# Answered! Question: A gas within a piston -cylinder assembly undergoes a thermodynamic cycle consisting of the follow……

process 1–> constant volume V= 0.028 m^3 , U2-U1= 26.4 KJ
process 2 ..>pV=constant U3=U2
process 3..>constant pressure p=1.4 bar ,

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Answered! Question: A gas within a piston -cylinder assembly undergoes a thermodynamic cycle consisting of the follow……
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there are no signifacnt changes in the kintic energy and potenial energy

A) sketch the cycle on the P-V digram

b) calculate the boundry work and the net heat transfer during Process 1–> 2 in KJ

c) calculate the boundry work and the net heat transfer during process 2–>3 in KJ

d) calculate the net heat transfer during process 3–> 1 , in KJ

e) calcualte the net work for the cycle , is this a power cycle or refrigrent cycle?

## General guidance

Concepts and reason

This question is based on the thermodynamics cyclic for a closed system.

Closed system:

It is the system, in which there is no exchange of matter with its surrounding such as mass transfer. In this system, energy (work or heat) can be exchange with it surrounding. In a closed system, the change in internal energy is zero. The example of the closed system is piston-cylinder.

First of all, draw the P-V diagram with the help of the three thermodynamic cycles which are provided.

Use the expressions of heat transfer and work done for a closed system to calculate the heat transfer and work done for each process.

Use the magnitude of the work done due to each process in order to find the net work done by adding all of them.

Fundamentals

Write the expression of the heat transfer for a closed system.

Here, is the heat transfer,is the change in internal energy and is the work done.

Write the expression of the work done at constant pressure.

Here, is the pressure, is the final volume and is the initial volume.

Write the expression of the work done at constant temperature.

Here, and are the initial and final pressure respectively, and are the initial and final volume respectively.

Write the expression of ideal gas law.

Here, is the number of moles and is the gas constant.

FIRST STEP | ALL STEPS | ANSWER ONLY

## Step-by-step

### Step 1 of 8

(A)

Draw the P-V diagram of the cycle.

From the above diagram, there are three cycles that can be explained as:

Process 1-2: It is a constant volume process.

Process 2-3: It is a constant volume and pressure process. It is also called constant temperature process. This is because pressure and volume of an ideal gas law are directly proportional to the temperature. That’s why temperature is a constant that can be explained mathematically as,

In the above expression, write the expression for one mole of gas.

In above expression, is the gas constant and is constant. Therefore, above expression can be written as,

Process 3-1: It is a constant pressure process.

Part AThe P-V diagram of the cycle is,

Explanation | Common mistakes | Hint for next step

The P-V diagram of the cycle has been drawn with the help of the provided three processes.

These three processes are constant volume process, constant pressure process, and constant pressure and volume process.

While drawing the P-V of the thermodynamic cycle, if the process is constant volume, draw the perpendicular straight line to the axis of the volume; if the process is constant pressure, draw the perpendicular straight line to the axis of the pressure; if the process is constant for both pressure and temperature, draw the curve as parabolic in P-V diagram.

### Step 2 of 8

(b.1)

Calculate the work done during the process 1-2.

The process 1-2 is the constant volume process. Therefore, change in volume is zero in a closed system. That’s why work done in the process 1-2 is zero, which can be mathematically written as,

Part b.1The work done during the process 1-2 is .

Explanation | Common mistakes | Hint for next step

The work done during the process 1-2 has been calculated by using the expression of the work done at constant pressure.

During this calculation, there is no change in volume of the piston cylinder. That’s why the work done in this process is zero.

### Step 3 of 8

(b.2)

Calculate the heat transfers during the process 1-2.

Here, is the heat transfer during the process 1-2, is the change in internal energy during the process 1-2.

Substitute for and for .

Part b.2The heat transfer during process 1-2 is .

Explanation | Common mistakes | Hint for next step

The heat transfers during process 1-2 have been calculated by using the expression of the heat transfer for a closed system.

During this calculation, there is no change in volume of the piston cylinder. That’s why the work done in this process is zero.

Therefore, heat transfer during process 1-2 is equal to change in the internal energy. The change in internal energy during process 1-2 is given. This value is substituted in the expression of the heat transfer for a closed system in order to find the heat transfer during process 1-2.

### Step 4 of 8

(c.1)

Calculate the volume at point 3 in the above cycle.

Here, is the work done during the process 3-1.

Substitute for , for and for .

The volume at point 1 is equal to point 2 because process 1-2 is a constant volume process. Therefore, the volume of point two can be mathematically written as:

Substitute for :

Calculate the work done during the process 2-3.

Here, is the work done during process 2-3, is the pressure at a point , is the volume at point 2, and is the volume at point 3.

Substitute for , for , and for .

Part c.1The work done during process 2-3 is .

Explanation | Common mistakes | Hint for next step

The work done during process 2-3 has been calculated by using the expression of the work done at a constant temperature (both pressure and volume are constant).

During this calculation, first of all, calculate the volume at point 3 by using the equation of work done during process 3-1, and then calculate the work done during process 2-3 by using the expression of work done at a constant temperature.

### Step 5 of 8

(c.2)

Calculate heat transfers during process 2-3.

Here, is the heat transfer during the process 2-3 and is the change in internal energy during process 2-3.

As mentioned in the question, the change in internal energy during process 2-3 is zero. Therefore, the work done is equal to the heat transfer, which can be mathematically written as,

Substitute for .

Part c.2The heat transfer during process 2-3 is .

Explanation | Hint for next step

The heat transfers during process 2-3 have been calculated by using the expression of the heat transfer for a closed system.

During this calculation, the change in internal energy is zero. Therefore, the work done is equal to the heat transfer which is calculated in part c.1.

### Step 6 of 8

(d)

The net change in internal energy in a closed cycle is zero. Therefore, the net change in internal energy can be mathematically written as,

Substitute for and for .

Calculate the heat transfers during the process 2-3.

Here, is the heat transfer during the process 3-1 and is the change in internal energy during the process 3-1.

Substitute for and for .

Part dThe heat transfer during process 3-1 is .

Explanation | Common mistakes | Hint for next step

The heat transfers during process 3-1 have been calculated by using the expression of the heat transfer for a closed system.

During this calculation, first of all, calculate the change in internal energy during process 3-1 by using the concepts of the net change in internal energy during the whole cycle. The change in internal energy for a closed system is zero.

Calculate the change in internal energy in the process 3-1 by substituting the known value of change in internal energy during process 2-3 and change in internal energy during process 2-1. Thus, the change in internal energy during process 3-1 is calculated.

Now, apply the expression of heat transfer for a closed system to find the net heat transfer during process 3-1 by substituting the value of the change in internal energy during process 3-1 and the value of the work done during process 3-1. Thus, net heat transfers during process 3-1 have been calculated.

### Step 7 of 8

(e.1)

Calculate the net work done during the cycle.

Substitute for , for , and for .

Part e.1The net work done during the cycle is .

Explanation | Common mistakes | Hint for next step

The net work done in this cycle has been calculated by adding the work done due to all the processes.

During the calculation, substitute the values of the work done for the process 1-2, the process 2-3 and the process 3-1 in the expression of net work done on the cycle. Thus, net work done in the whole cycle has been calculated.

### Step 8 of 8

(e.2)

When net work done in the cycle is positive, the positive value of the network done shows that the cycle is power cycle.

When the network done of the cycle is negative, the negative values of the cycle shows that the cycle is a refrigerant cycle.

Here, in this cycle, the net work done during this cycle is positive. Therefore, this cycle is a power cycle.

Part e.2This thermodynamic cycle is a power cycle.

Explanation | Common mistakes

The cycle is decided as either power or refrigeration on the basis of the work production and absorption. If the cycle is work producing, then the cycle is power, and if the cycle is work absorbing, then the cycle is a refrigeration cycle. In the cycle, work is produced. That’s why it is power cycle.

Part AThe P-V diagram of the cycle is,

Part b.1The work done during the process 1-2 is .

Part b.2The heat transfer during process 1-2 is .

Part c.1The work done during process 2-3 is .

Part c.2The heat transfer during process 2-3 is .

Part dThe heat transfer during process 3-1 is .

Part e.1The net work done during the cycle is .

Part e.2This thermodynamic cycle is a power cycle.