Final Volume of Gas Expansion Examples

Understanding Final Volume of Gas After Expansion

The Ideal Gas Law, expressed as PV = nRT, allows us to determine the relationships between pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. When a gas expands, its volume changes while other conditions may also vary. In this guide, we will look at three practical examples of calculating the final volume of a gas after expansion, allowing readers to grasp the application of the Ideal Gas Law in real-world scenarios.

Example 1: Expanding Air in a Balloon

A common scenario involves a balloon filled with air. Suppose a balloon containing 2 liters of air at a pressure of 1 atm and a temperature of 298 K is taken outside where the pressure drops to 0.5 atm. We want to calculate the final volume of the balloon after expansion.

To solve this, we can use the Ideal Gas Law:

1. Initially, we have:

   • P1 = 1 atm
   • V1 = 2 L
   • T1 = 298 K

2. After moving outside:

   • P2 = 0.5 atm
   • T2 = 298 K (assuming temperature remains constant)

Using the formula derived from the Ideal Gas Law (P1V1/T1 = P2V2/T2), we can rearrange it to find V2:

• V2 = (P1 * V1 * T2) / (P2 * T1)
• V2 = (1 atm * 2 L * 298 K) / (0.5 atm * 298 K)
• V2 = (2 atm * 2 L) / 0.5 atm = 4 L

The final volume of the balloon after expansion is 4 liters.

Notes:

• This example assumes that the temperature remains constant (isothermal expansion).

Example 2: Compressed Gas in a Cylinder

In industrial applications, gases are often compressed for storage. Imagine a gas cylinder containing nitrogen at a pressure of 5 atm and a volume of 10 liters at 300 K. If the gas is allowed to expand to a pressure of 1 atm, what will be the new volume?

To calculate this, we apply the Ideal Gas Law again:

1. Initial Conditions:

   • P1 = 5 atm
   • V1 = 10 L
   • T1 = 300 K

2. Final Conditions:

   • P2 = 1 atm
   • T2 = 300 K (temperature is constant)

Using our rearranged formula:

• V2 = (P1 * V1 * T2) / (P2 * T1)
• V2 = (5 atm * 10 L * 300 K) / (1 atm * 300 K)
• V2 = (50 atm * 10 L) / 1 atm = 50 L

The final volume of the gas after expansion is 50 liters.

Notes:

• This scenario is also an isothermal process, and in practical applications, we often consider efficiency loss due to heat exchange.

Example 3: Helium Balloon Rising in the Atmosphere

Consider a helium balloon released from ground level, where it has a volume of 1 liter at a pressure of 1 atm and a temperature of 293 K. As the balloon rises, the pressure decreases to 0.8 atm. We want to find the volume of the balloon at this new pressure.

1. Initial Conditions:

   • P1 = 1 atm
   • V1 = 1 L
   • T1 = 293 K

2. Final Conditions:

   • P2 = 0.8 atm
   • T2 = 293 K (assuming constant temperature)

Using the Ideal Gas Law:

• V2 = (P1 * V1 * T2) / (P2 * T1)
• V2 = (1 atm * 1 L * 293 K) / (0.8 atm * 293 K)
• V2 = (1 atm * 1 L) / 0.8 atm = 1.25 L

The final volume of the helium balloon after rising is 1.25 liters.

Notes:

• As the balloon ascends, the temperature might decrease, but for simplicity, we assume it remains constant in this example.