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Home/ Questions/Find W12, the work done on the gas as it expands from state 1 to state 2.
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lyytutoria
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lyytutoriaExpert
Asked: April 13, 20222022-04-13T05:44:42+00:00 2022-04-13T05:44:42+00:00In: Physics

Find W12, the work done on the gas as it expands from state 1 to state 2.

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A) Find W12, the work done on the gas as it expands from state 1 to state 2. Express the work done in terms of Po and Vo
B) Find W23 , the work done on the gas as it cools from state 2 to state 3. Express your answer in terms of Po and Vo.
C) Find W34, the work done on the gas as it is compressed from state 3 to state 4. Express your answer in terms of Po and Vo.
D) Find W41, the work done on the gas as it is heated from state 4 to state 1. Express your answer in terms of Po and Vo .
E)What is Wnet, the total work done on the gas during one cycle? Express your answer in terms of Po and Vo .
F) When the gas is in state 1, its temperature is T1. Find the temperature T3 of the gas when it is in state 3. (Keep in mind that this is an ideal gas.) Express T3 in terms of T1.
The diagram shows the pressure and volume of an ideal gas during one cycle of an engine. As the gas proceeds from state 1 to state 2, it is heated at constant pressure. It is then cooled at constant volume, until it reaches state 3. The gas is then cooled at constant pressure to state 4. Finally, the gas is heated at constant volume until it returns to state 1. 

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♦ Relevant knowledge
The increase in gas can indicate this system may have performed work. This is an example of a function called a path which describes properties whose value depend on the path that is derived from two values. In contrast state functions are property values that are not dependent on the way it took to get there value. These kinds of functions are used in thermodynamics. Internal energy as well as volume and pressure represent examples of state-related functions, while heat and work are paths functions.

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    lyytutoria Expert
    2022-04-27T01:47:48+00:00Added an answer on April 27, 2022 at 1:47 am
    Concepts and Reason

    This problem was solved by the use of a thermodynamic process as well as an ideal gas equation.

    Find the area below the P– V curve to find the work done on gas from states 1 through 2. State 2 to state 3. State 3 to sate 4. And state 4 to state 1. Add all the work done, i.e. 1->2->3->4 to find the total amount of work that was done on the gas in one cycle.

    The last part of the equation is ideal gas equation. Find temperature for state 3.

    Fundamentals

    Here is the equation of state for ideal gas:

    PV = nRT

    Here, V is volume and P is pressure. N is the number moles. R is universal gas constant. T is ideal gas temperature.

    Calculate the work that gas does in a thermodynamic process by using:

    W = P(dv)

    Gas’s work depends on its initial and final states, as well as the path.

     

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  2. Daniel Singh
    2022-04-13T05:45:16+00:00Added an answer on April 13, 2022 at 5:45 am
    Concepts and Reason

    This problem was solved by the use of a thermodynamic process as well as an ideal gas equation.

    Find the area below the P– V curve to find the work done on gas from states 1 through 2. State 2 to state 3. State 3 to sate 4. And state 4 to state 1. Add all the work done, i.e. 1→2→3→4
to find the total amount of work that was done on the gas in one cycle.

    The last part of the equation is ideal gas equation. Find temperature for state 3.

    Fundamentals

    Here is the equation of state for ideal gas:

    PV = nRT

    Here, V is volume and P is pressure. N is the number moles. R is universal gas constant. T is ideal gas temperature.

    Calculate the work that gas does in a thermodynamic process by using:

    W = P(dv)

    Gas’s work depends on its initial and final states, as well as the path.

    (A)

    Calculated as follows:

    W2 = P(V2 -V)

    Substitute for and for . is required to replace in the equation.

    W2 = (3p.)(4V, -V.)
=9p.

    (B)

    Calculated as follows:

    apa S=M

    During the transition from state 2 to 3 the volume of the gas remains constant. Substitute 0 for DV.

    0=
(0)2 S=M

    The gas is therefore cooled from state 2 to 3 at a temperature of 0.

    (C)

    According to the following calculation, the work on the gas from states 3 and 4 was:

    W34 = P (V-13)

    Substitute for and for . is required to replace in the equation.

    W34 =(p.)(1.-4V)
=-3p.V

    (D)

    According to the following calculation, the work on the gas from states 4 to 1 was:

    apa S=M

    During the transition from state 4 to 1., the volume of the gas remains constant. Substitute 0 for DV.

    W, =S P(0)
= 0

    The gas cools down from state 4 to 1 at a temperature of 0.

    (E)

    One cycle of gas work is:

    W =W2+W33 +W34 +W

    Substitute for, -3p.V
for, 0 for and.

    W =9p.V, +0-3p.V+0
=6p.V.

    (F)

    The following formula calculates the temperature at which the gas is in its state 1:

    PV = nRT,
nR

    Substitute [katex]3p_0[/katex] for or [katex]V_0[/katex] for .

    T: 3p V
nR

    The following formula calculates the temperature at which the gas is in 3 state:

    Substitute [katex]p_0[/katex] for , and [katex]4V_0[/katex] for .

    T, P. (41)
nR

    Rearrange equation T: 3p V
nR
to get the value of nR.

    nR = 3pV

    Use nR = 3pV
in equationnR
.

    Т; - P, 4,
Зри,
Part A – Ans

    The gas expansion from state 1 into state 2 was .

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