Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In an experiment to simulate conditions inside an automobile engine, 0.185 mol of air at 780 K and 3.00 \(\times\) 10\(^6\) Pa is contained in a cylinder of volume 40.0 cm\(^3\). Then 645 J of heat is transferred to the cylinder. (a) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, and use the data in Table 19.1 even though the pressure is not low. Draw a \(pV\)-diagram for this process. (b) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, repeat part (a).

Short Answer

Expert verified
Final temperature at constant volume: higher than initial; at constant pressure: higher but less than constant volume case.

Step by step solution

01

Understand the Problem

We are given that air behaves like nitrogen gas and we need to calculate the final temperature under two conditions: constant volume and constant pressure. We are also given the initial temperature, pressure, number of moles, and heat added.
02

Analyze Constant Volume Process

When volume is constant, no work is done on the system; hence the heat added (\(Q\)) changes the internal energy. For a mole of nitrogen gas, given \(C_v = \frac{5}{2}R\), use: \( Q = nC_v\Delta T \). Calculate \(\Delta T\) and then find the final temperature.
03

Calculate Final Temperature at Constant Volume

\( Q = nC_v\Delta T \rightarrow 645 J = 0.185 \times \frac{5}{2} \times 8.31 \times (T_{final} - 780) \). Solving this equation, we find: \( T_{final} = T_{initial} + \frac{645}{0.185 \times \frac{5}{2} \times 8.31} \).
04

Substitute and Solve

First, evaluate the quantity \( \frac{645}{0.185 \times \frac{5}{2} \times 8.31} \), which results in an increase in temperature. Adding this to the initial 780 K gives the final temperature at constant volume.
05

Analyze Constant Pressure Process

For constant pressure, the equation \( Q = nC_p\Delta T \) applies, where \( C_p = \frac{7}{2}R \). Calculate \(\Delta T\) and find the final temperature under constant pressure condition.
06

Calculate Final Temperature at Constant Pressure

\( Q = nC_p\Delta T \rightarrow 645 J = 0.185 \times \frac{7}{2} \times 8.31 \times (T_{final} - 780) \). Solve for \( T_{final} = T_{initial} + \frac{645}{0.185 \times \frac{7}{2} \times 8.31} \).
07

Substitute and Solve

Evaluate the expression \( \frac{645}{0.185 \times \frac{7}{2} \times 8.31} \) and add this to the initial temperature of 780 K, to determine the final temperature at constant pressure.
08

Draw p-V Diagram

Plot a p-V diagram: For constant volume, plot a vertical line, and for constant pressure, plot a horizontal line moving to the right as volume increases due to temperature increase.

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 Vaia!

Key Concepts

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

Constant Volume Process
A constant volume process, also known as an isochoric process, occurs when the volume of the system remains unchanged. In this scenario:
  • No work is done since work is defined by pressure changes over volume, and here, the volume does not change.
  • All the heat added to the system goes into changing the internal energy.
To describe this mathematically, we use the equation for heat transfer in a constant volume process: \( Q = nC_v\Delta T \).
  • \(Q\) is the heat added,
  • \(n\) is the amount of substance in moles,
  • \(C_v\) is the molar heat capacity at constant volume,
  • \(\Delta T\) is the change in temperature.
Knowing this, we can predict how the temperature will rise when a certain amount of heat, here 645 J, is added to air mimicked as nitrogen gas, while its volume remains constant. More heat translates directly into increased temperature since no energy does work in moving molecules against external pressures.
Constant Pressure Process
A constant pressure process, or isobaric process, is distinguished by a system that maintains its pressure constant as other conditions change. In the case of heat addition:
  • Some of the heat energy goes into doing work by expanding the volume.
  • The remaining energy increases the internal temperature.
This is mathematically expressed by the equation \( Q = nC_p\Delta T \), where
  • \(Q\) is the total heat added to the system,
  • \(C_p\) is the molar heat capacity at constant pressure, which is higher than \(C_v\) due to the work done by the system.
Since work is done in this process, for the same amount of heat transfer, the temperature change \(\Delta T\) is less compared to a constant volume process. This is because a portion of the energy goes into doing external work (expanding the gas), besides increasing the internal energy.
Heat Transfer
Heat transfer is a fundamental concept in thermodynamics referring to the movement of energy from one body or medium to another due to thermal interactions. This transfer can result in changes in temperature or phase of a substance.
  • In a constant volume process, all the transferred heat changes the internal energy, resulting in a temperature increase.
  • In a constant pressure process, heat goes into both changing the internal energy and performing work as the system expands its volume.
The amount of temperature change from a given heat transfer depends on whether the process occurs at constant volume or constant pressure and on the specific heat capacities of the substance. Both processes rely on caloric pathways to harness or redirect energy to determine temperature outcomes.
Ideal Gas Law
The Ideal Gas Law is a cornerstone equation in thermodynamics, encapsulating the behavior of ideal gases under various conditions. It is given by the formula: \[ PV = nRT \]
  • \(P\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant (8.31 J/mol·K),
  • \(T\) is the absolute temperature.
Using the ideal gas law, we can derive relationships crucial to predicting behaviors in varied thermodynamic processes. For example, when volume stays constant, any change in temperature directly translates to a change in pressure. Alternatively, for constant pressure processes, volume and temperature changes are directly proportional, allowing expansions to mitigate temperature increases. Understanding the interplay between these variables through the Ideal Gas Law is critical for analyzing and predicting the outcomes of heat transfer processes in controlled environments.

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 gas in a cylinder is held at a constant pressure of 1.80 \(\times\) 10\(^5\) \(Pa\) and is cooled and compressed from 1.70 m\(^3\) to 1.20 m\(^3\). The internal energy of the gas decreases by 1.40 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [\(Q\)] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

You have a cylinder that contains 500 L of the gas mixture pressurized to 2000 psi (gauge pressure). A regulator sets the gas flow to deliver 8.2 \(L\)/min at atmospheric pressure. Assume that this flow is slow enough that the expansion is isothermal and the gases remain mixed. How much time will it take to empty the cylinder? (a) 1 h; (b) 33 h; (c) 57 h; (d) 140 h.

During an isothermal compression of an ideal gas, 410 J of heat must be removed from the gas to maintain constant temperature. How much work is done by the gas during the process?

The power output of an automobile engine is directly proportional to the mass of air that can be forced into the volume of the engine's cylinders to react chemically with gasoline. Many cars have a \(turbocharger\), which compresses the air before it enters the engine, giving a greater mass of air per volume. This rapid, essentially adiabatic compression also heats the air. To compress it further, the air then passes through an \(intercooler\) in which the air exchanges heat with its surroundings at essentially constant pressure. The air is then drawn into the cylinders. In a typical installation, air is taken into the turbocharger at atmospheric pressure (1.01 \(\times\) 10\(^5\) Pa), density \(\rho\) = 1.23 kg/m\(^3\), and temperature 15.0\(^\circ\)C. It is compressed adiabatically to 1.45 \(\times\) 10\(^5\) Pa. In the intercooler, the air is cooled to the original temperature of 15.0\(^\circ\)C at a constant pressure of 1.45 \(\times\) 10\(^5\) Pa. (a) Draw a \(pV\)-diagram for this sequence of processes. (b) If the volume of one of the engine's cylinders is 575 cm\(^3\), what mass of air exiting from the intercooler will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger and intercooler? (c) If the intercooler is not used, what mass of air exiting from the turbocharger will fill the cylinder at 1.45 \(\times\) 10\(^5\) Pa? Compared to the power output of an engine that takes in air at 1.01 \(\times\) 10\(^5\) Pa at 15.0\(^\circ\)C, what percentage increase in power is obtained by using the turbocharger alone?

A large research balloon containing \(2.00 \times 10^{3} \mathrm{~m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. \(\mathbf{P} 19.50\) ). Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

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.

Sign-up for free