Question

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

Short answer

The number of moles of helium in the balloon is approximately 0.000807 mol. At an altitude with atmospheric pressure 500 mmHg, the volume of the balloon becomes approximately 3.04 L. Since this volume is greater than the manufacturer's maximum allowable volume of 2.5 L, the balloon will burst.

Step-by-step solution

Determine the moles of Helium in the Balloon

We are given the volume of helium in the balloon: V = 2.0 L. The pressure P at sea level is 760 mmHg. In terms of atmospheres, P = (760 mmHg) (1 atm/760 mmHg) = 1 atm. We also need to convert the volume from L to m^3: V = (2.0 L) (10^-3 m^3/L) = 2.0 x 10^-3 m^3. We are given that the temperature is constant, so we can approximate the temperature as the standard temperature, which is T = 298 K. We need to find the number of moles of helium, n, using the ideal gas law: \(PV=nRT\). Rearrange the equation to solve for n: n = PV/RT.

Calculate the Number of Moles of Helium

Using the given values P = 1 atm, V = 2.0 x 10^-3 m^3, R = 8.314 J/mol.K, and T = 298 K n = (1 atm x 2.0 x 10^-3 m^3) / (8.314 J/mol.K x 298 K) ≈ 0.000807 mol

Calculate the Volume of the Balloon at 500 mmHg

The pressure at the altitude is given as 500 mmHg. We will need to convert this to atm: P2 = (500 mmHg) (1 atm/760 mmHg) ≈ 0.658 atm. We assume that the temperature remains constant, so we can use the ideal gas law to find the new volume V2. Since n and T are constant, we can rearrange the ideal gas law as\(\frac{P_1V_1}{P_2} = V_2\). Now, we can calculate the volume at the altitude: V2 = (1 atm × 2.0 x 10^-3 m^3) / 0.658 atm ≈ 3.04 x 10^-3 m^3 Convert this value back to liters: V2 = 3.04 x 10^-3 m^3 x (10^3 L/m^3) = 3.04 L.

Determine If the Balloon Will Burst

Since the volume of the balloon at the given altitude exceeds the manufacturer's maximum volume (3.04 L > 2.5 L), the balloon will burst.

Key concepts

Gas Laws

Understanding the behavior of gases in various conditions is essential for many scientific and practical applications. The gas laws provide this understanding by describing how the variables of pressure, volume, and temperature relate to the amount of gas present.

One fundamental equation that brings these variables together is the ideal gas law, represented as \(PV = nRT\). In this equation, \(P\) stands for pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) represents the temperature in Kelvins.

The ideal gas law assumes that gases are composed of particles that are infinitely small and do not interact with each other. While real gases don't always perfectly follow these assumptions, the ideal gas law is a good approximation under many conditions, particularly at low pressure and high temperature.

In the context of the exercise provided, the ideal gas law is applied to calculate the number of moles of helium initially in the balloon, and then to find out how the volume changes as the atmospheric pressure decreases when the balloon rises. This direct application of the ideal gas law is crucial in predicting the behavior of the balloon under different atmospheric pressures while keeping the temperature constant.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the Earth's atmosphere on any surface it comes into contact with. It is usually measured in units such as millimeters of mercury (mmHg) or atmospheres (atm), among others. At sea level, standard atmospheric pressure is defined as 760 mmHg or 1 atm.

Atmospheric pressure is not constant; it decreases as altitude increases because the atmosphere becomes less dense and there are fewer air molecules to exert pressure. This is a critical concept when considering the expansion of a helium balloon released from sea level, as experienced in our exercise.

As the balloon rises to higher altitudes, it encounters lower atmospheric pressure, which allows the gas inside the balloon to expand because less external pressure is opposing the gas's expansion. This principle is integral to understanding why the balloon's volume increases as it ascends and ultimately helps us predict whether or not the balloon will burst at a specific altitude given the maximum volume specified by its manufacturer.

Helium Balloon Expansion

Helium balloon expansion is a result of the delicate balance between the internal pressure exerted by the helium gas and the external atmospheric pressure pressing against the balloon's surface. When a balloon filled with helium rises in the atmosphere, it expands because the atmospheric pressure outside decreases, reducing the force on the balloon's exterior.

Following the ideal gas law as the balloon's ascent leads to lower atmospheric pressures, the same amount of gas (in moles) will occupy a larger volume if the temperature remains constant, as demonstrated in the exercise provided. This expansion phenomenon is based on Boyle's Law, one of the specific gas laws, which states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with the pressure.

The importance of understanding helium balloon expansion cannot be understated, especially in contexts such as meteorological data collection using weather balloons, which have to withstand significant changes in atmospheric pressure without bursting. The insights from the ideal gas law also have broader implications in engineering, environmental studies, and even in medical applications such as the use of breathing gases in healthcare.