Question

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled with \(2.0 \mathrm{L}\) of helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only \(500 .\) mm Hg, will the balloon burst?

Short answer

Using the combined gas law, we calculate the final volume of the helium-filled balloon after rising to an altitude with atmospheric pressure of 500 mm Hg to be 3.04 L. Since this volume exceeds the maximum allowed volume of 2.5 L given by the manufacturer, the balloon will burst at that altitude.

Step-by-step solution

Identity the relevant information and equations

In this problem, we are given the initial volume (V1 = 2.0 L) of the helium-filled balloon, the maximum volume it can have (Vmax = 2.5 L), the initial pressure at sea level (P1 = 760 mm Hg) and the pressure at the higher altitude (P2 = 500 mm Hg). We will use the combined gas law, which states that the ratio of the initial pressure, volume, and temperature of a gas to its final pressure, volume, and temperature is constant as long as the amount of gas is unchanged. The combined gas law is given by: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] Since we are not given any temperature values, we can assume that the temperature remains constant throughout the process, which simplifies the equation to: \[P_1V_1 = P_2V_2\] In this step, we identified the relevant information and equations we will use to solve the problem.

Solve for the final volume (V2)

Now, we will rearrange the simplified combined gas law equation to solve for V2, the final volume of the balloon: \[V_2 = \frac{P_1V_1}{P_2}\] We can now plug in the given values and solve for V2: \[V_2 = \frac{(760\;\text{mm}\;\text{Hg})(2.0\;\text{L})}{(500\;\text{mm}\;\text{Hg})}\] \[V_2 = \frac{1520}{500}\] \[V_2 = 3.04\;\text{L}\] In this step, we solved for the final volume (V2) of the balloon after rising to the higher altitude.

Compare the final volume (V2) to the maximum volume (Vmax)

We found that the final volume of the balloon is 3.04 L. Now we need to compare this to the maximum volume the balloon can have without bursting, which is 2.5 L: \[3.04\;\text{L} > 2.5\;\text{L}\] Since the final volume (3.04 L) is greater than the maximum allowed volume (2.5 L), the balloon will burst when it reaches the altitude where the atmospheric pressure is 500 mm Hg. In this final step, we compared the final volume of the balloon to its maximum volume to determine whether it would burst at the higher altitude.

Key concepts

Balloon Volume Calculation

In the scenario of a helium-filled balloon, one essential calculation is determining how its volume changes as it rises. Initially, the balloon has a volume of 2.0 liters at sea level. Our task is to find out what happens to this volume when the balloon rises to an altitude where the atmospheric pressure is lower. This involves using the combined gas law. By calculating the final volume of the balloon using \[V_2 = \frac{P_1V_1}{P_2}\]we substitute in the values:

• Initial pressure, \(P_1 = 760\) mm Hg
• Initial volume, \(V_1 = 2.0\) L
• Final pressure, \(P_2 = 500\) mm Hg

The resulting calculation shows that the balloon's volume at the higher altitude increases to 3.04 liters. It's crucial to compare this with the balloon's maximum designed capacity, which is 2.5 liters. Since 3.04 liters exceeds the maximum, it signifies that the balloon will burst at this altitude.

Atmospheric Pressure Effects

Atmospheric pressure has a significant effect on balloon volume, especially when the balloon rises to higher altitudes. At sea level, standard atmospheric pressure is 760 mm Hg. As one ascends higher in altitude, the atmospheric pressure decreases because there is less air pressing down from above. In our problem, the pressure decreases to 500 mm Hg at the higher altitude.
Understanding how pressure affects the volume of a gas is key; as atmospheric pressure decreases, the volume of gas expands if the balloon can stretch. This is due to less force being exerted on the balloon from outside. It's as if the balloon "pushes" back to occupy more space.
However, a balloon has a limit on the amount it can expand before it bursts. In practice, these changes in atmospheric pressure must be considered when inflating a balloon to ensure it can withstand the conditions it may encounter, such as varying pressures with altitude changes.

Gas Laws

Gas laws are the foundation for understanding the behavior of gases under different conditions of temperature, pressure, and volume. In this exercise, the combined gas law is particularly helpful. This law establishes a proportional relationship between pressure, volume, and temperature for a given amount of gas when the latter remains constant.
The formula for the combined gas law is \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]. By assuming constant temperature, it simplifies the equation to \[P_1V_1 = P_2V_2\].This simplification allows us to solve for the unknown volume or pressure given the other parameters. It illustrates how, with a fixed amount of gas, any change in the product of pressure and volume on one side of the equation must result in an equivalent change on the other side, which is what determines whether the balloon bursts. Gas laws like this one provide an understanding of the physics behind everyday phenomena, such as how a balloon behaves as it rises through the atmosphere.