Work done in an adiabatic process and isothermals

  • Thread starter Amith2006
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
Mute
Homework Helper
1,388
10

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
Andrew Mason
Science Advisor
Homework Helper
7,662
385

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
 
Last edited:
  • #4
427
2
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?
 

Attachments

  • untitled.PNG
    untitled.PNG
    842 bytes · Views: 705
  • #5
Andrew Mason
Science Advisor
Homework Helper
7,662
385
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
427
2
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?
 
Last edited:
  • #7
Andrew Mason
Science Advisor
Homework Helper
7,662
385
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
 
Last edited:
  • #8
427
2
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
2,967
5
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?
 

Related Threads on Work done in an adiabatic process and isothermals

Replies
2
Views
959
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
8
Views
24K
Replies
5
Views
3K
Replies
2
Views
15K
Replies
2
Views
3K
Replies
3
Views
5K
Replies
1
Views
936
Replies
3
Views
2K
  • Last Post
Replies
2
Views
7K
Top