the new volume?

Solution:

Identify Variables:
- $P_1 = 755 \, \text{mmHg}$, $V_1 = 534 \, \text{L}$, $T_1 = 14.9 + 273.15 = 288.05 \, \text{K}$
- $P_2 = 278 \, \text{mmHg}$, $T_2 = -36.1 + 273.15 = 237.05 \, \text{K}$
- Unknown: $V_2$
Rearrange the Combined Gas Law for $V_2$:
$$ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} $$
Substitute and Solve:
$$ V_2 = 534 \, \text{L} \times \frac{755}{278} \times \frac{237.05}{288.05} $$
$$ V_2 \approx 534 \times 2.716 \times 0.823 \approx 1194 \, \text{L} $$

Result: The balloon expands significantly to approximately 1,190 L due to the drop in pressure, despite the cooling effect.

Mixtures of Gases: Dalton's Law and Gas Collection Over Water

In the real world, we rarely deal with pure gases. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual component:

$$ P_{total} = P_A + P_B + P_C + ... $$

Collecting Gas Over Water

A common laboratory technique involves collecting gas produced by a reaction by bubbling it through water into an inverted container. The gas collected is not pure; it is a mixture of the product gas and water vapor.

To find the pressure of the "dry" gas, you must subtract the vapor pressure of water at that specific temperature:

$$ P_{gas} = P_{total} - P_{H_2O} $$

Case Study: Reaction of Mg and HCl

Problem: Hydrogen gas is produced by the reaction $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$ and collected over water. The total pressure is $740.0 \, \text{mmHg}$ at $24.5^\circ\text{C}$. What is the pressure of the dry hydrogen gas?

Analysis:

Total Pressure ($P_{total}$) = 740.0 mmHg
Temperature = 24.5°C
Water Vapor Pressure at 24.5°C ($P_{H_2O}$) $\approx 23.8 \, \text{mmHg}$ (from standard tables)

Calculation:

$$ P_{H_2} = 740.0 \, \text{mmHg} - 23.8 \, \text{mmHg} = 716.2 \, \text{mmHg} $$

Insight: If you neglected to subtract the water vapor pressure, you would overestimate the amount of hydrogen produced by about 3%.

Kinetic Molecular Theory and Real Gas Behavior

The Ideal Gas Law assumes that gas particles have zero volume and experience no intermolecular forces. Kinetic Molecular Theory (KMT) explains why gases exert pressure (collisions) and how temperature relates to kinetic energy.

Root Mean Square Velocity ($u_{rms}$)

Temperature is a measure of average kinetic energy. However, lighter molecules move faster than heavier ones at the same temperature. The root mean square velocity is calculated as:

$$ u_{rms} = \sqrt{\frac{3RT}{M}} $$

Crucial Note on Units: Because energy involves Joules ($kg \cdot m^2/s^2$), you must use $R = 8.314 \, \text{J/mol K}$ and the molar mass $M$ must be in kg/mol (not g/mol).

Example: Speed of Hydrogen ($H_2$) at 30°C

$T = 303.15 \, \text{K}$
$M_{H_2} = 2.016 \, \text{g/mol} = 2.016 \times 10^{-3} \, \text{kg/mol}$
$u_{rms} = \sqrt{\frac{3(8.314)(303.15)}{2.016 \times 10^{-3}}} \approx 1,936 \, \text{m/s}$

Real Gases: The Van der Waals Equation

At high pressures or low temperatures, gases behave "non-ideally." Particles are crowded together (volume becomes significant) and move slowly enough to attract one another (intermolecular forces reduce pressure).

The Van der Waals equation corrects for these factors:

$$ \left( P + \frac{an^2}{V^2} \right) (V - nb) = nRT $$

Correction for Attraction ($a$): The term $\frac{an^2}{V^2}$ is added to the measured pressure. It accounts for the fact that attractive forces effectively reduce the pressure the gas exerts on the walls.
Correction for Volume ($b$): The term $nb$ is subtracted from the volume. It represents the "excluded volume"—the physical space the gas molecules occupy.

Comparison: Ideal vs. Real (n-Hexane)
Consider $n$-hexane at high pressure ($91$ bar) and high temperature ($660$ K).
Using the Ideal Gas Law, the molar volume is calculated as $0.603 \, \text{L/mol}$. However, using the Van der Waals equation, the molar volume is $0.391 \, \text{L/mol}$.
Why the difference? At 91 bar, the gas molecules are pushed close together. The attractive forces (represented by the large '$a$' value for hexane) pull the molecules tighter than predicted by ideal laws, significantly reducing the volume.

Key Takeaways

Always check your units: Temperature must be in Kelvin. For energy calculations (velocity), mass must be in kg.
Density depends on P and T: Gases are denser at high pressures and low temperatures. Formula: $\rho = PM/RT$.
Dalton's Law: When collecting gas over water, remember to subtract the vapor pressure of water from the total pressure.
Real vs. Ideal: The Ideal Gas Law is an approximation. It works best at low pressures and high temperatures. Under extreme conditions, use Van der Waals corrections.