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A sample of a gas has a pressure of \(100 . \mathrm{mmHg}\) in a sealed \(125-\mathrm{mL}\). flask. This gas sample is transferred to another flask with a volume of \(200 . \mathrm{mL}\). Calculate the new pressure. Assume that the temperature remains constant.

Short Answer

Expert verified
The new pressure is 62.5 mmHg.

Step by step solution

01

Identify Given Values

Note the initial and final volumes and the initial pressure:- Initial Pressure \( P_1 = 100 \, \text{mmHg} \)- Initial Volume \( V_1 = 125 \, \text{mL} \)- Final Volume \( V_2 = 200 \, \text{mL} \)We need to find the final pressure \( P_2 \).
02

State Boyle's Law

Boyle's Law explains the relationship between the pressure and volume of a gas at constant temperature. It states that the product of the initial pressure and volume is equal to the product of the final pressure and volume:\[ P_1 \times V_1 = P_2 \times V_2 \]
03

Rearrange Boyle's Law Formula

To find the final pressure \( P_2 \), rearrange the formula:\[ P_2 = \frac{P_1 \times V_1}{V_2} \] This formula allows us to calculate the new pressure given the other known values.
04

Insert Known Values

Substitute the given values into the rearranged Boyle's Law formula:\[ P_2 = \frac{100 \, \text{mmHg} \times 125 \, \text{mL}}{200 \, \text{mL}} \]
05

Calculate the Final Pressure

Perform the calculation to find the final pressure:\[ P_2 = \frac{12500 \, \text{mmHg} \, \text{mL}}{200 \, \text{mL}} = 62.5 \, \text{mmHg} \] So, the new pressure in the 200 mL flask is 62.5 mmHg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boyle's Law
Boyle's Law is a fundamental principle in chemistry and physics that describes how the pressure of a gas tends to decrease as the volume increases, provided the temperature remains constant. This law, discovered by Robert Boyle in the 17th century, is crucial in understanding gas behavior.According to Boyle's Law, the product of the initial pressure and the volume of a gas remains constant during any transformation. This can be mathematically described by the equation:\[ P_1 \times V_1 = P_2 \times V_2 \]where:- \( P_1 \) is the initial pressure of the gas- \( V_1 \) is the initial volume- \( P_2 \) is the final pressure after a change in volume- \( V_2 \) is the final volumeBoyle's Law helps predict how a change in volume will affect the pressure of a gas in a sealed container, assuming no other variables such as temperature change.
Pressure and Volume Relationship
The relationship between pressure and volume in gases, as noted by Boyle's Law, is inversely proportional. This means that as the volume of a gas increases, the pressure decreases, and vice versa. When the gas in a sealed container is compressed, the molecules are forced into a smaller space, increasing the pressure. Conversely, when the container's volume increases, the molecules have more room to move, resulting in reduced pressure. Understanding this pressure-volume relationship is key to solving problems involving gases. For example, in the given exercise, a gas stored initially in a 125 mL flask at 100 mmHg is transferred to a 200 mL flask. Since the volume increases, the pressure decreases, as shown by the calculation of the new pressure (62.5 mmHg). This relationship is pivotal in many real-world applications, such as designing medical syringes and engines where control over gas pressure is essential.
Sealed Gas Samples
Sealed gas samples refer to gases that are contained within a closed system, preventing any exchange of matter with their surroundings. In such a system, factors like pressure, volume, and temperature become critical in determining the behavior of the gas. In the context of Boyle's Law, working with sealed gas samples implies that we can assume the number of gas molecules remains constant. Thus, any changes in pressure or volume must result from internal adjustments within the contained system, rather than external influences. Sealed gas environments are often used in scientific experiments to maintain controlled conditions and ensure that the results are consistent and reliable. By knowing the initial conditions and applying Boyle's Law, we can easily predict how a gas will behave when transferred from one sealed container to another, as demonstrated in the exercise by calculating the new pressure as the gas was moved to a larger volume.

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Most popular questions from this chapter

The gas burner in a stove or furnace admits enough air so that methane gas can react completely with oxygen in the air according to the equation $$\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ Air is one-fifth oxygen by volume. Both air and methane gas are supplied to the flame by passing them through separate small tubes. Compared with the tube for the methane gas, determine how much bigger the cross section of the tube for the air needs to be. Assume that both gases are at the same \(T\) and \(P\).

Why can't a hand-driven pump on a water well pull underground water from depths more than \(33 \mathrm{ft}\) ? Would it help to have a motor-driven vacuum pump?

The effects of intermolecular interactions on gas properties depend on \(T\) and \(P .\) Do these effects become more or less significant when each change occurs? Why? (a) A sealed container of gas is compressed to a smaller volume at constant temperature. (b) A container of gas has more gas added into the same volume at constant temperature. (c) The gas in a container of variable volume is heated at constant pressure.

Convert these pressure values. (a) \(120 . \mathrm{mmHg}\) to atm (b) \(2.00 \mathrm{~atm}\) to \(\mathrm{mmHg}\) (c) \(100 . \mathrm{kPa}\) to \(\mathrm{mmHg}\) (d) \(200 . \mathrm{kPa}\) to \(\mathrm{atm}\) (e) \(36.0 \mathrm{kPa}\) to atm (f) \(600 . \mathrm{kPa}\) to \(\mathrm{mmHg}\)

The mean fraction by mass of water vapor and cloud water in Earth's atmosphere is about 0.0025 . Assume that the atmosphere contains two components: "air," with a molar mass of \(29.2 \mathrm{~g} / \mathrm{mol}\), and water vapor. Determine the mean mole fraction of water vapor in Earth's atmosphere. Determine the mean partial pressure of water vapor. Why is this so much smaller than the typical partial pressure of water vapor at Earth's surface on a rainy summer day ( \(25 \mathrm{mmHg}\) )?

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