/*! 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 39 One mole of neon, a monatomic ga... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

Short Answer

Expert verified
Total heat added is approximately 13620 J.

Step by step solution

01

Understand the Process

The process consists of two stages: first, the neon gas at constant volume is heated until the pressure is tripled; next, the gas at this new pressure is heated at constant pressure until the volume is doubled. Both processes assume the gas behaves ideally.
02

Ideal Gas Law Application

For an ideal gas undergoing change, the ideal gas law \( PV = nRT \) can be used where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. Since \( n = 1 \) for this exercise, simplify as \( PV = RT \).
03

Calculate Heat for Constant Volume

In a constant volume process, \( \Delta Q = nC_v\Delta T \), where \( C_v = \frac{3}{2}R \) for a monatomic ideal gas. Initial temperature \( T_1 = 273K \). After tripling the pressure, \( P_2 = 3P_1 \), the new temperature is \( T_2 = 3T_1 = 819K \). The heat is \( Q_1 = 1 \times \frac{3}{2}R(819K - 273K) \).
04

Calculate Heat for Constant Pressure

For constant pressure, \( \Delta Q = nC_p\Delta T \), where \( C_p = \frac{5}{2}R \). The initial pressure here is \( 3P_1 \) and temperature \( T_2 = 819K \). Volume is doubled, so new temperature \( T_3 = 2T_2 = 1638K \). The heat is \( Q_2 = 1 \times \frac{5}{2}R(1638K - 819K) \).
05

Calculate Total Heat Added

Sum the heats from both processes: \( Q_{total} = Q_1 + Q_2 = \frac{3}{2}R(546K) + \frac{5}{2}R(819K) \). Substitute \( R = 8.314 \ J/mol\cdot K \) to find the total heat in Joules.

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 fundamental principle in thermodynamics that describes the behavior of gases under various conditions. It is given by the equation \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
In this exercise, we are dealing with one mole of a gas (\( n=1 \)), simplifying our equation to \( PV = RT \).
This law helps predict how changes in one of these properties affect the others. For example, when we heat neon gas while keeping the volume steady, the pressure rises, showing how tightly linked these properties are.
Understanding this relationship is crucial for predicting gas behavior in different thermodynamic processes.
Monatomic Gas
Monatomic gases are composed of single-atom molecules like helium or neon. These gases often behave like ideal gases, especially at standard conditions. One special characteristic of monatomic gases is their heat capacity, which distinguishes them from diatomic or polyatomic gases. When considering their heat capacity values, monatomic gases are simple systems. Their specific heat capacity at a constant volume is \( C_v = \frac{3}{2}R \), where \( R \) is the ideal gas constant.
This characteristic is particularly important in thermodynamic calculations, such as determining the energy absorbed or released during heat transfer.
Heat Capacity
Heat Capacity is a crucial property in thermodynamics, representing the amount of heat needed to change a substance's temperature by a certain amount. For gases, it varies depending on how they are held during heating, either at constant volume or constant pressure.
  • Constant Volume Heat Capacity \( (C_v) \): In monatomic gases, \( C_v \) is typically \( \frac{3}{2}R \). This value is used when the gas is heated without any change in volume.
  • Constant Pressure Heat Capacity \( (C_p) \): For monatomic gases, \( C_p \) equals \( \frac{5}{2}R \). This applies when the gas is allowed to expand, like when volume changes at a constant pressure.
The distinction between \( C_v \) and \( C_p \) forms the basis for many calculations, such as those determining the total heat transferred to or from the gas.
Constant Volume Process
In a Constant Volume Process, the volume stays the same while other properties like pressure and temperature can vary. For an ideal gas like neon, this situation is common when the gas is heated inside a sealed, non-expandable container.
Since no work is done on or by the gas (work is done by volume changes), the energy transferred as heat results in temperature and pressure changes. In our specific scenario, as the neon is heated, its pressure triples, and the temperature rises accordingly from \( T_1 \) to \( T_2 \).
The heat added, \( Q \), can be calculated using the formula \( Q = nC_v\Delta T \), reinforcing the significance of heat capacity in thermal processes.
Constant Pressure Process
During a Constant Pressure Process, the gas pressure remains fixed while the volume and temperature can change. This is typical when a gas is allowed to expand or is compressed in an environment where pressure doesn't change, such as in an open container or system connected to a constant pressure reservoir.
In our exercise, after the initial constant volume heating, neon experiences an expansion, doubling its volume. During this process, the temperature elevates from \( T_2 \) to \( T_3 \), and the heat transferred is calculated using \( Q = nC_p\Delta T \).
This process illustrates how heat capacity at constant pressure (\( C_p \)) is critical in assessing the energy required for temperature changes under constant pressure 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 fifteen-watt heater is used to heat a monatomic ideal gas at a constant pressure of \(7.60 \times 10^{5} \mathrm{~Pa}\) During the process, the \(1.40 \times 10^{-3} \mathrm{~m}^{3}\) volume of the gas increases by \(25.0 \%\). How long was the heater on?

The drawing (not to scale) shows the way in which the pressure and volume change for an ideal gas that is used as the working substance in a Carnot engine. The gas begins at point a (pressure \(=P_{\mathrm{a}}\), volume \(\left.=V_{\mathrm{a}}\right)\) and expands isothermally at temperature \(T_{\mathrm{H}}\) until point \(\mathrm{b}\) (pressure \(=P_{\mathrm{b}}\), volume \(=V_{\mathrm{b}}\) ) is reached. During this expansion, the input heat of magnitude \(\left|Q_{\mathrm{H}}\right|\) enters the gas from the hot reservoir of the engine. Then, from point \(\mathrm{b}\) to point \(\mathrm{c}\) (pressure \(=P_{\mathrm{c}}\), volume \(\left.=V_{\mathrm{c}}\right)\), the gas expands adiabatically. Next, the gas is compressed isothermally at temperature \(T_{\mathrm{C}}\) from point \(\mathrm{c}\) to point \(\mathrm{d}\) (pressure \(=P_{\mathrm{d}}\), volume \(=V_{\mathrm{d}}\) ). During this compression, heat of magnitude \(\left|Q_{\mathrm{C}}\right|\) is rejected to the cold reservoir of the engine. Finally, the gas is compressed adiabatically from point \(d\) to point a, where the gas is back in its initial state. The overall process a to \(b\) to \(\mathrm{c}\) to \(\mathrm{d}\) is called a Carnot cycle. Prove for this cycle that \(|Q \mathrm{c}| /|Q \mathrm{H}|=T_{\mathrm{C}} / T_{\mathrm{H}}\)

Two Camot air conditioners, \(\mathrm{A}\) and \(\mathrm{B}\), are removing heat from different rooms. The outside temperature is the same for both, but the room temperatures are different. The room serviced by unit \(\mathrm{A}\) is kept colder than the room serviced by unit B. The heat removed from both rooms is the same, (a) Which unit requires the greater amount of work? (b) Which unit deposits the greater amount of heat outside? Explain. The outside temperature is \(309.0 \mathrm{~K}\). The room serviced by unit \(\mathrm{A}\) is kept at a temperature of \(294.0 \mathrm{~K}\), while the room serviced by unit \(\mathrm{B}\) is kept at \(301.0 \mathrm{~K}\). The heat removed from either room is \(4330 \mathrm{~J}\). For both units, find the magnitude of the work required and the magnitude of the heat deposited outside. Verify that your answers are consistent with your answers to the Concept Questions.

Concept Questions A system does \(4.8 \times 10^{4} \mathrm{~J}\) of work, and \(7.6 \times 10^{4} \mathrm{~J}\) of heat flows into the system in the process. (a) Considered by itself, does the work increase or decrease the internal energy of the system? (b) Considered by itself, does the heat increase or decrease the internal energy? (c) Considering the work and heat together, does the internal energy of the system increase, decrease, or remain the same? Explain. Problem Find the change in the internal energy of the system. Make sure that your answer is consistent with your answers to the Concept Questions.

A Carnot engine operates with an efficiency of \(27.0 \%\) when the temperature of its cold reservoir is \(275 \mathrm{~K}\). Assuming that the temperature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to \(32.0 \% ?\)

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.