Question

A weather balloon is filled with helium to a volume of \(0.275 \mathrm{~m}^{3}\) at \(22^{\circ} \mathrm{C}\) and \(10^{5} \mathrm{~Pa}\). The balloon ascends to an altitude where the pressure is \(0.64 \times 10^{5} \mathrm{~Pa}\) and the temperature is \(241 \mathrm{~K}\left(-32{ }^{\circ} \mathrm{C}\right) .\) What is the volume of the balloon at this altitude?

Short answer

The volume of the balloon at altitude is approximately 0.351 m³.

Step-by-step solution

Understand the Problem

We need to find the new volume of the weather balloon when it ascends to a higher altitude. The initial conditions are given by the initial volume, temperature, and pressure of the balloon. The final conditions include a new temperature and pressure at the altitude. We'll use the ideal gas law to find the new volume, assuming the amount of gas remains constant.

Apply the Combined Gas Law

The Combined Gas Law is \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]where- \(P_1 = 10^5 \mathrm{~Pa}\), \(V_1 = 0.275 \mathrm{~m}^3\), and \(T_1 = 22^{\circ} \mathrm{C} = 273 + 22 = 295 \mathrm{~K}\)- \(P_2 = 0.64 \times 10^5 \mathrm{~Pa}\), and \(T_2 = 241 \mathrm{~K}\)We need to solve for \(V_2\).

Rearrange the Equation and Solve for V2

Rearrange the Combined Gas Law to solve for \(V_2\):\[ V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} \]Substitute in the values:\[ V_2 = \frac{(10^5 \mathrm{~Pa})(0.275 \mathrm{~m}^3)(241 \mathrm{~K})}{(295 \mathrm{~K})(0.64 \times 10^5 \mathrm{~Pa})} \]

Calculate the Final Volume V2

Carry out the calculation:\[ V_2 = \frac{(10^5)(0.275)(241)}{(295)(0.64 \times 10^5)} \]Simplify:\[ V_2 = \frac{27500 \times 241}{295 \times 64000} \]\[ V_2 = \frac{6627500}{18880000} \] \[ V_2 \approx 0.351 \mathrm{~m}^3 \]So, the new volume of the balloon is approximately \(0.351 \mathrm{~m}^3\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is one of the fundamental concepts in chemistry and physics that relates the pressure, volume, and temperature of a gas to the number of molecules it contains. This law is commonly expressed by the equation \( PV = nRT \), where:

• \( P \) stands for pressure.
• \( V \) represents volume.
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant.
• \( T \) denotes temperature in Kelvin.

This equation gives us an understanding of how changes in one property, such as temperature, will impact the other properties like volume. The Ideal Gas Law is an extension of simple gas laws combining Boyle's Law, Charles's Law, and Avogadro's Law under one equation. In practice, if you have constant amount of a gas and either temperature or pressure changes, you can predict changes in volume as long as the gas behaves ideally—meaning gas particles do not interact and occupy no volume.
In the context of our exercise, once we know the initial conditions of the gas in the balloon and then account for changes in pressure and temperature during ascent, we can calculate its new volume using the Combined Gas Law, a derivative calculation of the Ideal Gas Law.

Gas Volume Calculation

Calculating gas volumes under changing conditions involves the application of the Combined Gas Law. This formulation is especially handy when transformations involve constant quantity of gas molecules. In calculations like these, we use:

• \( P_1, V_1, T_1 \): initial pressure, volume, and temperature.
• \( P_2, V_2, T_2 \): final pressure, volume, and temperature.

The Combined Gas Law formula \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) comes about by rearranging the Ideal Gas Law for constant moles \( n \). This equation allows us to solve for the unknown variable—in this case, the final volume \( V_2 \).
In our exercise, after identifying known parameters \((P_1 = 10^5 \text{ Pa}, V_1 = 0.275 \text{ m}^3, T_1 = 295 \text{ K})\) and \((P_2 = 0.64 \times 10^5 \text{ Pa}, T_2 = 241 \text{ K})\), we solve for \( V_2 \).
The process of substitution and calculation reveals our final result of \( V_2 \approx 0.351 \text{ m}^3 \), indicating how much the balloon expands when ascending to the new conditions.

Altitude Effects on Gas

As a weather balloon ascends into the higher altitudes of Earth's atmosphere, several changes occur, primarily in pressure and temperature. High altitudes generally mean lower atmospheric pressure and temperature.
As pressure decreases with altitude, a gas will expand if the temperature stays constant, according to Boyle's Law. However, since both pressure and temperature change with altitude, we must utilize the Combined Gas Law to compute these changes accurately.

• Reduced pressure can lead to increased volume, assuming temperature does not decrease drastically.
• Temperature also influences volume; lower temperatures tend to decrease gas volume unless compensated by an equivalent reduction in pressure.

In more scientific terms, the particles of gas in the balloon are less compressed and spread out more as the surrounding pressure decreases. The exercise example, with a calculated increase in balloon volume with reduced pressure and lower temperatures, helps illustrate these principles. Understanding altitude effects on gases is crucial for applications in meteorology, aviation, and even biology where gas volumes affect bodily functions at high altitudes.