Question

A weather balloon filled with He gas has a volume of \(2.00 \times 10^{3} \mathrm{m}^{3}\) at ground level, where the atmospheric pressure is 1.000 atm and the temperature \(27^{\circ}\) C. After the balloon rises high above Earth to a point where the atmospheric pressure is 0.340 atm, its volume increases to \(5.00 \times 10^{3} \mathrm{m}^{3} .\) What is the temperature of the atmosphere at this altitude?

Short answer

The temperature of the atmosphere at the altitude where the atmospheric pressure is 0.34 atm is approximately \( -18.04^{\circ}C\)

Step-by-step solution

Initial Temperature Conversion

Convert the initial temperature from Celsius to Kelvin by using the conversion equation \(K = C + 273.15\). Thus, the initial temperature in Kelvin is \(27^{\circ}C + 273.15K = 300.15K\).

Apply the Ideal Gas Law

Since the number of moles and the gas constant don't change, and we can equate the initial and final states of the balloon using the equation \(P_1V_1/T_1 = P_2V_2/T_2\), where 1 refers to the initial state and 2 refers to the final.

Solve for Final Temperature

We can rearrange the equation from step 2 to solve for \(T_2\) giving us the equation \(T_2 = P_2V_2T_1/(P_1V_1)\). Substituting the given values into this equation gives us the final temperature in Kelvin as \(T_2 = (0.340\, atm \times 5000\,m^3 \times 300.15\,K) / (1.000\, atm \times 2000\,m^3) = 255.1125\,K\).

Final Temperature Conversion

Convert the final temperature from Kelvin back to Celsius using the conversion equation \(C = K - 273.15\), giving a final temperature of \(255.1125\,K - 273.15 = -18.04^{\circ}C.\)

Key concepts

Gas Laws

Gas Laws help us understand the behavior of gases under various conditions. These laws form the backbone of many scientific calculations involving gases. The

• Ideal Gas Law equation is expressed as: \(PV = nRT\), where:
• \(P\) stands for pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
• The Ideal Gas Law helps predict changes in pressure, volume, and temperature.

When solving problems involving a fixed amount of gas as seen in the original exercise, it is often useful to use the rearranged form of the Ideal Gas Law: \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\). This helps in understanding how gases expand or contract due to changes in atmospheric conditions. In real-life applications, such as a weather balloon rising into the atmosphere, changes in external pressure and temperature drastically affect the balloon's volume. Understanding these concepts can aid in accurately predicting such changes.

Temperature Conversion

Temperature Conversion is crucial when dealing with gas laws because gas law equations require temperatures to be in Kelvin.

• Celsius to Kelvin conversion: \(K = C + 273.15\)
• Kelvin to Celsius conversion: \(C = K - 273.15\)

This is necessary because Kelvin is an absolute scale, allowing for a uniform platform to measure thermal energy changes. For instance, in the original exercise, the initial temperature is converted from 27°C to 300.15 K to align with the requirements of the gas law formula. The final temperature is also converted back to Celsius to provide a more relatable measure for everyday use. Paying attention to these conversions ensures accuracy and consistency in all gas-related calculations.

Atmospheric Pressure

Atmospheric Pressure refers to the force exerted by the weight of the air above us. This force changes with altitude and impacts how gases behave.

• Standard Atmospheric Pressure at sea level is 1.000 atm, or 101.3 kPa.
• Pressure decreases as altitude increases because there is less air above exerting weight.

In the weather balloon problem, atmospheric pressure is a critical variable, causing the balloon to expand as it rises to higher altitudes with lower pressure, namely 0.340 atm in the final state. This concept helps us comprehend how pressure differences encountered by ascending balloons can trigger significant changes in the volume and eventually affect the gas temperatures inside them. Atmospheric pressure is one of the driving factors affecting weather conditions and thus important in meteorology as well.

Helium Gas

Helium Gas is a lighter-than-air, noble gas often used in balloons due to its low density and non-reactive properties. Noble gases like helium are notable for their stability because of their full outer electron shells.

• Helium is non-flammable and safe, making it ideal for applications such as weather balloons.
• Its low density compared to air allows helium-filled balloons to easily rise when released.

In the context of the original problem, helium acts as the filling gas in the weather balloon, which expands as it reaches higher altitudes where the pressure is lower. This property facilitates the use of helium in atmospheric studies and weather forecasting, providing vital data for understanding climate patterns. Its behavior under changing pressure and temperature is reliably predictable using the Ideal Gas Law, enhancing its utility in scientific exploration.