/*! 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 58 (a) How much heat must be added ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) How much heat must be added to raise the temperature of 1.5 mol of air from \(25.0^{\circ} \mathrm{C}\) to \(33.0^{\circ} \mathrm{C}\) at constant volume? Assume air is completely diatomic. (b) Repeat the problem for the same number of moles of xenon, Xe.

Short Answer

Expert verified
(a) The heat required to raise the temperature of 1.5 mol of diatomic air from 25°C to 33°C is \( 249.39 \, J \). (b) The heat required to raise the temperature of 1.5 mol of xenon from 25°C to 33°C is \( 149.64 \, J \).

Step by step solution

01

Determine the specific heat capacity of diatomic air and xenon

For diatomic air, we use the specific heat capacity at constant volume (Cv) formula for diatomic gas: \( Cv = \frac{5}{2} R \), where R is the universal gas constant, R = 8.314 J/molK. For xenon, which is a monoatomic gas, we use the specific heat capacity at constant volume (Cv) formula for monoatomic gas: \( Cv = \frac{3}{2} R \).
02

Calculate the heat required for diatomic air

Using the heat capacity formula \( q = mcΔT \), where m is the number of moles (1.5 mol), c is the specific heat capacity and ΔT is the change in temperature (33 - 25 = 8°C or 8K), we can find the heat required to raise the temperature of diatomic air: \( q_{air} = (1.5 \, mol) \times \left(\frac{5}{2} \times 8.314 \, \frac{J}{mol \cdot K}\right) \times (8 \, K) \) Calculate the value of \(q_{air}\): \( q_{air} = 1.5 \times 20.785 \times 8 \) \( q_{air} = 249.39 \, J \)
03

Calculate the heat required for xenon

Now, we can find the heat required to raise the temperature of xenon using the same heat capacity formula: \( q_{Xe} = (1.5 \, mol) \times \left(\frac{3}{2} \times 8.314 \, \frac{J}{mol \cdot K}\right) \times (8 \, K) \) Calculate the value of \(q_{Xe}\): \( q_{Xe} = 1.5 \times 12.471 \times 8 \) \( q_{Xe} = 149.64 \, J \) (a) The heat required to raise the temperature of 1.5 mol of diatomic air from 25°C to 33°C is 249.39 J. (b) The heat required to raise the temperature of 1.5 mol of xenon from 25°C to 33°C is 149.64 J.

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.

Specific Heat Capacity
Understanding how much energy is required to change the temperature of a substance is key in thermodynamics. Specific heat capacity is the quantity of heat required to change the temperature of one mole of a substance by one degree Kelvin. Each substance has its own specific heat capacity, influenced by its structure and bonding.
The heat capacity at constant volume for a substance is denoted as \( C_v \). The formula \( q = mc\Delta T \) allows us to calculate the energy needed to change temperature, where \( q \) is the heat added, \( m \) is the number of moles, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
In our exercise, this formula helps calculate the energy for both diatomic air and xenon, illustrating differences based on their molecular nature.
Diatomic Gas
Diatomic gases, like oxygen and nitrogen in air, consist of two atoms per molecule. These gases exhibit specific heat capacities that differ from monoatomic gases.
Under constant volume conditions, the specific heat capacity \( C_v \) for a diatomic gas is given by \( C_v = \frac{5}{2} R \). This accounts for both translational and rotational movements of the molecules. Diatomic gases have more degrees of freedom compared to monoatomic gases, allowing them to store more heat energy, hence a larger \( C_v \).
In our calculation, we used this value for air, considering it as a diatomic gas, to determine that 249.39 Joules of heat is needed to increase its temperature.
Monoatomic Gas
Monoatomic gases, like xenon, are composed of single atoms. These gases have a simplified structure, influencing their specific heat capacity.
The heat capacity at constant volume for a monoatomic gas is \( C_v = \frac{3}{2} R \). This lower value arises due to fewer degrees of freedom; monoatomic gases only possess translational motion as opposed to rotational or vibrational.
For xenon, this principle demonstrated why it requires less energy—149.64 Joules—to achieve the same temperature increase as diatomic air, reflecting its simpler atomic nature.
Universal Gas Constant
The universal gas constant \( R \) is a crucial element in thermodynamic equations. It represents the molar energy scale for gases and connects the microscopic world of molecules to macroscopic properties like volume and pressure. In energy units, \( R \) is valued at 8.314 Joules per mole per Kelvin.
This constant is integrated into various gas equations, including those for specific heat calculations \( C_v = \frac{f}{2} R \), where \( f \) is the degrees of freedom.
By using \( R \) in our heat calculations, we ensure accuracy when determining how much energy is needed to change the temperature of different gases. The uniformity offered by \( R \) allows us to compare different gas behaviors under similar conditions.

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) Hydrogen molecules (molar mass is equal to 2.016 g/mol) have \(v_{\text {rms }}\) equal to 193 m/s. What is the temperature? (b) Much of the gas near the Sun is atomic hydrogen (H rather than \(\mathrm{H}_{2}\) ). Its temperature would have to be \(1.5 \times 10^{7} \mathrm{K}\) for the rms speed \(v_{\mathrm{rms}}\) to equal the escape velocity from the Sun. What is that velocity?

Calculate the depth to which Avogadro's number of table tennis balls would cover Earth. Each ball has a diameter of \(3.75 \mathrm{cm} .\) Assume the space between balls adds an extra \(25.0 \%\) to their volume and assume they are not crushed by their own weight.

A sealed, perfectly insulated container contains 0.630 mol of air at \(20.0^{\circ} \mathrm{C}\) and an iron stirring bar of mass 40.0 g. The stirring bar is magnetically driven to a kinetic energy of \(50.0 \mathrm{J}\) and allowed to slow down by air resistance. What is the equilibrium temperature?

88\. Find the total number of collisions between molecules in \(1.00 \mathrm{s}\) in \(1.00 \mathrm{L}\) of nitrogen gas at standard temperature and pressure \(\left(0^{\circ} \mathrm{C}, 1.00 \mathrm{atm}\right) .\) Use \(1.88 \times 10^{-10} \mathrm{m}\) as the effective radius of a nitrogen molecule. (The number of collisions per second is the reciprocal of the collision time.) Keep in mind that each collision involves two molecules, so if one molecule collides once in a certain period of time, the collision of the molecule it hit cannot be counted.

. In car racing, one advantage of mixing liquid nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) with air is that the boiling of the "nitrous" absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, \(-88^{\circ} \mathrm{C},\) is mixed with \(4.0 \mathrm{mol}\) of air (assumed diatomic) at \(30^{\circ} \mathrm{C}\). What is the final temperature of the mixture? Use the measured heat capacity of \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C},\) which is \(30.4 \mathrm{J} / \mathrm{mol}^{\circ} \mathrm{C} .\) (The primary advantage of nitrous oxide is that it consists of \(1 / 3\) oxygen, which is more than air contains, so it supplies more oxygen to bum the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.)

See all solutions

Recommended explanations on Physics 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.