/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 An ideal gas at a pressure of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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An ideal gas at a pressure of \(152 \mathrm{kPa}\) 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 \(92.66 \mathrm{kPa}\), what is the volume of the bulb that was originally filled with gas?

Short Answer

Expert verified
The volume of the originally filled bulb is approximately \(1.249\, \mathrm{L}\).

Step by step solution

01

Understand the Initial and Final States

In this problem, the gas initially at a pressure of \(152 \mathrm{kPa}\) is contained in a bulb of unknown volume \(V_1\). When it expands into a previously empty bulb of known volume \(V_2 = 0.800\, \mathrm{L}\), the final pressure becomes \(92.66 \mathrm{kPa}\). The temperature remains constant throughout the process.
02

Use Boyle's Law

Boyle's Law states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional: \( P_1 \cdot V_1 = P_2 \cdot (V_1 + V_2) \). This is because the initial product \(P_1V_1\) must equal the final product of pressure and volume \(P_2(V_1 + V_2)\) when temperature is constant.
03

Substitute Known Values

Substitute the known pressures and volume into the Boyle's Law equation: \(152 \cdot V_1 = 92.66 \cdot (V_1 + 0.800)\). This substitution uses \(P_1 = 152\, \mathrm{kPa}\), \(P_2 = 92.66\, \mathrm{kPa}\), and \(V_2 = 0.800\, \mathrm{L}\).
04

Solve for Unknown Volume \(V_1\)

Rearrange the equation from step 3 to solve for \(V_1\):\[152V_1 = 92.66V_1 + 92.66 \times 0.800\]First, distribute the \(92.66\) into the bracket:\[152V_1 = 92.66V_1 + 74.128\]Rearrange to solve for \(V_1\):\[59.34V_1 = 74.128\]Thus, \(V_1 = \frac{74.128}{59.34} = 1.249\, \mathrm{L}\).

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Key Concepts

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

Ideal Gas
An ideal gas is a simplified model that helps us understand the behavior of gases under various conditions. It assumes that the gas particles are point particles with no volume and no intermolecular forces acting between them. This model is very useful for studying gas laws because it allows us to make predictions about gas behavior in different scenarios. In the context of this exercise, recognizing the gas as an ideal gas means we can apply simple relations like Boyle's Law without worrying about these additional complexities.

To consider gases as ideal, they should be at relatively high temperatures and low pressures, where the actual gas molecules are far enough apart that their size and interaction forces become negligible. While real gases deviate from this ideal behavior at very high pressures and low temperatures, many gases under normal conditions behave almost like ideal gases, making this concept extremely useful in simplifying calculations without introducing significant error.
Constant Temperature
The condition of constant temperature is one of the cornerstones of Boyle's Law, which is instrumental in solving this exercise. When we say the temperature is held constant, we refer to an isothermal process. In such processes, although pressure and volume can change, the temperature doesn't. This condition means any change in the volume and pressure of the gas doesn't result in a change in temperature, hence energy input or removal to maintain temperature isn't accounted for.

In our given problem, when the gas expands into a new bulb, the temperature remains constant, which allows us to directly apply Boyle's Law: \[ P_1 \times V_1 = P_2 \times V_2 \] Thus, knowing the temperature does not change assures us that it doesn't affect our equations and calculations for volumes and pressures in this scenario. Isothermal conditions are quite practical for theoretical studies since other factors influencing gas behavior can be ignored.
Volume Calculation
Volume calculation in an ideal gas scenario under constant temperature hinges on Boyle's Law: the inverse relationship between pressure and volume at a steady temperature. Our objective here is to find the unknown initial volume, denoted as \( V_1 \), of the bulb containing the gas.

Given the problem data, we know the initial pressure and final pressure as well as the volume the gas expands into:- P_1 = 152 \, \mathrm{kPa} - P_2 = 92.66 \, \mathrm{kPa}- V_2 = 0.800 \, \mathrm{L} Substituting these values into Boyle's Law gives us the equation:\[ 152V_1 = 92.66(V_1 + 0.800) \]By solving this equation, you can find:\[ 59.34V_1 = 74.128 \] Finally, the calculations yield the initial volume as:\[ V_1 = \frac{74.128}{59.34} \approx 1.249 \, \mathrm{L} \]This calculation reveals the original volume of the bulb before the gas expanded, illustrating the power of Boyle's Law under ideal conditions and constant temperature.

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Most popular questions from this chapter

Rank the following gases from least dense to most dense at \(101.33 \mathrm{kPa}\) and \(298 \mathrm{~K}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{NH}_{3}, \mathrm{HCl}\)

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 30 minutes the average cockroach (running at \(0.08 \mathrm{~km} / \mathrm{h})\) consumed \(1.0 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at \(101.33 \mathrm{kPa}\) pressure and \(20^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 day by a 6.3 -g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 2.0-L fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, how much of the available \(\mathrm{O}_{2}\) will the cockroach consume in 1 day? (Air is \(21 \mathrm{~mol} \% \mathrm{O}_{2}\).)

A plasma-screen TV contains thousands of tiny cells tilled with a mixture of Xe, Ne, and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, \(0.900 \mathrm{~mm} \times 0.300 \mathrm{~mm} \times 10.0 \mathrm{~mm}\), contains \(4 \%\) Xe in a 1:1 Ne:He mixture at a total pressure of \(66.66 \mathrm{kPa}\). Calculate the number of Xe, Ne, and He atoms in the cell and state the assumptions you need to make in your calculation.

Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at \(\mathrm{STP}\), the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, \((\mathbf{d})\) rate of effusion through a pinhole leak?

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 20 -L container at \(709.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL}\). (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

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