Question

A balloon occupies a volume of \(1.25 \mathrm{~L}\) at sea level where the ambient pressure is 1 atm. What volume would the balloon occupy at an altitude of \(10,700 \mathrm{~m}\), where the air pressure is only \(220 \mathrm{mmHg}\) ?

Short answer

The balloon's volume at 10,700 m is approximately 4.32 L.

Step-by-step solution

Understand the Problem

We have a balloon with an initial volume of \(1.25\,\text{L}\) at sea level, where the pressure is \(1\,\text{atm}\). We need to find out what the volume of the balloon would be at an altitude of \(10,700\,\text{m}\) where the pressure is \(220\,\text{mmHg}\).

Convert Pressures

Convert the air pressure at \(10,700\,\text{m}\) from mmHg to atm. Knowing that \(1\,\text{atm} = 760\,\text{mmHg}\), we have:\[P_2 = \frac{220\,\text{mmHg}}{760\,\text{mmHg/atm}} \approx 0.2895\,\text{atm}.\]

Apply Boyle's Law

Use Boyle's Law which states that \(P_1V_1 = P_2V_2\), where \(P\) is pressure and \(V\) is volume. Given \(P_1 = 1\,\text{atm}\), \(V_1 = 1.25\,\text{L}\), and \(P_2 = 0.2895\,\text{atm}\), solve for \(V_2\):\[V_2 = \frac{P_1V_1}{P_2} = \frac{1 \times 1.25}{0.2895} \approx 4.32\,\text{L}.\]

Conclusion

Based on Boyle's Law, the balloon will expand as the pressure decreases. The volume of the balloon at an altitude of \(10,700\,\text{m}\) will be approximately \(4.32\,\text{L}\).

Key concepts

Pressure Conversion

When comparing gas behavior at different altitudes or conditions, pressure conversion is crucial. In the exercise, we deal with two pressure values: 1 atm at sea level and 220 mmHg at an altitude of 10,700 m. As pressure directly impacts a gas's volume, converting units to a common measure, like atmospheres (atm), is necessary for accurate calculations.

Here's how you perform the conversion: atmospheric pressure, measured in mmHg, can be converted to atm using the relationship where 1 atm is equivalent to 760 mmHg. Therefore, we divide the pressure at altitude in mmHg by 760:

• Given: 220 mmHg
• Conversion: \(P_2 = \frac{220\,\text{mmHg}}{760\,\text{mmHg/atm}}\approx 0.2895\,\text{atm}\)

By converting to atm, we ensure consistency for further calculations involving Boyle's Law. This method of conversion is standard in facilitating our understanding of gas behaviors under varying pressure conditions.

Volume Calculation

Volume calculation plays a key role in understanding how gases expand or contract under different pressure. Boyle's Law, which states that the product of pressure and volume remains constant for an ideal gas when temperature is held constant, is essential in predicting these changes.

Using the initial conditions of the balloon (1.25 L at 1 atm), we computed the future volume at lower pressure (0.2895 atm) using the formula:

• Bole's Law formula: \(P_1V_1 = P_2V_2\)
• Reorganized for new volume: \(V_2 = \frac{P_1V_1}{P_2}\)

Applying the known values:

• \(V_2 = \frac{1\,\text{atm} \times 1.25\,\text{L}}{0.2895\,\text{atm}} \approx 4.32\,\text{L}\)

This result, approximately 4.32 L, shows how a decrease in pressure increases the volume, thanks to the inverse relationship stated in Boyle’s Law.

Gas Laws

Gas laws describe the behavior of gases in response to changes in pressure, volume, and temperature. Boyle's Law, one of the fundamental principles, emphasizes the inverse relationship between pressure and volume at constant temperature.

Boyle's Law is expressed as \(P_1V_1 = P_2V_2\). This equation shows how, if pressure decreases, the volume increases, provided temperature remains constant. It underscores the concept that gases are compressible and their molecules take up more space when pressure is reduced.

In this exercise, the balloon's expansion from 1.25 L to approximately 4.32 L illustrates Boyle's Law in action. By decreasing the external pressure to 0.2895 atm, the balloon adapts by expanding its volume. Understanding such principles is vital in fields ranging from meteorology to engineering, where gas behavior under varying conditions is frequently evaluated.