Question

Two gas bulbs \(\mathrm{A}\) and \(\mathrm{B}\) are connected by a tube having a stopcock. Bulb A has a volume of \(100 \mathrm{~mL}\) and contains hydrogen. After opening the gas from \(\mathrm{A}\) to the evacuated bulb B, the pressure falls down by \(40 \%\). The volume \((\mathrm{mL})\) of B must be: (a) 75 (b) 150 (c) 125 (d) 200

Short answer

The volume of B closest to the calculated value is (a) 75 mL.

Step-by-step solution

Define the Problem

We have two bulbs, A and B, connected by a stopcock. Initially, only bulb A is filled with hydrogen gas, and bulb B is evacuated. When the stopcock is opened, the gas from A expands into both bulbs and causes the pressure in bulb A to fall by 40%.

Apply Boyle's Law

According to Boyle's Law, which states that the pressure times the volume of a gas remains constant for a given amount of gas at constant temperature, the initial pressure-volume product should equal the final pressure-volume product. Let the initial pressure in bulb A be \(P_i\) and the initial volume be \(V_A = 100 \, \text{mL}\). The final pressure \(P_f\) is 60% of \(P_i\) which is 0.6\(P_i\). Let the volume of bulb B be \(V_B\).

Set Up the Equation

Initially, the gas was only in bulb A, so the initial pressure-volume product is \(P_i \times V_A = P_i \times 100\). After the stopcock is opened, the gas fills both bulbs, so the new volume is \(V_A + V_B\), and the pressure is 0.6\(P_i\). Therefore, the pressure-volume product becomes \(0.6 \times P_i \times (100 + V_B)\).

Solve for the Volume of Bulb B

Equate the initial and final pressure-volume products:\[ P_i \times 100 = 0.6 \times P_i \times (100 + V_B) \]Dividing both sides by \(P_i\) and simplifying:\[ 100 = 0.6 \times (100 + V_B) \]\[ 100 = 60 + 0.6V_B \]\[ 40 = 0.6V_B \]Divide both sides by 0.6:\[ V_B = \frac{40}{0.6} = 66.67 \approx 67 \, \text{mL} \]

Determine the Closest Option

The calculated volume of bulb B is approximately 67 mL. However, none of the given options exactly match this value. Given the options (a) 75, (b) 150, (c) 125, (d) 200, the closest option would be (a) 75 mL.

Key concepts

Boyle's Law

Boyle's Law is a fundamental concept in gas laws that describes how the pressure of a gas tends to decrease as the volume of the container increases, as long as the temperature and the amount of gas remain constant. This relationship is expressed mathematically as \( P_1V_1 = P_2V_2 \).
- **Pressure and Volume:** In this formula, \( P_1 \) and \( P_2 \) represent the initial and final pressures of the gas, while \( V_1 \) and \( V_2 \) represent the initial and final volumes, respectively.
- **Constant Product:** The key takeaway here is that the product of the pressure and the volume is constant throughout this process.The core idea is quite simple: if you decrease the volume of a container (making it smaller), the gas particles will have less space to move around, thus increasing the pressure. Conversely, if the volume increases, the pressure decreases because the gas particles are distributed over a larger space, as we see in our given exercise when the gas expands from bulb A into bulb B.

Pressure-Volume Relationship

The pressure-volume relationship in gases is not just a theoretical concept but a very practical one, as it applies to many real-world scenarios. This relationship is notably described by Boyle’s Law, and can be observed when a gas is allowed to spread out in a larger volume, as in our exercise.
**Understanding Changes:**- In the problem, the pressure in bulb A falls by 40% after opening the stopcock, illustrating a direct decrease in pressure as the gas volume increases (since it now occupies both bulbs). The final pressure is 60% of the initial pressure.- Mathematically, this relationship translates into \( P_i \times V_A = P_f \times (V_A + V_B) \) as derived in the exercise solution.This means that if the initial pressure of the gas was \( P_i \), after it expands into a combined space of bulbs A and B, the pressure becomes \( 0.6 \times P_i \), fitting into the relationship. The exercise exemplifies how understanding this relationship allows us to solve for variables such as the volume of the second bulb.

Expansion of Gases

The expansion of gases is a fascinating process where gas particles spread out to fill a larger volume. This process happens spontaneously when the constraints on a gas are removed, such as when a stopcock is opened, allowing the gas to move into an initially evacuated space.
**Key Characteristics:** - **Spontaneous Spreading:** When a gas expands, it's not due to an external force; rather, the gas naturally occupies as much volume as possible. This is due to the random motion of gas particles, which are in constant, rapid movement. - **Volume Increase, Pressure Drop:** In our exercise, gas from bulb A expanded into bulb B, leading to a reduction in pressure to 60% of its original value as the combined volume increased. Consider the way gas expands when you let air out of a balloon—it rushes out to fill the space around it. Similarly, when the stopcock opens, the hydrogen gas from bulb A naturally fills both spaces, doubling the overall volume that the gas occupies, which in turn decreases the pressure in bulb A. Understanding how gases behave during expansion is key to solving problems related to real-life applications of gas laws.