Work done in an adiabatic process and isothermals

In summary: T2.In summary, it is stated that the work done between two isothermals is independent of the adiabatic path between them. This means that for any two points on different isothermals, the work done along any adiabatic path between them will be the same. However, there is only one reversible adiabatic path between any two points on the same isothermal curve. This means that the work done along this path will be the same for any two points on the same isothermal curve, but different from the work done along any other adiabatic path between those two points.
  • #1
Amith2006
427
2

Homework Statement



It is said that the work done between 2 isothermals is independent of the adiabatic.

Homework Equations



The work done during an adiabatic process is given by,
W = C(v){T1 – T2)


The Attempt at a Solution



From the above equation it is clear that W depends only on the initial and final temperatures. So W is the same along any adiabatic curve. But work is a path function. Then the work done along different adiabatic curves should be different. It is contradictory to each other. Could somebody please clear my doubt? Here C(v) is the specific heat of the gas at constant volume.
 
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  • #2
Amith2006 said:

Homework Statement



It is said that the work done between 2 isothermals is independent of the adiabatic.

Homework Equations



The work done during an adiabatic process is given by,
W = C(v){T1 – T2)


The Attempt at a Solution



From the above equation it is clear that W depends only on the initial and final temperatures. So W is the same along any adiabatic curve. But work is a path function. Then the work done along different adiabatic curves should be different. It is contradictory to each other. Could somebody please clear my doubt? Here C(v) is the specific heat of the gas at constant volume.

Recall that the first law states [itex]dE = \delta W + \delta Q = -P dV + T dS[/itex]. During an adiabatic process, [itex]\delta Q = 0[/itex], and similarly, [itex]dS = 0[/itex] when the process is carried out reversibly. So, you then have that [itex]\delta W = -P dV[/itex] - i.e., for a reversible adiabatic process the work does not depend on the path because the work differential can be equated to an exact differential.

Does this explain it?
 
  • #3
Amith2006 said:

Homework Statement



It is said that the work done between 2 isothermals is independent of the adiabatic.

Homework Equations



The work done during an adiabatic process is given by,
W = C(v){T1 – T2)

The Attempt at a Solution



From the above equation it is clear that W depends only on the initial and final temperatures. So W is the same along any adiabatic curve. But work is a path function. Then the work done along different adiabatic curves should be different. It is contradictory to each other. Could somebody please clear my doubt? Here C(v) is the specific heat of the gas at constant volume.
There are many adiabatic paths. But there is only one reversible adiabatic path. The temperature difference depends on the adiabatic path. For example, an adiabatic free expansion (P=0) does no work, so there is no temperature change.

For the reversible adiabatic path the temperature change is:

[tex]\Delta T = \frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{nC_v (1-\gamma)}[/tex]

So you are right that Work is a path function. But so is the temperature change in an adiabatic process.

AM
 
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  • #4
I didn't get you. Could you please explain it in a more simple way? I am herewith attaching a P-V diagram in which I have drawn two adiabatics 1 & 2 between 2 isothermals at temperatures T1 & T2. As we move from A to B in adiabatic 1 and from D to C in adiabatic 2, will the change in temperature be the same i.e. T2 - T1?
 

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  • #5
Amith2006 said:
I didn't get you. Could you please explain it in a more simple way? I am herewith attaching a P-V diagram in which I have drawn two adiabatics 1 & 2 between 2 isothermals at temperatures T1 & T2. As we move from A to B in adiabatic 1 and from D to C in adiabatic 2, will the change in temperature be the same i.e. T2 - T1?
Yes, because [itex]W = nC_v\Delta T[/itex] and the temperatures are the same.

If you take some path other than the adiabatic path from A to B the work done will be different. (which simply means that heat is flowing into or out of the system, so it would not be adiabatic).

AM
 
  • #6
Andrew Mason said:
Yes, because [itex]W = nC_v\Delta T[/itex] and the temperatures are the same.

If you take some path other than the adiabatic path from A to B the work done will be different. (which simply means that heat is flowing into or out of the system, so it would not be adiabatic).

AM

Do u mean that between 2 isothermals, u can have only one adiabatic? If at we draw another adiabatic between T1 and T2, it will not be adiabatic. But the statement says that the work done between 2 isothermals is independent of the adiabatic. Doesn't this statement mean that the work done for all adiabatics whose initial and final temperatures are same, are equal?
 
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  • #7
Amith2006 said:
Do u mean that between 2 isothermals, u can have only one adiabatic?
Between two states there is only one reversible adiabatic path. Using the adiabatic condition:

[tex]PV^{\gamma} = K \text{and} PdV = - dU = -nC_vdT[/tex],

the work done in this process is:

[tex]W = \int_{V_i}^{V_f}PdV = \int_{V_i}^{V_f}KV^{-\gamma}dV =
\frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{(1-\gamma)} = - \int_{T_i}^{T_f} nC_vdT = - nC_v(T_f - T_i)[/tex]

If at we draw another adiabatic between T1 and T2, it will not be adiabatic.
??
If you draw another reversible adiabatic path from A to T2 it will be AB because there is only one reversible adiabatic path from A to T2. If you draw a reversible adiabatic path between some other point on T1 to T2, the work done will be the same. But it is not the same path because the initial and final volumes and pressures are different.

But the statement says that the work done between 2 isothermals is independent of the adiabatic. Doesn't this statement mean that the work done for all adiabatics whose initial and final temperatures are same, are equal?
Yes, exactly. It is independent of the adiabatic path between those two isothermals. But for each point on T1 there is only one adiabatic path to T2.

AM
 
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  • #8
Thank you very much for your guidance.
 
  • #9
In the relationship
[tex]W = - n C_{v} (T_{f} - T_{i})[/tex]

If the volume changes in an adiabatic process why is [tex]C_{v}[/tex] used?
 
  • #10
The work done during an adiabatic process, according to the First Law of Thermodynamics is equal to the change of what? If this is not dependent on which adiabatic we go along, but only on the initial and final temperatures, then what can we say that the quantity in the first question only depends on? Do you know of any physical systems with this property?
 
  • #11
In an adiabatic process dQ = 0 so that dU = dW.

So then you use,

[tex] C_{v} = ( \frac{\Delta U}{\Delta T})_{v}[/tex]

to make

[tex]
W = - n C_{v} (T_{f} - T_{i})
[/tex]

Is that right?
 

1. What is an adiabatic process?

An adiabatic process is a thermodynamic process in which there is no transfer of heat or matter between the system and its surroundings. This means that the system experiences no change in temperature, allowing for the work done to be calculated using only changes in volume and pressure.

2. How is work done in an adiabatic process calculated?

The work done in an adiabatic process can be calculated using the formula W = -PΔV, where W represents work, P represents pressure, and ΔV represents the change in volume. This formula assumes that the process is reversible and that the gas behaves ideally.

3. What is an isothermal process?

An isothermal process is a thermodynamic process in which the temperature remains constant throughout. This means that any changes in the system's internal energy are balanced by the transfer of heat, resulting in no change in temperature.

4. How is work done in an isothermal process calculated?

The work done in an isothermal process can be calculated using the formula W = -nRTln(V2/V1), where W represents work, n represents the number of moles of gas, R represents the gas constant, T represents the temperature, and V1 and V2 represent the initial and final volumes, respectively.

5. How do adiabatic and isothermal processes differ?

The main difference between adiabatic and isothermal processes is the change in temperature. In an adiabatic process, the temperature remains constant, while in an isothermal process, the temperature remains constant. Additionally, work is done in an adiabatic process due to changes in pressure and volume, while work is done in an isothermal process due to the transfer of heat.

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