# Work done in an adiabatic process and isothermals

1. Dec 22, 2006

### Amith2006

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Dec 22, 2006

### Mute

Recall that the first law states $dE = \delta W + \delta Q = -P dV + T dS$. During an adiabatic process, $\delta Q = 0$, and similarly, $dS = 0$ when the process is carried out reversibly. So, you then have that $\delta W = -P dV$ - 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. Dec 22, 2006

### Andrew Mason

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:

$$\Delta T = \frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{nC_v (1-\gamma)}$$

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

AM

Last edited: Dec 22, 2006
4. Dec 23, 2006

### Amith2006

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. Dec 23, 2006

### Andrew Mason

Yes, because $W = nC_v\Delta T$ 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. Dec 24, 2006

### Amith2006

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: Dec 24, 2006
7. Dec 24, 2006

### Andrew Mason

Between two states there is only one reversible adiabatic path. Using the adiabatic condition:

$$PV^{\gamma} = K \text{and} PdV = - dU = -nC_vdT$$,

the work done in this process is:

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

??
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.

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: Dec 24, 2006
8. Dec 24, 2006

### Amith2006

Thank you very much for your guidance.

9. Oct 10, 2010

### sugar_scoot

In the relationship
$$W = - n C_{v} (T_{f} - T_{i})$$

If the volume changes in an adiabatic process why is $$C_{v}$$ used?

10. Oct 10, 2010

### Dickfore

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. Oct 10, 2010

### sugar_scoot

In an adiabatic process dQ = 0 so that dU = dW.

So then you use,

$$C_{v} = ( \frac{\Delta U}{\Delta T})_{v}$$

to make

$$W = - n C_{v} (T_{f} - T_{i})$$

Is that right?