/*! 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 5 The volume occupied by a gas dep... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The volume occupied by a gas depends linearly on degrees Celsius at constant pressure, but it is not directly proportional to degrees Celsius. However, it is directly proportional to kelvins. What is the difference between a linear relationship and a direct proportion?

Short Answer

Expert verified
Direct proportion implies a constant ratio; a linear relationship includes a constant term.

Step by step solution

01

Understanding Direct Proportion

In a direct proportion, two quantities have a constant ratio. If one quantity doubles, the other also doubles. Mathematically, if quantity A is directly proportional to quantity B, then \( A = k \times B \), where \( k \) is a constant.
02

Understanding Linear Relationship

A linear relationship between two quantities is represented by the equation \( y = mx + c \). Here, \( m \) is the slope of the line, and \( c \) is the y-intercept. Unlike direct proportion, it allows for an addition of a constant \( c \), meaning the two quantities do not have a constant ratio.
03

Applying Definitions to the Problem

The problem states that volume is directly proportional to temperature in kelvin but only linearly related to temperature in Celsius. This means when dealing with Celsius, there is an additional constant, similar to the \( c \) in \( y = mx + c \), because Celsius is not an absolute scale (0°C ≠ 0 Kelvin).

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.

Direct Proportion
Direct proportion is a fundamental concept in mathematics, especially when dealing with relationships between two quantities. When two quantities are in direct proportion, their ratio remains constant. This means that if one quantity increases, the other one increases at a consistent rate, maintaining the same ratio. For example:- If we consider two quantities, say, quantity A and quantity B, that are directly proportional, then we can express this relationship as: \( A = k \times B \).- Here, \( k \) is a constant of proportionality. Therefore, if quantity A doubles, quantity B also doubles.Direct proportionality is a straightforward way to see how two quantities relate to each other, often symbolizing a harmonious and simple relationship where outputs scale linearly with inputs, given that the constant is not zero.Understanding direct proportion is crucial, especially in physics and chemistry where many phenomena rely on these predictable relationships.
Linear Relationship
Linear relationships describe how two variables relate to each other through a straight line when graphed. This relationship can be expressed in the equation:\( y = mx + c \) Where:- \( y \) is the dependent variable- \( x \) is the independent variable- \( m \) is the slope of the line, indicating the rate of change- \( c \) is the y-intercept, the value where the line crosses the y-axis.A key difference from direct proportion is the y-intercept. In direct proportion, there is no additional constant \( c \). This baggage of "c" allows the line to start from anywhere on the y-axis rather than zero. Having a linear relationship means that the relationship progresses at a constant rate of change, but it may start anywhere vertically on the graph, unlike direct proportion, which always starts from the origin. This flexibility makes linear relationships broadly applicable to real-world situations.
Temperature Scales
Temperature scales are systems of measuring temperature where specific standard points, like the freezing and boiling points of water, are used as references. The two most commonly used temperature scales are Celsius and Kelvin: - **Celsius**: - It is a scale where water freezes at 0°C and boils at 100°C under normal atmospheric pressure. - It's a relative scale, which means 0°C doesn't represent absolute zero. - **Kelvin**: - This is the absolute temperature scale, so absolute zero (the coldest possible temperature) is marked as 0 Kelvin. - The increments are the same as on the Celsius scale (i.e., a change in temperature by 1 Kelvin equals a change by 1°C), but it starts at absolute zero. The Kelvin scale is widely used in scientific contexts because it allows for direct proportionality with thermodynamic equations, providing a more accurate and absolute measure of thermal energy.
Celsius and Kelvin
The Celsius and Kelvin scales, while related, serve different scientific and practical purposes. They are vital in understanding temperature measurements in scientific studies. - **Celsius (°C):** - Commonly used in everyday situations and weather forecasting. - Its zero point is based on the freezing point of water. - However, it does not represent the absence of thermal energy. - **Kelvin (K):** - Used extensively in scientific research because it starts at absolute zero, a physical reality where no thermal energy exists. - The direct proportionality of temperature-related calculations with Kelvin makes it crucial in gas laws, like the one where volume is directly proportional to temperature when measured in Kelvin. Understanding both scales allows scientists and students to appropriately apply them according to the situation, especially when working with laws of thermodynamics and gases.

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 of Ne and Ar gases at \(350 \mathrm{~K}\) contains twice as many moles of Ne as of Ar and has a total mass of \(50.0 \mathrm{~g}\). If the density of the mixture is \(4.00 \mathrm{~g} / \mathrm{L},\) what is the partial pressure of \(\mathrm{Ne}\) ?

Chloroform, \(\mathrm{CHCl}_{3},\) is a volatile (easily vaporized) liquid solvent. Calculate the density of chloroform vapor at \(98^{\circ} \mathrm{C}\) and \(797 \mathrm{mmHg}\). Give the answer in grams per liter.

The combustion method used to analyze for carbon and hydrogen can be adapted to give percentage \(\mathrm{N}\) by collecting the nitrogen from combustion of the eompound as \(\mathrm{N}_{2}\). A sample of a compound weighing \(8.75 \mathrm{mg}\) gave \(1.59 \mathrm{~mL} \mathrm{~N}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(749 \mathrm{mmHg} .\) What is the percent- age \(\mathrm{N}\) in the compound?

A sample of neon gas is placed into a \(22.4-\mathrm{L}\) rigid container that already contains \(1.00 \mathrm{~mol}\) of nitrogen gas. Indicate which statement is correct in describing the change that occurs when the neon gas is added to the container. Assume ideal gas behavior. The partial pressure of the nitrogen gas in the container would double. The partial pressure of the nitrogen gas would increase by some amount, but we cannot know the exact amount without more information. ci The partial pressure of the nitrogen gas in the container would decrease by \(1 / 2(50 \%)\). d) The partial pressure of the nitrogen gas in the container would decrease by some amount, but we cannot know-the exat amount without more information. e The partial pressure of the nitrogen gas would not change.

In an experiment, you fill a heavy-walled 6.00 -L flask with methane gas, \(\mathrm{CH}_{4}\). If the flask contains \(7.13 \mathrm{~g}\) of methane at \(19^{\circ} \mathrm{C}\), what is the gas pressure?

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.