/*! 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 23 In a polytropic process, \(P V^{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a polytropic process, \(P V^{n}=\) constant : (a) If \(n=1\), process is isothermal (b) If \(n=\infty\), process is isochoric (c) If \(n=0\), process is isobaric (d) all the above

Short Answer

Expert verified
The answer is (d) all of the above.

Step by step solution

01

Understanding the Polytropic Process Equation

The given equation for the polytropic process is \(P V^{n} = \text{constant}\). This equation indicates that for a polytropic process, the pressure \(P\) and the volume \(V\) are related by the exponent \(n\), which characterizes the nature of the process.
02

Analyzing n=1 Case

When \(n=1\), the equation becomes \(P V = \text{constant} \). This is the definition of an isothermal process for an ideal gas, which keeps temperature constant during the process. Therefore, statement (a) is true.
03

Analyzing n=∞ Case

For \(n = \infty\), the equation transforms to \(PV^{ ext{large}} \rightarrow P = 0\), essentially implying no change in volume occurs because the pressure becomes extremely low and remains unchanged with volume. Thus, this describes an isochoric process, where the volume is constant. Therefore, statement (b) is true.
04

Analyzing n=0 Case

When \(n=0\), the equation becomes \(P = \text{constant} \), indicating that pressure does not change, which defines an isobaric process. Hence, statement (c) is true.
05

Conclusion of the Process Conditions

Since all three conditions (isothermal for \(n=1\), isochoric for \(n=\infty\), and isobaric for \(n=0\)) are satisfied by the equations' transformations, the correct answer to the exercise is (d) all of the above.

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.

Isothermal Process
The isothermal process is a thermodynamic process that occurs at a constant temperature. In this type of process, energy is added or removed from a system so precisely that its internal temperature remains unchanged throughout. The key aspect here is the maintenance of thermal equilibrium.
In an isothermal process involving an ideal gas, the relationship between pressure and volume is given by Boyle's Law: \(PV = \text{constant}\). Thus, when the volume of the gas increases, its pressure decreases and vice versa, provided that the temperature is held constant.
  • This process can typically be visualized when a gas is compressed or expanded slowly, allowing the heat exchange with the environment to maintain the same temperature.
  • Common practical applications include compressors and heat engines where maintaining a stable temperature is crucial.
To summarize, during an isothermal process, the change in state of the system is isovolumetric but occurs without a change in internal energy since the temperature remains the same.
Isochoric Process
An isochoric process is one where the volume remains constant throughout. In physics, it is often referred to as an isovolumetric or constant volume process. This means that the pressure of the system can change, but the volume does not.
In the context of the polytropic process equation, if we set \(n = \infty\), the pressure becomes negligible in terms of volume effect, translating into a constant volume scenario because no expansion or contraction is considered possible.
  • Characteristics of isochoric processes include fixed mounted containers where the gas or liquid cannot change volume at all.
  • When heat is added to a system during an isochoric process, all added energy will be used to increase the system's internal energy and, consequently, its temperature.
These processes are particularly significant in calorimetry and engines, where understanding heat exchange without volume change is necessary.
Isobaric Process
In an isobaric process, the pressure within the system remains constant during the thermodynamic process. This means any change in volume or temperature does not affect the pressure level.
According to the polytropic process equation, if \(n = 0\), the variable affecting the pressure calculation nullifies, indicating that the pressure remains constant while the volume may change.
  • Isobaric conditions are typically seen in open systems where pressure is equal to the atmospheric level and do not vary as conditions change.
  • Practical examples include the heating or cooling of air in an open room and can also be observed in pistons where the gas expands or contracts freely while maintaining constant pressure.
Understanding isobaric processes is fundamental in designing systems like pressure vessels and in various applications such as welding and drying processes where consistent pressure is crucial.

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

In an adiab-tic expansion, a gas does \(25 \mathrm{~J}\) of work while in an adiabatic compression 100J of work is done on a gas. The change of internal energy in the two processes respectively are: (a) \(25 \mathrm{~J}\) and \(-100 \mathrm{~J}\) (b) \(-25 \mathrm{~J}\) and \(100 \mathrm{~J}\) (c) \(-25 \mathrm{~J}\) and \(-100 \mathrm{~J}\) (d) \(25 \mathrm{~J}\) and \(100 \mathrm{~J}\)

A closed system undergoes a change of state by process \(1 \rightarrow 2\) for which \(Q_{12}=10 \mathrm{~J}\) and \(W_{12}=-5 \mathrm{~J}\). The system is now returned to its initial state by a different path \(2 \rightarrow 1\) for which \(Q_{21}\) is \(-3 \mathrm{~J}\). The total energy for the cycle is : (a) \(-8 \mathrm{~J}\) (b) zero (c) \(-2 \mathrm{~J}\) (d) \(+5 \mathrm{~J}\)

One mole of an ideal gas is enclosed in a conducting vertical cylinder under a light piston. If isothermally the volume of the gas is increased \(n\) times, then the work done in increasing the volume is : [Assume atmospheric pressure is \(P_{0}\) and temperature is \(\left.T_{0}\right]\) (a) \(R T_{0} \log _{e} n\) (b) \(-R T_{0} \log _{t} n\) (c) \(R T_{0} \log _{c} n-(n-1) R T_{0}\) (d) \((n-1) R T_{0}-R T_{0} \log _{e} n\)

A given quantity of an ideal gas is at the pressure \(P\) and the absolute temperature \(T\). The isothermal bulk modulus of the gas is : (a) \(\frac{2}{3} P\) (b) \(P\) (c) \(\frac{3}{2} P\) (d) \(2 P\)

An adiabatic cylinder closed at both ends consists of a freely moving non- coducting thin piston which divides the cylinder into two equal parts and each part contains \(28 \mathrm{~g}\) of \(\mathrm{N}_{2}\). Initially \(1 / 3 \mathrm{rd}\) molecules of nitrogen in the right part are dissociated into atoms. The length of the cylinder is \(1 \mathrm{~m}\) and area of cross-section is \(10^{-2} \mathrm{~m}^{2}\). The natural length of the spring connected to the piston and right wall of the cylinder is \(l-50 \mathrm{~cm}\) and \(k=\sqrt{2} \times 10^{3} \mathrm{~N} / \mathrm{m}\). If the initial pressure in each part is \(P_{0}=\sqrt{2} \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) that what work must be done by the gas in the right part? (a) \(-1414 \mathrm{~J}\) (b) \(1414 \mathrm{~J}\) (c) \(-1515 \mathrm{~J}\) (d) \(1515 \mathrm{~J}\)

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.