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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

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)$$ If 159 \(\mathrm{mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ At a certain temperature and pressure, 1.2 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) reacts with 3.6 \(\mathrm{Lof} \mathrm{H}_{2} .\) If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3},\) at the same temperature and pressure, will be produced?

A deep-sea diver uses a gas cylinder with a volume of 10.0 \(\mathrm{L}\) and a content of 51.2 \(\mathrm{g}\) of \(\mathrm{O}_{2}\) and 32.6 \(\mathrm{g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\) .

The density of a gas of unknown molar mass was measured as a function of pressure at \(0^{\circ} \mathrm{C},\) as in the table that follows. (a) Determine a precise molar mass for the gas. [Hint: Graph \(d / P\) versus \(P . ](\mathbf{b})\) Why is \(d / P\) not a constant as a function of pressure? $$\begin{array}{lllll}{\text { Pressure (atm) }} & {1.00} & {0.666} & {0.500} & {0.333} & {0.250} \\ \hline \text { Density (g/L) } & {2.3074} & {1.5263} & {1.1401} & {0.7571} & {0.5660}\end{array}$$

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? \begin{equation}\begin{array}{l}{\text { (a) Because Xe is a noble gas, there is less tendency for the Xe }} \\ {\text { atoms to repel one another, so they pack more densely in }} \\ {\text { the gaseous state. }} \\ {\text { (b) Xe atoms have a higher mass than } \mathrm{N}_{2} \text { molecules. Because }} \\ {\text { both gases at STP have the same number of molecules per }} \\ {\text { unit volume, the Xe gas must be denser. }}\\\\{\text { (c) The Xe atoms are larger than } \mathrm{N}_{2} \text { molecules and thus take }} \\ {\text { up a larger fraction of the space occupied by the gas. }} \\\ {\text { (d) Because the Xe atoms are much more massive than the }} \\\ {\mathrm{N}_{2} \text { molecules, they move more slowly and thus exert }} \\\ {\text { less upward force on the gas container and make the gas }} \\ {\text { appear denser. }}\end{array}\end{equation}
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