Gas Laws: Boyle's, Charles', and Combined Gas Law – Study Notes

Gas Laws

Introduction

Gas laws describe the relationships between pressure, volume, and temperature for a given amount of gas. Understanding these laws is fundamental in General Chemistry, as they allow prediction and calculation of gas behavior under varying conditions.

Unit Conversions for Pressure

• Atmosphere (atm): Standard unit of pressure.

• Torr: 1 atm = 760 torr.

• Kilopascal (kPa): 1 atm = 101.325 kPa.

• Example Conversion: To convert 648 torr to atm:

$648\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}} = 0.853\ \text{atm}$

Boyle's Law

Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature for a fixed amount of gas.

• Formula: $P_1V_1 = P_2V_2$

• Key Points:

  • If pressure increases, volume decreases (and vice versa), provided temperature and amount of gas remain constant.

• Example: A 3.50 L sample of hydrogen gas has a pressure of 0.750 atm. What is its volume if the pressure is increased to 1.50 atm at constant temperature?

$P_1V_1 = P_2V_2$ $(0.750\ \text{atm})(3.50\ \text{L}) = (1.50\ \text{atm})(V_2)$ $V_2 = 1.75\ \text{L}$

Charles' Law

Charles' Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure for a fixed amount of gas.

• Formula: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

• Key Points:

  • As temperature increases, volume increases (and vice versa), provided pressure and amount of gas remain constant.

  • Temperature must be in Kelvin for calculations.

• Example: A 0.684 L sample of helium gas, initially at 0°C (273 K), is heated at constant pressure. The final volume of the gas is 0.986 L. What is its final temperature in °C?

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ $\frac{0.684\ \text{L}}{273\ \text{K}} = \frac{0.986\ \text{L}}{T_2}$ $T_2 = 393\ \text{K}$ $T_2 = 393\ \text{K} - 273 = 120\ ^\circ\text{C}$

Combined Gas Law

The Combined Gas Law combines Boyle's and Charles' laws to relate pressure, volume, and temperature for a fixed amount of gas.

• Formula: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

• Key Points:

  • Used when a gas sample undergoes changes in pressure, volume, and temperature simultaneously.

  • Amount of gas (number of moles) must remain constant.

• Example: Calculate the volume of 0.896 g hydrogen gas that originally occupies 908 mL at 717 torr and 20°C. After its temperature and pressure are changed to 20°C and 1.07 atm, respectively.

$P_1 = 717\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}} = 0.943\ \text{atm}$ $V_1 = 908\ \text{mL} = 0.908\ \text{L}$ $T_1 = 20^\circ\text{C} = 293\ \text{K}$ $P_2 = 1.07\ \text{atm}$ $T_2 = 293\ \text{K}$ $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ $\frac{0.943\ \text{atm} \times 0.908\ \text{L}}{293\ \text{K}} = \frac{1.07\ \text{atm} \times V_2}{293\ \text{K}}$ $V_2 = \frac{0.943 \times 0.908}{1.07} = 0.801\ \text{L}$

Summary Table: Gas Law Relationships

Law	Relationship	Formula	Constant
Boyle's Law	$P \propto \frac{1}{V}$	$P_1V_1 = P_2V_2$	Temperature, moles
Charles' Law	$V \propto T$	$\frac{V_1}{T_1} = \frac{V_2}{T_2}$	Pressure, moles
Combined Gas Law	All variables	$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$	Moles

Additional info: For all gas law calculations, temperature must be in Kelvin, and pressure and volume units must be consistent.