/*! 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 27 A 93-L sample of dry air cools f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 93-L sample of dry air cools from \(145^{\circ} \mathrm{C}\) to \(-22^{\circ} \mathrm{C}\) while the pressure is maintained at 2.85 atm. What is the final volume?

Short Answer

Expert verified
The final volume is approximately 55.8 L.

Step by step solution

01

- Write Down the Initial Parameters

Given:Initial Volume, \( V_1 = 93 \text{ L} \)Initial Temperature, \( T_1 = 145^{\circ}\text{C} \)Converted to Kelvin: \( T_1 (K) = 145 + 273.15 = 418.15 \text{ K} \)Final Temperature, \( T_2 = -22^{\circ}\text{C} \)Converted to Kelvin: \( T_2 (K) = -22 + 273.15 = 251.15 \text{ K} \)Pressure, \( P = 2.85 \text{ atm} \).
02

- Use the Combined Gas Law

The Combined Gas Law relates pressure, volume, and temperature: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \].Since the pressure remains constant, we can simplify it to Charles's Law which states: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \].
03

- Rearrange for the Final Volume

Rearrange Charles's Law to solve for \( V_2 \):\[ V_2 = V_1 \times \frac{T_2}{T_1} \].
04

- Substitute Known Values

Substitute the known values into the equation:\[ V_2 = 93 \text{ L} \times \frac{251.15 \text{ K}}{418.15 \text{ K}} \].
05

- Calculate the Final Volume

Perform the calculation:\[ V_2 \approx 55.8 \text{ L} \].

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.

Gas Laws
Gas laws are a set of scientific principles that describe the behavior of gases. These laws help us understand how variables such as pressure, volume, and temperature interact with each other in a gas.
The various laws you might encounter include Boyle's Law, Charles's Law, and Avogadro's Law. Each law isolates one or two variables while holding the others constant.
When all gas laws are combined, they form the Ideal Gas Law represented as \textbf{PV = nRT}, where P stands for pressure, V for volume, T for temperature, n is the number of moles of gas, and R is the gas constant. For this exercise, we'll focus on Combined Gas Law derived from these fundamental principles. It combines Boyle’s, Charles’s, and Gay-Lussac’s laws into one formula:
  • \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
This equation is useful to solve problems involving changes in more than one factor – in this case, both temperature and volume.
Charles's Law
Charles's Law is a specific gas law that highlights the direct relationship between the volume and temperature of a gas held at constant pressure. Mathematically, it's represented as:
  • \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
This relationship implies that when the temperature of a gas increases, its volume increases proportionally, and vice versa, provided the gas pressure is unchanged.
In the given exercise, since the pressure remains constant, we use Charles's Law to find the final volume of the gas after a change in temperature. This simplification helps us transition from the more general Combined Gas Law to a more straightforward problem-solving approach.
Temperature Conversion
Temperature conversion is crucial when dealing with gas laws because temperatures must be in absolute terms, i.e., Kelvin (K). The Kelvin scale is used since it starts at absolute zero, which is the theoretical point where particle motion stops.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:
  • \[ T(K) = T(^{\text{C}}) + 273.15 \]
In this exercise, the initial temperature was \(145^{\text{C}}\), which converts to \[418.15 \text{ K} \] and the final temperature was \(-22^{\text{C}}\), converting to \[251.15 \text{ K} \].
Converting correctly ensures that we apply gas laws accurately without discrepancies due to incompatible units.
Pressure-Volume Relationship
The pressure-volume relationship in gases is generally concerned with how the volume of a gas changes with pressure. According to Boyle's Law, for a fixed amount of gas at constant temperature, the volume is inversely proportional to pressure:
  • \[ P_1 V_1 = P_2 V_2 \]
However, in our exercise, the pressure remains constant at 2.85 atm throughout the process, so we do not need to compute changes in pressure.
Simplifying the Combined Gas Law under constant pressure conditions allows us to use Charles's Law directly. Changing either volume or temperature necessitates consistent units and accurate calculations for practical outcomes.
Understanding these interrelationships gives us better insight into gas behavior in various practical and theoretical applications.

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

Sulfur dioxide emissions from coal-burning power plants are removed by flue- gas desulfurization. The flue gas passes through a scrubber, and a slurry of wet calcium carbonate reacts with it to form carbon dioxide and calcium sulfite. The calcium sulfite then reacts with oxygen to form calcium sulfate, which is sold as gypsum. (a) If the sulfur dioxide concentration is 1000 times higher than its mole fraction in clean, dry air \(\left(2 \times 10^{-10}\right),\) how much calcium sulfate \((\mathrm{kg})\) can be made from scrubbing \(4 \mathrm{GL}\) of flue gas ( \(\left.1 \mathrm{GL}=1 \times 10^{9} \mathrm{~L}\right)\) ? A state-of-the-art scrubber removes at least \(95 \%\) of the sulfur dioxide. (b) If the mole fraction of oxygen in air is \(0.209,\) what volume \((\mathrm{L})\) of air at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is needed to react with all the calcium sulfite?

The lunar surface reaches \(370 \mathrm{~K}\) at midday. The atmosphere consists of neon, argon, and helium at a total pressure of only \(2 \times 10^{-14} \mathrm{~atm} .\) Calculate the \(\mathrm{rm}\) speed of each component in the lunar atmosphere. [Use \(R=8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})\) and express \(\mathscr{A}\) in \(\mathrm{kg} / \mathrm{mol} .]\)

Canadian chemists have developed a modern variation of the 1899 Mond process for preparing extremely pure metallic nickel. A sample of impure nickel reacts with carbon monoxide at \(50^{\circ} \mathrm{C}\) to form gaseous nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\) (a) How many grams of nickel can be converted to the carbonyl with \(3.55 \mathrm{~m}^{3}\) of CO at \(100.7 \mathrm{kPa} ?\) (b) The carbonyl is then decomposed at \(21 \mathrm{~atm}\) and \(155^{\circ} \mathrm{C}\) to pure \((>99.95 \%)\) nickel. How many grams of nickel are obtained per cubic meter of the carbonyl? (c) The released carbon monoxide is cooled and collected for reuse by passing it through water at \(35^{\circ} \mathrm{C}\). If the barometric pressure is 769 torr, what volume (in \(\mathrm{m}^{3}\) ) of \(\mathrm{CO}\) is formed per cubic meter of carbonyl?

When \(35.6 \mathrm{~L}\) of ammonia and \(40.5 \mathrm{~L}\) of oxygen gas at \(\mathrm{STP}\) burn, nitrogen monoxide and water form. After the products return to STP, how many grams of nitrogen monoxide are present? $$\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad[\text { unbalanced }]$$

For each of the following, which shows the greater deviation from ideal behavior at the same set of conditions? Explain. (a) Argon or xenon (b) Water vapor or neon (c) Mercury vapor or radon (d) Water vapor or methane

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.