The Combined Gas Law: Relationships Between Pressure, Volume, and Temperature

The Combined Gas Law

Introduction to the Combined Gas Law

The Combined Gas Law is a fundamental principle in chemistry that relates the pressure, volume, and temperature of a fixed amount of gas. It is derived by combining three individual gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. This law is essential for understanding how gases behave under varying conditions of pressure, volume, and temperature.

• Boyle's Law: Pressure is inversely proportional to volume (at constant temperature).

• Charles's Law: Volume is directly proportional to temperature (at constant pressure).

• Gay-Lussac's Law: Pressure is directly proportional to temperature (at constant volume).

Mathematical Representation

The Combined Gas Law can be expressed mathematically as:

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

• P = Pressure (in atm, Pa, or any consistent unit)

• V = Volume (in L, mL, or any consistent unit)

• T = Temperature (in Kelvin, always use Kelvin for gas law calculations)

• Subscripts 1 and 2 refer to the initial and final states of the gas, respectively.

Derivation from Individual Gas Laws

The Combined Gas Law is derived by combining the following relationships:

Boyle's Law	Charles's Law	Gay-Lussac's Law
$P \propto \frac{1}{V}$ (at constant T)	$V \propto T$ (at constant P)	$P \propto T$ (at constant V)

By combining these proportionalities, we obtain the Combined Gas Law equation above.

Worked Example

Example: A sample of gas initially has a volume of 900 mL at 520 K and 1.85 atm. What is the pressure of the gas if the volume decreases to 330 mL, while the temperature increases to 770 K?

• Given: $V_1 = 900$ mL, $T_1 = 520$ K, $P_1 = 1.85$ atm

• Find: $P_2$ when $V_2 = 330$ mL, $T_2 = 770$ K

Using the Combined Gas Law:

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

Solving for $P_2$:

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

Plug in the values:

$P_2 = \frac{1.85 \times 900 \times 770}{520 \times 330} = 7.47$ atm

Practice Problems

1. Practice 1: A 4.30 L gas has a pressure of 7.0 atm when the temperature is 60.0 °C. What will be the temperature of the gas mixture if the volume and pressure are decreased to 2.45 L and 4.03 kPa respectively?

   • Answer: 1,783,154.1 K

2. Practice 2: A sealed container with a movable piston contains a gas with a pressure of 1380 torr, a volume of 820 mL, and a temperature of 31°C. What would the volume be if the new pressure is now 2.83 atm, while the temperature decreased to 25°C?

   • Answer: 2,481,881,700 mL

Key Points to Remember

• Always convert temperatures to Kelvin when using gas laws: $T(K) = T(°C) + 273.15$

• Units for pressure and volume must be consistent throughout the calculation.

• The Combined Gas Law is useful when the amount of gas is constant but pressure, volume, and temperature all change.