/*! 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 49 A vessel containing \(39.5 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A vessel containing \(39.5 \mathrm{~cm}^{3}\) of helium gas at \(25^{\circ} \mathrm{C}\) and \(106 \mathrm{kPa}\) was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If this required \(7.5 \mathrm{~cm}^{3}\) of ethanol, what was the final temperature of the helium?

Short Answer

Expert verified
The final temperature of the helium gas is approximately 241 K.

Step by step solution

01

Identify Known Variables

We are given that the initial volume \( V_1 \) of the helium gas is \( 39.5 \mathrm{~cm}^3 \) and the pressure \( P \) is constant at \( 106 \mathrm{kPa} \). The initial temperature \( T_1 \) is \( 25^{\circ} \mathrm{C} \), which we convert to Kelvin by adding 273 (i.e., \( 25 + 273 = 298 \mathrm{K} \)).Additionally, when the gas contracts, ethanol enters the vessel, leading to a new volume \( V_2 = 39.5 \mathrm{~cm}^3 - 7.5 \mathrm{~cm}^3 = 32 \mathrm{~cm}^3 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Helium Gas
Helium gas is a colorless, odorless, and tasteless noble gas that is incredibly light and non-reactive. It is the second lightest element in the universe, right after hydrogen. Helium's non-reactive nature makes it an ideal gas for various practical applications like balloons, airships, and cooling systems for nuclear reactors.
Helium is used in this exercise as it is one of the stable inert gases, making the calculations primarily dependent on its compressibility and expansion characteristics according to gas laws. Understanding the behavior of helium gas can aid in solving similar problems by applying the principles of gas laws effectively.
Temperature Conversion
Temperature conversion is essential for many scientific calculations, especially when dealing with gas laws. Most gas law equations require temperature to be expressed in Kelvin, as this absolute scale allows for calculations without negative numbers, which Celsius and Fahrenheit scales might involve.
To convert Celsius to Kelvin, we use the formula:
  • \( T(K) = T(°C) + 273.15 \)
In the exercise, the initial temperature is given as \(25^{\circ} \mathrm{C}\). Converting it into Kelvin gives \(298 \mathrm{~K}\). This step is crucial, as failing to convert temperatures to Kelvin can lead to incorrect results in gas law calculations. Always remember to add 273.15 to Celsius temperatures for accuracy.
Understanding temperature conversion ensures that all measurements are in the correct unit before proceeding with any calculations.
Boyle's Law
Boyle's Law is an important principle of gas behavior that examines the relationship between pressure and volume, holding the temperature constant. It states that the pressure of a gas is inversely proportional to its volume, given the temperature remains unchanged. This relationship can be expressed as:
  • \( P_1V_1 = P_2V_2 \)
This means if the volume decreases, the pressure increases, and vice versa, provided the amount of gas and temperature remain constant. While Boyle's Law isn't directly applied in the context of the given exercise since the pressure remains constant while the temperature changes, it lays the groundwork for understanding how changes in physical conditions like volume and pressure influence gas behavior.
By mastering Boyle's Law, you can accurately predict how a gas will behave if its container size or pressure is altered while maintaining a certain temperature.
Pressure and Volume Relationship
The pressure and volume relationship in gases is a cornerstone of understanding their behavior under different conditions, prominently illustrated by Boyle's Law and other gas laws. In the scenario given, while employing Boyle's Law directly isn't required due to the constant pressure condition, grasping the general relationship aids in conceptualizing how gases respond to changes in volume.
Usually, in a closed system, if the volume reduces, the pressure should increase, and vice versa. However, in this exercise, we observe that the volume reduces due to the entry of ethanol into the vessel while maintaining a constant pressure. This implies an adjustment in temperature instead of pressure, illustrating the interplay between multiple variables (volume, temperature, and pressure) in gas behaviors.
Understanding this intricate balance helps in solving real-world problems efficiently, where factors such as temperature shifts and pressure changes yield accurate predictions of gas behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mixture contained zinc sulfide, \(\mathrm{ZnS},\) and lead sulfide, PbS. A sample of the mixture weighing \(6.12 \mathrm{~g}\) was reacted with an excess of hydrochloric acid. The reactions are $$ \mathrm{ZnS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g) $$ \(\mathrm{PbS}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\) If the sample reacted completely and produced \(1.049 \mathrm{~L}\) of hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), at \(23^{\circ} \mathrm{C}\) and \(762 \mathrm{mmHg},\) what were the percentages of \(\mathrm{ZnS}\) and \(\mathrm{PbS}\) in the mixture?

If \(456 \mathrm{dm}^{3}\) of krypton at \(101 \mathrm{kPa}\) and \(21^{\circ} \mathrm{C}\) is compressed into a \(30.1-\mathrm{dm}^{3}\) tank at the same temperature, what is the pressure of krypton in the tank?

A \(275-\mathrm{mL}\) sample of CO gas is collected over water at \(31^{\circ} \mathrm{C}\) and \(755 \mathrm{mmHg}\). If the temperature of the gas collection apparatus rises to \(39^{\circ} \mathrm{C}\), what is the new volume of the sample? Assume that the barometric pressure does not change.

The reaction \(8 \mathrm{H}_{2}(g)+\mathrm{S}_{8}(l) \longrightarrow 8 \mathrm{H}_{2} \mathrm{~S}(g)\) is run at \(125^{\circ} \mathrm{C}\) and a constant pressure of \(12.0 \mathrm{~atm}\). Assuming complete reaction, what mass of \(\mathrm{S}_{8}\) would be required to produce \(5.00 \times 10^{2} \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{~S}\) gas under these conditions?

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.