Gases and the Ideal Gas Law: Pressure and Volume Changes in the Atmosphere

Chapter 8: Gases

The Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and amount of a gas. It is commonly used to predict the behavior of gases under different conditions.

• Equation: The Ideal Gas Law is expressed as:

$PV = nRT$

• P = Pressure (in atmospheres, atm)

• V = Volume (in liters, L)

• n = Amount of gas (in moles, mol)

• R = Universal gas constant ($0.0821\ \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}$)

• T = Temperature (in Kelvin, K)

Pressure and Volume Changes: Weather Balloon Example

As a weather balloon rises through the atmosphere, the external pressure decreases because atmospheric pressure is lower at higher altitudes. According to the gas laws, if the temperature remains constant, the volume of the balloon will increase as the pressure decreases.

• Boyle's Law: For a fixed amount of gas at constant temperature, pressure and volume are inversely related:

$P_1V_1 = P_2V_2$

• As P decreases, V increases (if T and n are constant).

Combined Gas Law

When both temperature and pressure change, the Combined Gas Law is used:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

• P1, V1, T1: Initial pressure, volume, and temperature

• P2, V2, T2: Final pressure, volume, and temperature

Calculation Example: Weather Balloon Rising

Problem: A weather balloon has a volume of 2.0 L at a pressure of 1.00 atm and a temperature of 300 K at ground level. As it rises, the pressure drops to 0.50 atm and the temperature drops to 250 K. What is the new volume of the balloon?

• Given:

  • $P_1 = 1.00$ atm

  • $V_1 = 2.0$ L

  • $T_1 = 300$ K

  • $P_2 = 0.50$ atm

  • $T_2 = 250$ K

  • $V_2 = ?$

Solution:

Use the combined gas law:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Solve for $V_2$:

$V_2 = \frac{P_1V_1T_2}{P_2T_1}$

Plug in the values:

$V_2 = \frac{(1.00\ \text{atm})(2.0\ \text{L})(250\ \text{K})}{(0.50\ \text{atm})(300\ \text{K})}$

$V_2 = \frac{500}{150} = 3.33\ \text{L}$

• Answer: The new volume of the balloon is 3.33 L.

Summary Table: Gas Law Relationships

Law	Variables Held Constant	Relationship	Equation
Boyle's Law	n, T	P ∝ 1/V	$P_1V_1 = P_2V_2$
Charles's Law	n, P	V ∝ T	$\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Gay-Lussac's Law	n, V	P ∝ T	$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
Combined Gas Law	n	Relates P, V, T	$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Ideal Gas Law	None	Relates P, V, n, T	$PV = nRT$

Additional info: Academic context and a sample calculation were added to provide a complete, self-contained explanation suitable for GOB Chemistry students.