/*! 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 29 Calculate the new temperature, i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 29

Calculate the new temperature, in degrees Celsius, for each of the following with \(n\) and \(V\) constant: a. A sample of xenon at \(25^{\circ} \mathrm{C}\) and \(740 \mathrm{mmHg}\) is cooled to give a pressure of \(620 \mathrm{mmHg}\). b. A tank of argon gas with a pressure of \(0.950\) atm at \(-18{ }^{\circ} \mathrm{C}\) is heated to give a pressure of 1250 torr.

Short Answer

Expert verified
a.\( -23.15^{\text{o}} \text{C} \) , b. \( 168.71^{\text{o}} \text{C} \)

Step by step solution

01

- Understanding the initial conditions and the formula

Use the ideal gas law in the form of combined gas law for processes where the temperature, pressure, and volume change: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \(P\) is the pressure, \(T\) is the temperature in Kelvin, and the subscripts 1 and 2 refer to the initial and final states, respectively.
02

- Convert the temperature to Kelvin (Part a)

The initial temperature is given as \(25^{\text{o}} \text{C}\). Convert it to Kelvin. \[ T_1 = 25^{\text{o}} \text{C} + 273.15 = 298.15 \text{ K} \]
03

- Identify initial and final pressures (Part a)

The initial pressure \( P_1 \) is \( 740 \text{ mmHg} \) and the final pressure \( P_2 \) is \( 620 \text{ mmHg} \).
04

- Apply the combined gas law (Part a)

Rearrange the formula to solve for \( T_2 \): \[ T_2 = T_1 \times \frac{P_2}{P_1} \] Substitute the known values: \[ T_2 = 298.15 \text{ K} \times \frac{620 \text{ mmHg}}{740 \text{ mmHg}} \] Calculate the final temperature: \[ T_2 = 250 \text{ K} \]
05

- Convert the temperature back to Celsius (Part a)

Convert the final temperature back to Celsius: \[ T_2 = 250 \text{ K} - 273.15 = -23.15^{\text{o}} \text{C} \]
06

- Convert the temperature to Kelvin (Part b)

The initial temperature is given as \(-18^{\text{o}} \text{C}\). Convert it to Kelvin. \[ T_1 = -18^{\text{o}} \text{C} + 273.15 = 255.15 \text{ K} \]
07

- Identify initial and final pressures (Part b)

The initial pressure \( P_1 \) is\( 0.950 \text{ atm} \) and the final pressure \( P_2 \) is \( 1250 \text{ torr} \). Convert the pressures to the same unit (atm or torr). 1 atm = 760 torr. \[ P_1 = 0.950 \text{ atm} \] \[ P_2 = 1250 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 1.645 \text{ atm} \]
08

- Apply the combined gas law (Part b)

Rearrange the formula to solve for \( T_2 \): \[ T_2 = T_1 \times \frac{P_2}{P_1} \] Substitute the known values: \[ T_2 = 255.15 \text{ K} \times \frac{1.645 \text{ atm}}{0.950 \text{ atm}} \] Calculate the final temperature: \[ T_2 = 441.86 \text{ K} \]
09

- Convert the temperature back to Celsius (Part b)

Convert the final temperature back to Celsius: \[ T_2 = 441.86 \text{ K} - 273.15 = 168.71^{\text{o}} \text{C} \]

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 91影视!

Key Concepts

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

ideal gas law
In order to understand the combined gas law, first, we need to grasp the basics of the ideal gas law. The ideal gas law is represented by the equation: \[ PV = nRT \] Here,
  • \(P\) = Pressure
  • \(V\) = Volume
  • \(n\) = Number of moles of gas
  • \(R\) = Ideal gas constant
  • \(T\) = Temperature in Kelvin
This law helps us understand how gases behave under different conditions. The combined gas law is derived from the ideal gas law and relates the initial and final states of a gas sample when the number of moles remains constant. In the combined gas law, we use: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] This equation allows us to solve problems where the pressure and temperature of a gas change, while its volume and amount remain constant.
temperature conversion
Temperature conversion is a crucial step in solving gas law problems. The combined gas law requires the temperature to be in Kelvin. The conversion formulas are:
  • 掳C to Kelvin: \( T(K) = T(掳C) + 273.15 \)
  • Kelvin to 掳C: \( T(掳C) = T(K) - 273.15 \)
For example, if we have a temperature of \( 25^{\circ} \text{C} \), it needs to be converted to Kelvin: \[ 25^{\circ} \text{C} + 273.15 = 298.15 \text{ K} \] This step is vital since calculations involving gas laws must use absolute temperature values to be accurate.
pressure units conversion
Different pressure units can be used in problems, and converting between them is key. Common units include atmospheres (atm), millimeters of mercury (mmHg), and torr.
  • 1 atm = 760 mmHg
  • 1 atm = 760 torr
If we need to convert pressure values for consistent units in equations, we use these conversions. For instance, to convert 1250 torr to atm: \[ 1250 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 1.645 \text{ atm} \] Using the same units helps avoid mistakes in calculations and ensures correct results.
Kelvin to Celsius
Once we have solved for the temperature in Kelvin using the combined gas law, the final step is often to convert this back to degrees Celsius, especially when the problem specifically requests it. The conversion is simple: \[ T(^{\circ} \text{C}) = T(K) - 273.15 \] For example, if the final temperature calculated is \( 441.86 \text{ K} \), convert it to degrees Celsius as: \[ 441.86 \text{ K} - 273.15 = 168.71^{\circ} \text{C} \] This conversion is important for providing the answer in a standard and understandable temperature scale, often preferred in real-world scenarios.

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

What is the total pressure, in \(\mathrm{mmHg}\), of a gas mixture containing argon gas at \(0.25\) atm, helium gas at \(350 \mathrm{mmHg}\), and nitrogen gas at 360 torr?

A 10.0-L balloon contains helium gas at a pressure of \(655 \mathrm{mmHg}\). What is the new pressure, in \(\mathrm{mmHg}\), of the helium gas at each of the following volumes, if there is no change in temperature or amount of gas? a. \(20.0 \mathrm{~L}\) b. \(2.50 \mathrm{~L}\) c. \(1500 . \mathrm{mL}\)

Explain each of the following observations: a. Boiling water at sea level is hotter than boiling water in the mountains. b. Water used to sterilize surgical equipment is heated to \(120^{\circ} \mathrm{C}\) at \(2.0 \mathrm{~atm}\) in an autoclave.

Why do scuba divers need to exhale air when they ascend to the surface of the water?

A balloon contains \(2500 \mathrm{~mL}\) of helium gas at \(75^{\circ} \mathrm{C}\). What is the new volume, in milliliters, of the gas when the temperature changes to each of the following, if \(n\) and \(P\) do not change? a. \(55^{\circ} \mathrm{C}\) b. \(680 . \mathrm{K}\) c. \(-25^{\circ} \mathrm{C}\) d. \(240 . \mathrm{K}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation
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.