/*! 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 168 (A): When temperature of an idea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(A): When temperature of an ideal gas increases from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\) at constant \(\mathrm{P}\), volume increases by \(100 \mathrm{~L}\). \((\mathbf{R}): \mathrm{V} \alpha \mathrm{T}\) at constant \(\mathrm{P}\)

Short Answer

Expert verified
The statement that \(V \propto T\) at constant pressure is correct, as shown by applying Charles's Law.

Step by step solution

01

Understanding the Problem

We are given that an ideal gas's temperature increases from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). During this temperature increase, at constant pressure \(\mathrm{P}\), the volume increases by \(100 \mathrm{~L}\). We must relate this change to the direct proportionality of volume and temperature at constant pressure, expressed as \(\mathrm{V} \propto \mathrm{T}\).
02

Convert Temperatures from Celsius to Kelvin

To use the gas law, we need temperatures in Kelvin. Convert the given temperatures \(27^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\) to Kelvin using the formula: \(T(K) = T(^{\circ}C) + 273.15\). Thus, \(27^{\circ} \mathrm{C} = 300.15 \mathrm{~K}\) and \(127^{\circ} \mathrm{C} = 400.15 \mathrm{~K}\).
03

Apply Charles's Law

Charles's law states that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) for an ideal gas at constant pressure. Let the initial volume be \(V_1\) and the final volume be \(V_2 = V_1 + 100\). We have \(T_1 = 300.15 \mathrm{~K}\), \(T_2 = 400.15 \mathrm{~K}\).
04

Set Up Equation and Solve for Initial Volume

From Charles's law, \(\frac{V_1}{300.15} = \frac{V_1 + 100}{400.15}\). Cross-multiply to solve for \(V_1\). This gives:\[400.15 \cdot V_1 = 300.15 \cdot (V_1 + 100)\]Simplify and solve accordingly.
05

Simplify and Calculate Initial Volume

Distribute and simplify: \[400.15V_1 = 300.15V_1 + 30015\]Rearrange:\[400.15V_1 - 300.15V_1 = 30015\]\[100V_1 = 30015\]\[V_1 = \frac{30015}{100} = 300.15 \mathrm{~L}\].
06

Determine Final Volume

The initial volume \(V_1\) was found to be \(300.15 \mathrm{~L}\). Thus, the final volume \(V_2 = V_1 + 100 = 300.15 + 100 = 400.15 \mathrm{~L}\).
07

Evaluate the Proportional Relationship

The calculated volumes and given temperatures confirm that \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), thus supporting the relationship \(V \propto T\) at constant pressure.

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.

Ideal Gas Law
The Ideal Gas Law is a foundational concept in chemistry and physics, offering a simple equation to describe the state of an ideal gas. It is often expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for the amount of gas in moles, \( R \) as the ideal gas constant, and \( T \) for temperature in Kelvin. This equation ties together the properties of the gas and can be used to determine unknown properties when others are known.

While the Ideal Gas Law is quite comprehensive, specific conditions allow for the simplification into other laws such as Boyle’s Law, Charles’s Law, and Avogadro’s Law. In the exercise, Charles’s Law comes into play. This law, also known as the law of volumes, states that the volume of an ideal gas is directly proportional to its temperature, provided the pressure is kept constant. It is a specific form of the Ideal Gas Law, emphasizing the relation between temperature and volume at constant pressure.
Temperature Conversion
Temperature conversion is crucial when dealing with gas laws, as these often require temperature values in Kelvin to properly apply the equations. Celsius is a common unit of temperature but converting it to Kelvin makes sure we are correctly following the prerequisites of the Ideal Gas Law.

To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This conversion is straightforward and essential for scientific calculations as Kelvin is an absolute scale beginning at absolute zero, \(-273.15^{\circ}C\), where all thermal motion ceases.
  • Formula: \( T(K) = T(^{\circ}C) + 273.15 \)
  • Example: \( 27^{\circ}C = 300.15 K \)
  • Example: \( 127^{\circ}C = 400.15 K \)
By converting to Kelvin, calculations use absolute temperature values, avoiding complications associated with Celsius.
Proportionality of Volume and Temperature
The concept of proportionality between volume and temperature under constant pressure is captured elegantly in Charles's Law. It emphasizes that for a given amount of gas, the volume will increase as the temperature rises, so long as the pressure remains unchanged.

This relationship is quantitatively given by \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Here, \( V_1 \) and \( V_2 \) are initial and final volumes and \( T_1 \) and \( T_2 \) are initial and final temperatures. It should be noted that temperatures must be in Kelvin for the calculation to hold.

Mathematically, if you increase the temperature from, say, \( 300.15 K \) to \( 400.15 K \) as in the exercise, the volume will increase proportionally if pressure is constant. This linear relationship is important in predicting how a gas will behave when subjected to changes in temperature, making it a vital part of gas law studies.

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

Which of the following statement(s) is/are true about the effect of an increase in temperature on the distribution molecular speeds in a gas? a. the most probable speed increases b. the fraction of the molecules with the most probable speed increases c. the distribution becomes broader d. the area under the distribution curve remains the same as the under the lower temperature

A gas described by van der Waal's equation a. Behaves similar to an ideal gas in the limit of large molar volumes b. Behaves similar to an ideal gas in the limit of large pressures c. Is characterized by van der Waal's coefficients that are dependent on the identity of the gas but are independent of the temperature d. Has the pressure that is lower than the pressure exerted by the same gas behaving ideally

\(4.4 \mathrm{~g} \mathrm{CO}_{2}\) gas and \(2.24 \mathrm{l}\) of \(\mathrm{H}_{2}\) gas are taken in \(\mathrm{a}\) one litre container at \(300 \mathrm{~K}\). The total pressure of the gas in container will be a. \(14.856\) atm b. \(4378.15 \mathrm{~mm}\) of \(\mathrm{Hg}\) c. \(4.926 \mathrm{~atm}\) d. \(3743.76 \mathrm{~mm}\) of \(\mathrm{Hg}\)

A solution has a \(1: 4\) mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at \(20^{\circ} \mathrm{C}\) are \(400 \mathrm{~mm} \mathrm{Hg}\) for pentane and 120 \(\mathrm{mm} \mathrm{Hg}\) for hexane. The mole fraction of pentane in the vapour phase would be a. \(0.200\) b. \(0.549\) c. \(0.786\) d. \(0.478\)

The rate of effusion of two gases 'a' and 'b' under identical conditions of temperature and pressure are in the ratio of \(2: 1\). What is the ratio of \(\mathrm{rms}\) velocity of their molecules if \(\mathrm{T}_{\mathrm{a}}\) and \(\mathrm{T}_{\mathrm{b}}\) are in the ratio of \(2: 1\) ? a. \(\sqrt{2}: 1\) b. \(2: 1\) c. \(1: \sqrt{2}\) d. \(2 \sqrt{2}: 1\)

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.