/*! 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 57 Consider two separate gas contai... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider two separate gas containers at the following conditions: $$ \begin{array}{|ll|} \text { Container A } & \text { Container B } \\ \text { Contents: } \mathrm{SO}_{2}(g) & \text { Contents: unknown gas } \\ \text { Pressure }=P_{\mathrm{A}} & \text { Pressure }=P_{\mathrm{B}} \\ \text { Moles of gas }=1.0 \mathrm{~mol} & \text { Moles of gas }=2.0 \mathrm{~mol} \\ \text { Volume }=1.0 \mathrm{~L} & \text { Volume }=2.0 \mathrm{~L} \\ \text { Temperature }=7^{\circ} \mathrm{C} & \text { Temperature }=287^{\circ} \mathrm{C} \\ \hline \end{array} $$ How is the pressure in container \(\mathrm{B}\) related to the pressure in container \(\mathrm{A}\) ?

Short Answer

Expert verified
In conclusion, the pressure in Container B is twice the pressure in Container A, or \(P_B = 2 P_A\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. The conversion is done using the formula K = °C + 273.15. For Container A: Temperature (in Kelvin) = \(7 + 273.15\), \(T_A = 280.15 K\) For Container B: Temperature (in Kelvin) = \(287 + 273.15\), \(T_B = 560.15 K\)
02

Applying the Ideal Gas Law to Container A and Container B

Apply the Ideal Gas Law (\(PV=nRT\)) to both containers: For Container A: \(P_A \times 1.0 L = 1.0 mol \times R \times 280.15 K\) For Container B: \(P_B \times 2 L = 2.0 mol \times R \times 560.15 K\)
03

Solve for the pressure in Container B in terms of pressure in Container A

Our main goal in this step is to determine the relationship between the pressure in Container B and the pressure in Container A. Notice that R is a constant, so we can write the equations as: \(P_A = R \times 280.15 K\) \(2 P_B = 2 R \times 560.15 K\) Divide the second equation by 2: \(P_B = R \times 560.15 K\) Now, we need to relate the equations to each other. Divide the second equation by the first equation: \(\frac{P_B}{P_A} = \frac{R \times 560.15 K}{R \times 280.15 K}\) Since R is a constant, it cancels out: \(\frac{P_B}{P_A} = \frac{560.15}{280.15}\)
04

Calculate the relationship between the pressures

We can now determine the relationship between the pressures: \(\frac{P_B}{P_A} = \frac{560.15}{280.15} = 2.0\)
05

Conclusion

The pressure in Container B is twice the pressure in Container A. That is, \(P_B = 2 P_A\).

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Key Concepts

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

Temperature Conversion
When dealing with gases, a common requirement is to convert temperatures from Celsius to Kelvin. This is essential because the Kelvin scale is the standard for scientific calculations involving gases.
The conversion between these two scales is quite straightforward. You use the formula:
  • Kelvin = Celsius + 273.15
This simple formula allows you to reframe a temperature given in Celsius into Kelvin, which aligns perfectly with the calculations done under the Ideal Gas Law.
Using Kelvin ensures a consistent baseline, as Kelvin starts at absolute zero—the point where all molecular motion theoretically stops.
In the given problem, for example, Container A's temperature at 7°C must be converted to 280.15 K, and Container B's temperature at 287°C translates to 560.15 K.
Pressure Relationship
Pressure plays a crucial role in understanding how gases behave under different conditions. In this exercise, the goal is to explore the relationship between pressures in two different containers using the Ideal Gas Law.
The law establishes that pressure is dependent on temperature, volume, and the number of moles of gas. It can be described with the equation:
  • \( PV = nRT \)
Where:\( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.
By isolating the pressure term, you can understand how changes in temperature, volume, or the amount of gas influence it.
In our example, we find that the pressure in Container B is twice that in Container A. This is derived by comparing the two systems via their respective Ideal Gas Law equations. Understanding these relationships helps in predicting how gases respond to environmental changes.
Gas Laws
Gas laws describe the behavior of gases in various scenarios and how variables such as pressure, volume, and temperature interact. The Ideal Gas Law, which we've used here, is a combination of simpler laws like Boyle's, Charles's, and Avogadro's Laws.
Knowing these fundamental principles helps to better understand the Ideal Gas Law:
  • Boyle's Law shows how pressure and volume are inversely related when temperature remains constant.
  • Charles's Law states that volume and temperature are directly proportional if pressure is constant.
  • Avogadro's Law describes how volume relates to the number of moles of gas at a constant temperature and pressure.
Each of these laws interconnects to form the more comprehensive equation that is the Ideal Gas Law.
In the exercise, these laws are implicitly applied to solve for the relationship between the pressures in two different gas setups.
Containers
Containers holding gases serve as the controlled environment where we can apply gas laws. The volume of these containers directly affects the pressure and the behavior of the gas within.
In our scenario, Container A holds 1.0 L of gas, while Container B holds 2.0 L. This difference directly contributes to how the same substance might react differently in each container.
Container size plays into how we apply the Ideal Gas Law equations:
  • Larger containers can hold gas molecules farther apart, affecting pressure.
  • The volume allows for predictions about how much pressure will increase or decrease when variables change.
Understanding how containers function in these calculations aids in making accurate predictions and assessments of gas behaviors in both scientific experiments and practical applications.

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Most popular questions from this chapter

A 5.0-L flask contains \(0.60 \mathrm{~g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\). What is the pressure (in atm) inside the flask?

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In the presence of nitric acid, \(\mathrm{UO}^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}{ }^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: \(\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2+}(a q) \longrightarrow\) \(\mathrm{NO}(g)+\mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) If \(2.55 \times 10^{2} \mathrm{~mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{~atm}\), what amount (moles) of \(\mathrm{UO}^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess HCl gas to produce \(3.22 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at \(373 \mathrm{~K}\) and \(0.951\) atm and a manganese chloride compound \(\left(\mathrm{MnCl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?

Sulfur trioxide, \(\mathrm{SO}_{3}\), is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$ \begin{aligned} \mathrm{S}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{SO}_{3}(g) \end{aligned} $$ What volume of \(\mathrm{O}_{2}(g)\) at \(350 .{ }^{\circ} \mathrm{C}\) and a pressure of \(5.25 \mathrm{~atm}\) is needed to completely convert \(5.00 \mathrm{~g}\) sulfur to sulfur trioxide?

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