Question

An aerosol can contains 400. mL of compressed gas at 5.20 atm. When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas in the plastic bag? Assume a constant temperature.

Short answer

The pressure of the gas inside the plastic bag is approximately 0.97 atm.

Step-by-step solution

Understand the problem and the given data

We have an initial volume (V1) of 400 mL, which is the volume of compressed gas inside the aerosol can. The initial pressure (P1) is 5.20 atm. When the gas is transferred to the plastic bag, it inflates to a final volume (V2) of 2.14 L. We need to find the final pressure (P2) of the gas inside the plastic bag. Note: It is important to convert all given volumes to the same units.

Convert initial volume to liters

To convert the initial volume from mL to L, we can use the following conversion factor: 1 L = 1000 mL Therefore, the initial volume (V1) in liters is: V1 = 400 mL × \( \frac{1 L}{1000 mL} \) = 0.4 L

Apply the ideal gas law

Since the temperature and the number of moles of gas remain constant, we can use the following equation, which is derived from the ideal gas law: \( \frac{P1 \times V1}{T1} = \frac{P2 \times V2}{T2} \) Since T1 = T2, the equation can be simplified to: P1 × V1 = P2 × V2 We know the values of P1, V1, and V2, so we can solve for P2: P2 = \( \frac{P1 \times V1}{V2} \)

Calculate the final pressure

Now, we can substitute the given values into the equation to find P2: P2 = \( \frac{5.20 \; atm \times 0.4 \; L}{2.14 \; L} \) ≈ 0.97 atm So, the pressure of the gas inside the plastic bag is approximately 0.97 atm.

Key concepts

Pressure

Pressure is the force exerted by gas particles when they collide with the walls of their container. In the context of the Ideal Gas Law, pressure is an essential variable to understand how gases behave under various conditions. In the given problem, we start with an initial pressure (P1) of 5.20 atm inside the aerosol can. As the gas expands into the plastic bag, the volume increases, and the pressure decreases accordingly, assuming temperature remains constant. This relationship is represented by Boyle's Law, which states that the pressure of a gas is inversely related to its volume when temperature is constant.

The final step is to calculate the new pressure (P2) in the plastic bag. This is done using the formula derived from the Ideal Gas Law, where the Pressure-Volume Relationship can be simplified to:

• P1 x V1 = P2 x V2

This formula helps us understand how changes in volume can affect pressure, given a constant temperature. By substituting known values, the pressure in the bag is found to be approximately 0.97 atm.

Volume Conversion

When working with gases, converting volumes into consistent units is crucial. In this exercise, the gas initially occupies a volume of 400 mL inside the aerosol can. But, when dealing with equations rooted in the Ideal Gas Law, we need these volumes in liters to work consistently with common pressure-volume equations.

1 liter (L) equals 1000 milliliters (mL). To make the conversion from milliliters to liters, divide the volume in milliliters by 1000:

• 400 mL = 0.4 L

By converting the initial volume to 0.4 L, all volumes in the equation are comparable. It ensures accurate calculations when determining the new pressure after expansion. Incorrect volume units can lead to inaccuracies in calculations, so always check units carefully.

Constant Temperature

A key assumption in this problem is that the temperature remains constant as the gas moves from the aerosol can to the plastic bag. This assumption is central to applying Boyle's Law, which allows us to link changes in pressure and volume directly, without worrying about temperature changes.

When temperature is constant, we can simplify the Ideal Gas Law to a form that relates only pressure and volume:

• T1 = T2 ➔ so, P1 x V1 = P2 x V2

No temperature terms appear in the equation because temperature's constancy ensures that it doesn't affect the relationship of pressure and volume. In real-life scenarios, especially when gases undergo quick expansions or compressions, temperature changes could impact outcomes. But in this controlled problem, focusing on pressure and volume gives clear results.