Gases and the Ideal Gas Law: Pressure and Volume Changes in Weather Balloons

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. According to the gas laws, if the temperature also changes, the volume of the balloon will adjust to maintain the relationship described by the combined gas law.

• Combined Gas Law: When the amount of gas is constant, the relationship between pressure, volume, and temperature is given by:

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

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

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

Application: Weather Balloon Rising

• As the balloon rises, atmospheric pressure decreases.

• If the temperature remains constant, the volume of the balloon increases to compensate for the lower pressure (Boyle's Law).

• If temperature also changes, the combined gas law must be used to predict the new volume.

Example Calculation

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

• Given: $P_1 = 1.00\ \text{atm}$, $V_1 = 2.0\ \text{L}$, $T_1 = 300\ \text{K}$

• $P_2 = 0.50\ \text{atm}$, $T_2 = 250\ \text{K}$, $V_2 = ?$

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

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. According to the gas laws, if the temperature also changes, the volume of the balloon will adjust to maintain the relationship described by the combined gas law. Combined Gas Law: When the amount of gas is constant, the relationship between pressure, volume, and temperature is given by: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ P1, V1, T1: Initial pressure, volume, and temperature P2, V2, T2: Final pressure, volume, and temperature Application: Weather Balloon Rising As the balloon rises, atmospheric pressure decreases. If the temperature remains constant, the volume of the balloon increases to compensate for the lower pressure (Boyle's Law). If temperature also changes, the combined gas law must be used to predict the new volume. Example Calculation Problem: A weather balloon has a volume of 2.0 L at a pressure of 1.00 atm and a temperature of 300 K. As it rises, the pressure drops to 0.50 atm and the temperature decreases to 250 K. What is the new volume of the balloon? Given: $P_1 = 1.00\ \text{atm}$, $V_1 = 2.0\ \text{L}$, $T_1 = 300\ \text{K}$ $P_2 = 0.50\ \text{atm}$, $T_2 = 250\ \text{K}$, $V_2 = ?$ 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