Question

A 10.0-L balloon contains helium gas at a pressure of \(655 \mathrm{mmHg}\). What is the new pressure, in \(\mathrm{mmHg}\), of the helium gas at each of the following volumes, if there is no change in temperature or amount of gas? a. \(20.0 \mathrm{~L}\) b. \(2.50 \mathrm{~L}\) c. \(1500 . \mathrm{mL}\)

Short answer

a. 327.5 mmHg, b. 2620 mmHg, c. 4366.67 mmHg

Step-by-step solution

- Understand Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, this is expressed as \( P_1 V_1 = P_2 V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume.

- Identify Initial Conditions

The initial volume \( V_1 \) is 10.0 L and the initial pressure \( P_1 \) is 655 mmHg.

- Solve for New Pressure at Volume 20.0 L

Using the equation \( P_1 V_1 = P_2 V_2 \), substitute \( P_1 = 655 \text{mmHg} \), \( V_1 = 10.0 \text{L} \), and \( V_2 = 20.0 \text{L} \). Then solve for \( P_2 \): \( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 20.0 \text{L} \)\( P_2 = \frac{655 \text{mmHg} \times 10.0 \text{L}}{20.0 \text{L}} = 327.5 \text{mmHg} \)

- Solve for New Pressure at Volume 2.50 L

Using the same method, substitute \( V_2 = 2.50 \text{L} \). Then:\( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 2.50 \text{L} \)\( P_2 = \frac{655 \text{mmHg} \times 10.0 \text{L}}{2.50 \text{L}} = 2620 \text{mmHg} \)

- Convert 1500 mL to L and Solve for New Pressure

First, convert 1500 mL to L since 1000 mL = 1 L. Thus, 1500 mL = 1.5 L. Then substitute \( V_2 = 1.5 \text{L} \): \( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 1.5 \text{L} \)\( P_2 = \frac{655 \text{mmHg} \times 10.0 \text{L}}{1.5 \text{L}} \)\( P_2 = 4366.67 \text{mmHg} \)

Key concepts

Gas Laws

Gas laws are fundamental principles in chemistry that describe the behavior of gases in different conditions. One of the most important gas laws is Boyle's Law. These laws help us understand how gases respond to changes in pressure, volume, and temperature. Common gas laws to be aware of include:
Boyle's Law: Describes the pressure-volume relationship at constant temperature.
Charles's Law: Relates volume and temperature at constant pressure.
Ideal Gas Law: Combines several gas laws to provide a comprehensive description.
Boyle's Law is particularly useful when dealing with situations where temperature remains constant. It tells us that the pressure of a gas is inversely proportional to its volume. This means when one goes up, the other comes down, and vice versa. For our specific problem, measurement units for pressure in millimeters of Mercury (mmHg) were used, which is common in gas law problems.

Pressure-Volume Relationship

The pressure-volume relationship is key to understanding Boyle's Law. This relationship can be written as:
\( P_1 \cdot V_1 = P_2 \cdot V_2 \)
Here, \( P_1 \) and \( P_2 \) represent the initial and final pressures, and \( V_1 \) and \( V_2 \) correspond to the initial and final volumes.
In simpler terms, if you have a balloon filled with helium at a certain pressure and volume, and if you change the volume of the balloon by either compressing or expanding it without changing the temperature, you can find the new pressure using this equation.
Let's relate this to the given problem steps:
- For a 20.0 L volume: \( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 20.0 \text{L} \), \( P_2 \) becomes 327.5 mmHg
- For a 2.50 L volume: \( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 2.50 \text{L} \), \( P_2 \) becomes 2620 mmHg
- For a 1.5 L volume: \( 655 \text{mmHg} \times 10.0 \text{L} = P_2 \times 1.50 \text{L} \), \( P_2 \) becomes 4366.67 mmHg
This demonstrates the inverse relationship: reducing the volume increases the pressure, and increasing the volume decreases the pressure.

Helium Gas

Helium is a noble gas, meaning it's very stable and non-reactive, making it perfect for uses where non-reactivity is essential. Common applications include balloons, airships, and as a cooling medium for nuclear reactors and MRI machines.
Helium is lighter than air, which is why helium balloons float. In the context of Boyle's Law and our problem, helium behaves like an ideal gas under conditions typical for such calculations. This means:
- The helium gas follows Boyle's Law accurately for the pressures and volumes given.
- Our calculations for pressure changes as a result of volume changes hold true for helium, assuming constant temperature.
Understanding these properties ensures that you can confidently apply Boyle's Law to find new pressures when dealing with helium gas in closed systems, like balloons or other containers.
Keep in mind the importance of the conditions stated in problems: constant temperature and a set amount of gas. Without these constants, Boyle's Law would not apply directly.