Question

A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is \(25^{\circ} \mathrm{C}\) and the barometric pressure is 730 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to 6000 ft?

Short answer

The change in volume of the weather balloon as it ascends from sea level to 6000 ft is 167.32 L.

Step-by-step solution

Convert temperatures into Kelvin

To work with the gas laws, we should convert the given temperatures in Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperatures: T1 = \(25^{\circ} \mathrm{C} + 273.15 = 298.15\ \mathrm{K}\) T2 = \(15^{\circ} \mathrm{C} + 273.15 = 288.15\ \mathrm{K}\)

Convert pressures into atmospheres

To simplify our calculations, we should convert the given pressures from torr to atmospheres (atm). To do this, we divide the torr values by 760: P1 = 730 torr / 760 = 0.9605 atm P2 = 605 torr / 760 = 0.7961 atm

Setup the combined gas law equation

Now, we have T1, T2, P1, and P2 in the appropriate units. We can now use the combined gas law equation, remembering that V1 = 855 L: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)

Solve the equation for the final volume (V2)

Now we need to solve for V2. We can do this by rearranging the combined gas law equation and plugging in the known values: \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) Plug in all values: \(V_2 = \frac{(0.9605\ \mathrm{atm})(855\ \mathrm{L})(288.15\ \mathrm{K})}{(0.7961\ \mathrm{atm})(298.15\ \mathrm{K})}\)

Calculate the final volume (V2)

Perform the above calculation to find V2: \(V_2 = 1022.32\ \mathrm{L}\)

Calculate the change in volume

The change in volume is the final volume minus the initial volume: ΔV = V2 - V1 = 1022.32 L - 855 L = 167.32 L As the balloon ascends from sea level to 6000 ft, its volume increases by 167.32 L.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that describes the behavior of gases. It relates the four important properties of a gas: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas, often represented as the number of moles (\(n\)). The ideal gas law formula is given by:\[PV = nRT\]where \(R\) is the universal gas constant.
The Ideal Gas Law helps us predict how a gas will behave under different conditions, such as changing temperature or pressure. In many practical situations, like in this problem involving a helium balloon, we assume that gases behave ideally, which grants us a simpler way to estimate changes in a gas' properties.
In scenarios where the number of moles does not change, and instead only the volume, temperature, and pressure are altered, the Combined Gas Law comes into play, maintaining the core concepts of the Ideal Gas Law.

Volume and Pressure Relationship

The relationship between pressure and volume of a gas is described by Boyle's Law, which states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. In simpler terms, as the pressure on a gas increases, its volume decreases, and vice versa.

• This can be mathematically expressed as: \[PV = ext{constant}\]
• When using the Combined Gas Law in our problem, we see that pressure and volume are linked through \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), meaning changes in pressure and temperature will affect volume.

When solving gas-related problems, understanding this relationship helps us manipulate and predict the final volume a gas occupies when pressure conditions change, such as when a weather balloon rises to a higher altitude.

Temperature and Gas Behavior

The behavior of gas molecules is significantly influenced by changes in temperature. When temperature increases, the kinetic energy of gas molecules also increases, causing them to move more vigorously, which can lead to an increase in the volume of the gas if the pressure is constant.
For calculations involving gases, it’s crucial to use Kelvin (\(K\)), the absolute temperature scale. The conversion from Celsius to Kelvin is straightforward: just add 273.15 to the Celsius temperature.

• For instance, in our balloon problem, we converted 25°C to 298.15 K.
• Temperature changes are directly handled in the Combined Gas Law: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).

Recognizing how temperature influences volume and pressure through kinetic energy allows us to predict changes in gaseous systems effectively.

Helium Gas Properties

Helium is a noble gas known for its light weight and non-reactive nature. These properties make it an ideal choice for filling balloons, especially weather and scientific balloons. When helium is used in a weather balloon, like in this exercise, it aids in illustrating a larger volume expansion compared to heavier gases.

• Unlike heavier gases, helium rises in the atmosphere because it is less dense than the surrounding air.
• This lower density helps with reaching higher altitudes, making helium balloons useful for high-altitude observations.
• Moreover, as a monatomic gas, helium exhibits ideal gas behavior more closely compared to polyatomic gases!

Thus, understanding helium's properties helps in comprehending why it demonstrates significant volume changes under differing atmospheric conditions.