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 ' \(\mathrm{C}\) and the barometric pressure is 730 torr. The balloon then rises to an altitude of \(6000 \mathrm{ft},\) 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 \(6000 \mathrm{ft} ?\)

Short answer

The change in volume of the balloon as it ascends from sea level to 6000 ft is \(20\ L\).

Step-by-step solution

Convert temperatures to Kelvin

To start off, we need to convert the given temperatures from degrees Celcius to Kelvin (K) because the Ideal Gas Law uses the absolute temperature scale (Kelvin). The temperature T in Kelvin is calculated as \(T(K) = T(^{\circ}\mathrm{C}) + 273\). For the initial temperature, \(T_1 = 25^{\circ}\mathrm{C} = 25 + 273 = 298\ K\). For the final temperature, \(T_2 = 15^{\circ}\mathrm{C} = 15 + 273 = 288\ K\).

Convert the pressures to atmospheres

The given pressures need to be converted from torr to atmospheres, since the gas constant R used in the Ideal Gas Law was defined using pressure in atmospheres. The pressure P in atmospheres is calculated as \(P(\mathrm{atm}) = P(\mathrm{torr}) / 760\). For the initial pressure, \(P_1 = 730\ \mathrm{torr} = 730 / 760 = 0.961\ \mathrm{atm}\). For the final pressure, \(P_2 = 605\ \mathrm{torr} = 605 / 760 = 0.796\ \mathrm{atm}\).

Apply Ideal Gas Law

According to the Ideal Gas Law, \(P_1V_1 / T_1 = P_2V_2 / T_2\), where 1 refers to the initial state at sea level, and 2 refers to the final state at 6000ft. The initial volume \(V_1 = 855\ L\). We want to solve for the final volume \(V_2\). Rearranging the equation gives us: \(V_2 = P_1V_1T_2 / (P_2T_1)\). Substituting the values gives: \(V_2 = 0.961 \times 855 \times 288 / (0.796 \times 298) = 875\ L\).

Calculate the change in volume

The change in volume is the final volume minus the initial volume: \(\Delta V = V_2 - V_1 = 875\ L - 855\ L = 20\ L\). This is the volume increase as the balloon ascends from sea level to 6000ft. So, the change in volume of the balloon as it ascends from sea level to 6000 ft is 20 L.

Key concepts

Gas Volume Changes

Understanding how gas volume changes under different conditions is essential in the field of chemistry and physics. According to the Ideal Gas Law, gas volume is directly proportional to its temperature (when measured in Kelvin) and inversely proportional to its pressure. This means if the temperature increases while the pressure remains constant, the volume will increase. Conversely, if the pressure increases while the temperature remains constant, the volume will decrease.

For instance, in our weather balloon scenario, as the balloon rises, the external pressure decreases. However, because the temperature also drops, we need to use the Ideal Gas Law to find out the overall effect on volume. This demonstrates that gas volume changes are not always straightforward and require calculation to understand fully.

Temperature Conversion

Temperature plays a crucial role in calculations involving gas laws. However, not all temperature scales are created equally for such tasks. The Ideal Gas Law, for instance, requires temperatures to be in the Kelvin scale, as it is an absolute temperature scale with no negative values. This is crucial because volume and pressure are not negative in nature.

To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This shift aligns the scale so that 0 Kelvin (absolute zero) is the point at which particles theoretically have minimal thermal motion. Remember, when working with the Ideal Gas Law, always conducting temperature conversion to Kelvin ensures that the resulting calculations are accurate and meaningful.

Pressure Conversion

Pressure conversion is key when working with gases because different systems of measurement can be used. In the Ideal Gas Law, the pressure should be consistent with the gas constant used, often requiring pressure to be in atmospheres (atm). As such, converting from other units, like torr or millimeters of mercury (mmHg), to atmospheres becomes necessary.

The conversion factor from torr to atm is approximately 1 atm per 760 torr, derived from the fact that normal atmospheric pressure at sea level can support a column of mercury approximately 760 mm high. Therefore, when you come across pressure in torr in exercises, divide it by 760 to convert it to atmospheres before using it in gas law equations.

Kelvin Temperature Scale

The Kelvin temperature scale is integral to understanding and calculating gas behaviors. It is an absolute scale with its zero point, 0 Kelvin, representing absolute zero, theoretically the lowest temperature possible where all kinetic motion of particles stops. This scale was designed precisely for thermodynamic calculations because it avoids negative numbers, providing a direct correlation between energy and temperature.

When dealing with gases, incrementing the temperature by 1 Kelvin is equivalent to incrementing by 1 degree Celsius, but the baseline (0 K) is different. The Kelvin scale is preferred in scientific calculations, like those using the Ideal Gas Law, because it simplifies the mathematical relationships between temperature, pressure, and volume. Thus, the accurate application of the Kelvin scale is paramount for predicting how a gas will behave under varying pressure and temperature conditions.