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Chapter 10: Problem 109

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 (P鈧�) = 1.50 atm - Initial volume (V鈧�) = Unknown volume - Temperature (T) = constant After the stopcock is opened, the final conditions are: - Final pressure (P鈧�) = 695 torr - Final volume (V鈧�) = 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 (V鈧�).
02

Convert given pressure from atm to torr

Since we need to have consistent units, we'll convert the initial pressure (P鈧�) from atm to torr using the conversion factor: \(1 \: \text{atm} = 760 \: \text{torr}\). P鈧� = 1.50 atm 脳 \(\frac{760 \: \text{torr}}{1 \: \text{atm}}\) 鈮� 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: \[P鈧� \cdot V鈧� = P鈧� \cdot(V鈧� + 0.800)\] We need to solve this equation for V鈧�.
04

Solve the equation to find V鈧�

To find V鈧�, we have the equation: \(1140 \cdot V鈧� = 695 \cdot (V鈧� + 0.800)\) First, we'll divide by the common factor. V鈧� (1140 - 695) = 695 脳 0.800 V鈧� 脳 445 = 556 Now, we'll solve for V鈧� by dividing by 445. V鈧� 鈮� \(\frac{556}{445}\) 鈮� 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 set of bookshelves rests on a hard floor surface on four legs, each having a cross-sectional dimension of \(3.0 \times 4.1 \mathrm{cm}\) in contact with the floor. The total mass of the shelves plus the books stacked on them is 262 kg. Calculate the pressure in pascals exerted by the shelf footings on the surface.

(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of glycerol is 1.26 \(\mathrm{g} / \mathrm{mL}\) , whereas that of mercury is 13.6 \(\mathrm{g} / \mathrm{mL}\) . (b) What pressure, in atmospheres, is exerted on the body of a diver if she is 15 ft below the surface of the water when the atmospheric pressure is 750 torr? Assume that the density of the water is \(1.00 \mathrm{g} / \mathrm{cm}^{3}=1.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) The gravitational constant is \(9.81 \mathrm{m} / \mathrm{s}^{2},\) and \(1 \mathrm{Pa}=1 \mathrm{kg} / \mathrm{m}-\mathrm{s}^{2} .\)

Determine whether each of the following changes will increase, decrease, or not affect the rate with which gas molecules collide with the walls of their container: (a) increasing the volume of the container, (b) increasing the temperature, (c) increasing the molar mass of the gas.

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.

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.
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