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An ideal gas at a pressure of 1.50 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.800 \(\mathrm{L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is 695 torr, what is the volume of the bulb that was originally filled with gas?

Short Answer

Expert verified
The volume of the bulb that was originally filled with gas is approximately 1.249 L.

Step by step solution

01

Write down the initial and final conditions of the system

Before the stopcock is opened, we know the initial conditions of the gas in the bulb are: - Initial pressure (P₁) = 1.50 atm - Initial volume (V₁) = Unknown volume - Temperature (T) = constant After the stopcock is opened, the final conditions are: - Final pressure (P₂) = 695 torr - Final volume (V₂) = Unknown volume + 0.800 L We will use the ideal gas law and the relation between atm and torr to solve for the unknown volume (V₁).
02

Convert given pressure from atm to torr

Since we need to have consistent units, we'll convert the initial pressure (P₁) from atm to torr using the conversion factor: \(1 \: \text{atm} = 760 \: \text{torr}\). P₁ = 1.50 atm × \(\frac{760 \: \text{torr}}{1 \: \text{atm}}\) ≈ 1140 torr
03

Apply the ideal gas law to both initial and final conditions

Remember that for an ideal gas, the following relation holds: \[P \cdot V = n \cdot R \cdot T\] where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. Since the temperature and number of moles remain constant in this case, we can simplify the relation as follows: \[P \cdot V = constant\] Applying this to the initial and final conditions, we have: \[P₁ \cdot V₁ = P₂ \cdot(V₁ + 0.800)\] We need to solve this equation for V₁.
04

Solve the equation to find V₁

To find V₁, we have the equation: \(1140 \cdot V₁ = 695 \cdot (V₁ + 0.800)\) First, we'll divide by the common factor. V₁ (1140 - 695) = 695 × 0.800 V₁ × 445 = 556 Now, we'll solve for V₁ by dividing by 445. V₁ ≈ \(\frac{556}{445}\) ≈ 1.249 L The volume of the bulb that was originally filled with gas is approximately 1.249 L.

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Key Concepts

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

Pressure-Volume Relationship
The pressure-volume relationship, also known as Boyle's Law, is a fundamental principle in understanding how gases behave under varying pressure and volume conditions while keeping temperature constant. This principle states that the pressure of a given amount of an ideal gas is inversely proportional to its volume when the temperature and number of moles remain unchanged.

In the context of our exercise, the initial pressure in the bulb was 1.50 atm. Since the ideal gas law asserts that pressure and volume are inversely related, upon opening the stopcock, the gas expands, causing the volume to increase and the pressure to decrease. This expansion is crucial because it adheres to the ideal gas behavior where increased volume allows the gas particles to spread out, decreasing the frequency of collisions and thus the pressure.

Application of Boyle's Law to the Exercise

When the stopcock is opened, the gas expands into the second bulb, displaying the pressure-volume relationship. By keeping the temperature constant, we ensured that we could apply Boyle's Law and derive the final pressure after expansion based solely on volume change. Therefore, the final pressure and the combined volumes can be used to calculate the original unknown volume, demonstrating this inverse relationship.
Gas Expansion
Gas expansion occurs when a gas's volume increases. In real-life scenarios, expansion can occur due to heating or because a gas is allowed to spread out into a larger area, as was the case in our exercise. With constant temperature, as per Charles's Law, the volume of a gas increases as pressure decreases, provided the amount of gas (in moles) remains constant.

The ideal gas law also describes this behavior, where the volume of the gas has a direct relationship with its temperature (when pressure and the number of moles are constant), and an inverse relationship with its pressure (when temperature and the number of moles are constant).

Illustration of Gas Expansion

In the presented scenario, the gas in the bulb expanded in response to the opening of the stopcock into an evacuated bulb. Understanding how the gas expands and how this impacts pressure is instrumental in solving for the original volume of the filled bulb. We observed gas expansion while keeping the overall temperature steady, allowing us to focus on how the volume change affected pressure.
Stoichiometry
Stoichiometry, in the context of gas laws, involves working with the quantitative relationships between the reactants and products in a chemical reaction. In scenarios involving gases, this often means dealing with volumes, pressures, and temperatures, as these properties can determine the amount of substance involved.

While our example doesn't involve a chemical reaction, the principles of stoichiometry are still relevant to solving the problem as they require an understanding of proportional relationships. By utilizing the ratio of pressure to volume in the ideal gas law, we employed stoichiometric principles to deduce the missing volume of the original bulb.

Stoichiometric Aspect of the Problem

The relationship between pressure, volume, and the number of moles (which remains constant in this case) serves as the stoichiometric foundation for being able to set up the equation from the ideal gas law. By maintaining a direct stoichiometric relationship, the change in pressure and volume can be calculated accordingly, ultimately allowing us to determine the unknown volume.

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Most popular questions from this chapter

A quantity of \(\mathrm{N}_{2}\) gas originally held at 5.25 atm pressure in a 1.00 -L container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\) . A quantity of \(\mathrm{O}_{2}\) gas originally at 5.25 atm and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

Table 10.3 shows that the van der Wals \(b\) parameter has units of L/mol. This means that we can calculate the sizes of atoms or molecules from the \(b\) parameter. Refer back to the discussion in Section \(7.3 .\) Is the van der Waals radius we calculate from the \(b\) parameter of Table 10.3 more closely associated with the bonding or nonbonding atomic radius discussed there? Explain.

Consider a mixture of two gases, \(A\) and \(B,\) confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas C affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

(a) Are you more likely to see the density of a gas reported in \(\mathrm{g} / \mathrm{mL}, \mathrm{g} / \mathrm{L},\) or \(\mathrm{kg} / \mathrm{cm}^{3} ?(\mathbf{b})\) Which units are appropriate for expressing atmospheric pressures, \(\mathrm{N}, \mathrm{Pa},\) atm, kg/m \(^{2} ?\) (c) Which is most likely to be a gas at room temperature and ordinary atmospheric pressure, \(\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{K}_{2} \mathrm{O} .\)

You have an evacuated container of fixed volume and known mass and introduce a known mass of a gas sample. Measuring the pressure at constant temperature over time, you are surprised to see it slowly dropping. You measure the mass of the gas-filled container and find that the mass is what it should be-gas plus container-and the mass does not change over time, so you do not have a leak. Suggest an explanation for your observations.

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