Question

A meteorological balloon contains 250.0 L He at 22°C and 740.0 mm Hg. If the volume of the balloon can vary according to external conditions, what volume would it occupy at an altitude at which the temperature is -52°C and the pressure is 0.750 atm?

Short answer

The volume of the balloon at the new conditions is approximately 243.4 L.

Step-by-step solution

Convert temperatures to Kelvin

Initially, convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \] For the initial temperature, 22°C: \[ T_1 = 22 + 273.15 = 295.15 \text{ K} \] For the final temperature, -52°C: \[ T_2 = -52 + 273.15 = 221.15 \text{ K} \]

Convert pressures to the same units

Ensure the pressures are in the same units. Here, convert the initial pressure from mm Hg to atm using the conversion: \[ 1 \text{ atm} = 760 \text{ mm Hg} \] Thus, initial pressure: \[ P_1 = \frac{740.0 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.974 \text{ atm} \] The final pressure is already given in atm: \[ P_2 = 0.750 \text{ atm} \]

Apply the combined gas law

Use the combined gas law to find the new volume: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Rearrange this to solve for the final volume, \( V_2 \): \[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \] Substitute the known values: \[ V_2 = \frac{0.974 \text{ atm} \times 250.0 \text{ L} \times 221.15 \text{ K}}{0.750 \text{ atm} \times 295.15 \text{ K}} \]

Calculate the final volume

Perform the calculation to determine \( V_2 \): \[ V_2 = \frac{0.974 \times 250.0 \times 221.15}{0.750 \times 295.15} \approx 243.4 \text{ L} \]

Key concepts

Gas Laws

Gas laws are crucial in helping us understand how gases behave under different conditions of temperature, pressure, and volume. Among the many gas laws, the Combined Gas Law integrates Charles' Law, Boyle's Law, and Gay-Lussac's Law. This combined formula helps predict how a gas will respond when any two of these three properties are changed.
Charles' Law states that the volume of a gas is directly proportional to its absolute temperature when pressure is constant. Boyle’s Law notes an inverse relationship between volume and pressure at a constant temperature. Lastly, Gay-Lussac’s Law shows that pressure is directly proportional to temperature when the volume is held constant.
The Combined Gas Law is expressed mathematically as:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] This formula makes it possible to compute an unknown variable when shifting between different conditions.

Temperature Conversion

Temperature conversion is a critical step when solving problems related to gas laws. Typically, temperatures need to be converted to Kelvin for these computations. Kelvin is a thermodynamic temperature scale where absolute zero causes all kinetic motion to halt.
To convert Celsius to Kelvin, the formula is simple:
\[ T(K) = T(\degree C) + 273.15 \]
In the given problem, the initial temperature is 22°C, hence when converted to Kelvin:
\[ T_1 = 22 + 273.15 = 295.15 \text{ K} \] And for the final temperature, -52°C:
\[ T_2 = -52 + 273.15 = 221.15 \text{ K} \]

Pressure Conversion

Converting pressure to the same unit across all measurements is crucial for accurate calculations. In the provided problem, pressures were given in different units. Converting mm Hg to atm ensures consistency. The conversion factor is:
\[ 1 \text{ atm} = 760 \text{ mm Hg} \]
The initial pressure provided in mm Hg is converted as follows:
\[ P_1 = \frac{740.0 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.974 \text{ atm} \]
The final pressure is already in atm (0.750 atm), and no further conversion is needed for that value.

Volume Calculations

Finally, let's focus on volume calculations using the Combined Gas Law. Once temperatures and pressures are in appropriate units (Kelvin and atm), we apply the formula
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Let's solve for the unknown final volume (\