A 1.00 -mol sample of a monatomic ideal gas is taken through the cycle shown in Figure P 2.65 .

A 1.00 -mol sample of a monatomic ideal gas is taken through...

Key Concepts

Monatomic Ideal Gas

A monatomic ideal gas is a gaseous system in which the molecules consist of single atoms and behave according to the ideal gas law without significant interactions between them. It simplifies the analysis of thermodynamic processes since its internal energy depends only on its temperature, and its heat capacities have fixed values based on degrees of freedom.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates pressure, volume, temperature, and the number of moles of an ideal gas. It is a fundamental equation that allows one to calculate the state variables of a gas system during various thermodynamic processes and cycles.

Thermodynamic Cycle

A thermodynamic cycle is a series of processes that return a system to its original state after undergoing a sequence of transformations. In such cycles, the energy added as heat and the work done by or on the system are balanced over one complete cycle, making it possible to analyze efficiency and energy transfers.

Heat and Work

Heat and work are the two primary ways energy can be transferred between a system and its surroundings. In the context of a thermodynamic cycle, the net heat added to the system is equal to the net work done by the system, as described by the first law of thermodynamics.

First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation in thermodynamic systems. It establishes that the change in a system’s internal energy is equal to the heat added to the system minus the work done by the system, forming the basis for analyzing energy flows in cycles.

Efficiency of Engines

Efficiency is a measure of how well an engine converts the heat input into useful work. It is defined as the ratio of the net work output to the heat input during a complete cycle. Understanding efficiency is key to evaluating the performance of thermodynamic engines.

Carnot Cycle

The Carnot cycle represents an idealized engine cycle that operates between two temperature extremes with maximum possible efficiency. It serves as a benchmark for real engines and provides a theoretical maximum efficiency, as no irreversible losses occur in a Carnot cycle.

Transcript

00:01 Part a of the question, the total energy entering in the system by heat per cycle is given by q total equals to q1 plus q.

00:11 So let us first find q1.

00:14 So q1 will be equals to ncv tb minus t from the given sketch.

00:19 So we are from the ideal gas and sketch we can determine that at b point we have 3 pi vi equals.

00:30 To n r t i and at a p i b i equals to n r t i okay so we can determine q1 equals to n cv d t that is equals to t b minus t a which can be written as n three by two r cv equals to and t b is equals to three t i minus t a is equal to t i so from here after solving we get q1 equals to three n r tl okay which can also be written as q1 equals to 3 p i vn okay so now for the q2 we can tell write that q2 equals to delta e bc minus w bc okay change in internal energy in the process bc and work done in the process bc so where delta e is equal to n c v tc minus t b so from the ideal gas equation and sketch we can write at b 3 p i v i equals to n r t i and at c 3 p i multiplied by 2 v i that is equals to n r t i okay so now we can write that delta e equals to n c cv equals to 3 by 2 r cv equals to 3 by 2 r cv equals to 3 b by 2r multiplied by t is equal to 6 t i minus 3 t i okay so from here we get 4 .5 n r t i so which can be written as also delta e equals to 4 .5 p i v i okay and work done in the process bc this will be equals to p b multiplied by v b minus vc okay so p b is equals to 3 pi and v be equals to vi and vc equals to 2 vi so from here we get minus 3 pi vi okay so now q2 will be equals to delta a bc but which is equals to 4 .5 pi vi minus wbc which is equal to minus 3 pi vi so from here we get 7 .5 pi vi finish so the total energy entering into the system q is equals to q1 plus q2 so q1 is equal to 3 pi vi as we obtained above and q2 is equal to 7 .5 pi vi so from here we get q entering into the system equals to 10 .5 pi v so this is the answer for the part a of the problem okay now moving to the part b in which we will determine the energy exiting from the system okay so this will be equals to q equals to q3 plus q4 magnitude because energy leaving the system will be negative.

03:36 So from here q3 will be equals to ncv td minus tc.

03:42 So at the point d we can write that 3 pi, vi equals to nr ti and at point c we have pi 2vi that is equals to n r t i so we can write q3 equals to n multiplied by cv will be equals to 3 by 2 r and 2 t i minus 6 t i so from here we get minus 6 n r t i so q 3 comes out to be minus 6 p i v okay now solving further calculating q 4 so q 4 will be equals to n cp t a minus t t so now substituting values as we have calculated so n multiplied by cp will be equals to 5 by 2 r and a t i is equal t a t a equals to t i and t d equals to 2 t i so from here we get q4 equals to minus 2 .5 n r t i which is equals to minus 2 .5 p i v i okay so the total energy leaving the system by heat per cycle which is will be equals to q q3 magnitude plus q4 magnitude as both values are in the negative.

05:07 So q3 is equal to minus 6 p .i.

05:11 Vi magnitude plus minus 2 .5 p .i .i.

05:16 Magnitude.

05:17 So from here we get energy leaving equals to 8 .5 p .i .v.

05:24 Okay...

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